Electrochemistry
Electrochemistry Oxidation-Reduction Reactions
Oxidation - refers to the loss of electrons by a molecule, atom or ion Reduction - refers to the gain of electrons by an molecule, atom or ion Oxidation number - the oxidation number of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom; it is the charge an atom would possess if the bonding were ionic
In oxidation numbers, the charge is written before the number (to distinguish them from actual electronic charges - where the number is placed in front of the charge)
HCl Hydrogen is less electronegative than Chlorine, therefore, Chlorine gets all the shared electrons in the HCl bond. In this compound, the H oxidation number is therefore +1, and the oxidation number of Cl is -1. Chemical reactions in which the oxidation state of one or more substances changes are called oxidationreduction reactions (or redox reactions)
Keeping track of the oxidation numbers of the elements involved in a reaction will identify whether the reaction is a redox reaction or not
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) The oxidation numbers for the above elements and ions are:
Zn = 0, Zn2+ = +2 H+ = +1, H2 = 0 Thus, the oxidation number of both the Zn(s) and H +(aq) change during the course of the reaction, and so, this must be a redox reaction
In this reaction an unambiguous transfer of electrons occurs:
The elemental form of zinc (i.e. zinc metal) loses two electrons (and becomes soluble zinc ion) Two H+ ions in solution gain two electrons and form a hydrogen (H2) molecule
In other types of reactions, we can identify changes in oxidation number, but it is not so clear that a particular atom has gained or lost an electron:
Hydrogen has been oxidized and oxygen has been reduced in the above reaction involving the production of water from elemental hydrogen and oxygen However, the H-O bonds are not ionic, they are polar covalent, so there is not a complete transfer of shared electrons
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Using oxidation numbers is just a convenient bookkeeping method to understand reduction/oxidation types of reactions. It does not necessarily mean that complete electron transfer has occurred.
In a redox reaction, both oxidation and reduction must occur. In other words, something must be oxidized for something else to be reduced
The compound that contributes the electrons is called the reducing agent. The reducing agent gives up electrons, causing another compound to be reduced (and is therefore oxidized in the process) The compound that accepts electrons is called the oxidizing agent. The oxidizing agent accepts electrons from another compound, causing the other compound to be oxidized (in the process, the oxidizing agent is reduced)
In the above reaction of hydrogen and oxygen:
Hydrogen is the reducing agent (it contributes electrons to oxygen, and is itself oxidized) Oxygen is the oxidizing agent (it accepts electrons from hydrogen - oxidizing it, and is itself reduced) Balancing Oxidation-Reduction Reactions
In balancing redox reactions, gains and losses of electrons must be balanced
In many cases the equation can be balanced without explicitly considering the transfer of electrons. For example: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
However, many redox reactions are more complicated and cannot be balanced without taking into account the number of electrons lost and gained during the reaction
Half-Reactions Although oxidation and reduction must take place simultaneously (i.e. for something to be reduced, something else must be oxidized) it is often convenient to consider them as separate processes
The following is the oxidation of Tin(II) by Iron(III) Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
This reaction can be considered as two separate processes: o The oxidation of Sn2+ o The reduction of Fe3+ Oxidation: Sn2+(aq) Sn4+(aq) + 2eReduction: 2Fe3+(aq) + 2e- 2Fe2+(aq)
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In the oxidation reaction the electrons are shown as products (i.e. the tin is oxidized and releases electrons) In the reduction reaction the electrons are shown as reactants (i.e. the iron ion combines with electrons as it is reduced) Equations that show either oxidation, or reduction, alone are called half-reactions
The overall electrons produced in the oxidation half-reaction must equal the number of electrons consumed in the reduction half-reaction o When the equations have been balanced in this way, they are combined to produce the balanced overall redox reaction
Balancing Equations By The Method Of Half-Reactions The use of half-reactions provides a general way to balance redox reactions The reaction between permanganate ion (MnO4-) and oxalate ion (C2O42-) in acidic solution:
When these ions combine in an acidic solution, they react to produce manganese ion (Mn2+) and CO2 gas Here is the unbalanced equation that describes this reaction:
MnO4- + C2O42- Mn2+(aq) + CO2(g) To balance this redox reaction using the method of half-reactions, begin by writing the incomplete oxidation and reduction half-reactions: MnO4- Mn2+(aq) C2O42- CO2(g) Which compound is being reduced and which one is being oxidized?
Now, balance the atoms undergoing oxidation or reduction by the appropriate addition of coefficients.
In the above half-reactions, the Mn appears to be balanced on both sides of its half-reaction However, the carbon is not balanced in the second half-reaction - we need another carbon in the products
MnO4- Mn2+(aq) C2O42- 2CO2(g) If the reaction is done under acidic aqueous solution, H+ and H2O can be added to reactants or products to balance H and O atoms. Likewise, if the reaction is done under basic aqueous solution, OH - and H2O can be added to balance H and O atoms
In the manganese half-reaction we have 4 O atoms on the left and none on the right. We can add 4 H2O molecules on the right to take care of this: MnO4- Mn2+(aq) + 4H2O(l)
We now have 8 H atoms on the right that need to be balanced. This reaction is done under acidic conditions, so we can balance the H atoms by adding H+(aq) ions on the left: MnO4- + 8H+ Mn2+(aq) + 4H2O(l)
The atoms are now balanced, but the charges are not. We have a net charge of 7+ on the left, and 2+ on the right. We want to balance the charge without altering the number of atoms (they are balanced already). We balance the charge by adding the appropriate number of electrons:
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MnO4- + 8H+ + 5e- Mn2+(aq) + 4H2O(l) (balanced atoms, balanced charge)
How about the other half-reaction? C2O42- 2CO2(g)
The atoms are balanced, but the charges are not. We have a net charge of 2- on the left and 0 on the right. We can balance this by adding the appropriate number of electrons: C2O42- 2CO2(g) + 2e(balanced atoms, balanced charge)
Are we done? No. By the original definition above, in a redox reaction the overall electrons produced in the oxidation half-reaction must equal the number of electrons consumed in the reduction half-reaction. We have not yet identified which is the oxidation half-reaction and which is the reduction half-reaction, but it is clear that the electrons are not balancing (we have one halfreaction with 5e- and the other with 2e-) o The half-reactions must be multiplied by some factor, such that the electrons present in the half-reactions balance each other o For the above half-reactions, the lowest common multiple is 10. Thus, the half-reaction with 5e- should be multiplied by 2, and the half-reaction with 2e- should be multiplied by 5: 2MnO4- + 16H+ + 10e- 2Mn2+(aq) + 8H2O(l) 5C2O42- 10CO2(g) + 10e-
We can now describe the redox reaction accurately: o The oxidation half-reaction is: 5C2O42- 10CO2(g) + 10eo
The reduction half-reaction is:
2MnO4- + 16H+ + 10e- 2Mn2+(aq) + 8H2O(l) In other words, the oxalate ion is oxidized by the permanganate ion (and the permanganate ion is reduced) o
The overall balanced redox reaction is the combination of the two balanced halfreactions: 2MnO4- + 16H+ + 10e- 2Mn2+(aq) + 8H2O(l) 5C2O42- 10CO2(g) + 10e+ 2MnO4 + 16H + 5C2O42- 10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Summary of balancing half-reactions in acidic solutions: 1. Divide reaction into two incomplete half-reactions 2. Balance each half-reaction by doing the following: a. Balance all elements, except O and H b. Balance O by adding H2O c. Balance H by adding H+ d. Balance charge by adding e-as needed
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3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to get a common multiple 4. The overall reaction is the sum of the half-reactions 5. The oxidation half-reaction is the one that produces electrons as products, and the reduction half-reaction is the one that uses electrons as reactants Balancing Redox Reactions That Occur in Basic Solution In a basic aqueous solution the half-reactions must be balanced using OH- and H2O (instead of H+ and H2O)
The procedure for balancing in this case is actually the same as for balancing in an acidic solution, with one small twist: o Since we are under basic conditions, we have to neutralize any H + ions in solution o We achieve this by adding the appropriate number of OH- ions o We keep things balanced by adding the same number of OH - ions to both sides of the half-reaction
Voltaic Cells A spontaneous redox reaction involves the transfer of electrons
In principle, this transfer or movement of electrons can be used to do a type of work - specifically, electrical work A type of device that makes use of redox reactions to produce electron flow, and to allow electrical work, is known as a Voltaic Cell (named after Count Alessandro Volta, 1745-1827, an Italian physicist)
The development of the voltaic cell begins as follows... A spontaneous redox reaction that we have previously discussed involves the metals Zinc (Zn) and Copper (Cu)
If you recall the Activity Series for metals and hydrogen ion:
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When comparing Zinc with Copper in the activity series, we expect that Zinc should be easier to oxidize (i.e. should give up electrons easier than Copper) The spontaneity of the reaction between zinc and copper ion reflects the Activity Series relationship: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) o o
The reaction as shown is spontaneous. The Zinc metal is oxidized, and the Copper ion is reduced The reverse reaction, the oxidation of Copper and the reduction of Zinc ion, is not spontaneous
The redox half-reactions for the above reaction would be: Oxidation: Zn(s) Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- Cu(s)
The oxidation of a Zinc atom releases 2 electrons The reduction of a Copper ion is achieved by the acceptance of 2 electrons
Thus, there would appear to be a movement, or flow, of electrons from the Zinc metal to the Copper ions According to the above equation, if Zinc metal is placed in an aqueous solution containing Cu2+ ions (e.g. a solution of copper sulfate salt), the following will occur:
At the surface of the Zinc metal, the oxidation of Zinc atoms will be coupled to the reduction of Cu2+ ions Electrons from the oxidation of the zinc will reduce the copper ions to elemental copper on the surface of the zinc Over time, the zinc metal will dissolve (as solid zinc is oxidized to zinc ion) and zinc ion will build up in the solution. Also over time, the solution will no longer contain copper ions, an the copper will be present in elemental form
In the above reaction, the zinc and copper metal/ions are in direct contact with each other and that is why the reduced form of the copper (i.e. the metal) builds up on the surface of the zinc
What if we separated the Cu(s)/Cu2+ from the Zn(s)/Zn2+?
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The reaction will not occur because there is no way for the electrons released by the oxidation of Zinc metal to get over to the Copper ions and reduce them to Copper metal
What if we provide a path of conductance for the electrons released by the oxidation of the zinc to get over to the copper?
Now that the electrons have a path to the Cu/Cu2+ side it would appear that the reaction could proceed However, we now have another problem ...
At the beginning of the redox reaction we have a neutral aqueous salt solution, e.g. ZnSO 4 (i.e. the concentration of cation equals the concentration of anion) As the redox reaction proceeds we build up Zn2+ ions in the solution where the Zinc is being oxidized. Conversely, we remove Cu2+ ions from solution where the Copper ion is being reduced. The anion concentration (sulfate ion in this case) does not change. Thus, we are building up a net positive charge in the zinc solution, and a net negative charge in the copper solution These charges will oppose the flow of electrons. The positive charge in the zinc solution will make it harder for the negative electrons to leave. Likewise, the negative charge in the copper solution will repel the electrons that are trying to come over from the zinc side
We need a way to neutralize the charge build-up in the solutions due to the change in soluble ion concentration What if we had a tube filled with aqueous solution that connected the two redox reactions?
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This would allow anion to move from the copper reduction side to the zinc oxidation side and keep the charges in solution neutral In turn, this would now allow the electrons to flow However, as the reaction proceeds, we have yet another problem...
As the reaction proceeds, although movement of anion allows the overall charges in solution to remain neutral, the net movement of anion produces a concentration gradient across the two solutions. In other words, after a while the net concentration of ions in the zinc oxidation side will be greater than in the copper reduction side. This concentration gradient will oppose movement of anion Actually, this is not really a problem that we will see. What it means is that we have to realize that charge neutralization can occur by either anions moving to the left or cations moving to the right
In consideration of keeping the overall concentration of ions in balance between the two sides, cations will also be moving to the right:
Thus, in the connecting tube of solution we have net movement of both types of ions: o anions are going into the oxidation side o cations going into the reduction side
The connecting tube of solution is called a Salt Bridge Summary of the movement of ions, electrons and the redox half-reactions in a voltaic cell:
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The two solid metals in the different half-reactions are called electrodes
The metal in the half-reaction where oxidation is occurring is called the Anode The metal in the half-reaction where reduction is occurring is called the Cathode The cathode is often labeled with a "+"; "this electrode attracts electrons" The anode is often labeled with a "-"; "this electrode repels electrons"
Cell EMF Here is a summary of a voltaic cell:
A key question is: what causes the electrons to flow from the anode to the cathode?
The electrons flow from the anode to the cathode because of a difference in the potential energy of the electrons at the anode compared to the cathode In particular, the potential energy of the electrons is higher at the anode than at the cathode Like water molecules in a waterfall, the electrons move in a direction from high potential energy to low potential energy (i.e. "downhill" in an energy landscape)
The potential difference between two electrodes is measured in Volts (V)
One volt is the potential difference, between an anode and a cathode, that would be required in order to impart 1 Joule of energy to a charge of 1 coulomb (note: 1 coulomb is the charge of a collection of 6.24 x 1018 electrons) 1 V = 1 J/C (i.e. energy per electron)
The potential difference between two electrodes is called the electromotive force, or emf The emf of a voltaic cell is called the cell potential, or the cell voltage
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The cell voltage of a voltaic cell will be positive value
The cell voltage will depend upon the specific half-reactions, concentrations of ions, and temperature
Common tabulations of cell voltages are performed at 25°C and standard conditions of concentrations of liquid (i.e. 1M) and gas components (i.e. 1 atm) Under standard conditions, the emf of a voltaic cell is called the standard emf or the standard cell potential (E0cell) By convention, the potential associate with an electrode is the potential for reduction The standard cell potential is given by the standard reduction potential of the cathode, minus the standard reduction potential of the anode
E0red(cathode)
E0cell = E0red(cathode) - E0red(anode) 0 E red(anode) then the electron flow from anode
If > to cathode is spontaneous Standard Reduction Potentials The cell potential of a voltaic cell depends upon the two redox half-reactions that comprise the functioning cell
There are many different combinations of redox half-reactions. Do we need to tabulate all possible unique combinations? As you might have guessed, the answer is no, we don't. In principle, we could assign a unique standard potential to each half-reaction and by comparing these values determine the cell potential for any combination of half-reactions
There is one problem, however: we have seen that an isolated half-reaction doesn't result in reduction or oxidation or electron flow. Therefore, how are we going to assign a standard potential to a half-reaction?
We are going to use a generally agreed upon reduction reaction as the reference half-reaction against which to compare and tabulate all possible half-reactions The reference reaction is the reduction of H+(aq) ions to produce H2(g): 2H+(aq, 1M) + 2e- H2(g, 1 atm) E0red = 0.0V This is a Standard Hydrogen Electrode (SHE)
How do we use this SHE to determine the standard reduction potential, E 0red, for some redox half-reaction?
Let's say we are interested in determining the standard reduction potential for Zn2+ We would set up a voltaic cell using a SHE as one half-cell, and a Zinc electrode as the other halfcell
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The half-cell reactions are: Zn(s) Zn2+(aq) + 2e- (Zinc electrode) 2H+(aq) + 2e- H2(g) (SHE)
The overall cell reaction: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
Note: this is performed under Standard Conditions, where concentrations = 1.0M, and p = 1 atm
For the reaction as shown, the zinc electrode is the anode (oxidation), and the SHE is the cathode (reduction) o The voltage for the overall cell is measured at 0.76 Volts (i.e. this is the standard cell potential) o From the previous definition of the standard cell potential: E0cell = E0red(cathode) - E0red(anode) o
We know the value for the standard cell potential, it is +0.76V. We also know the value for the reduction potential for the SHE, it is 0V. Thus, 0.76V = E0red(cathode) - E0red(anode) 0.76V = 0 - E0red(anode) E0red(anode) = -0.76V
In other words, in reference to the SHE, the standard reduction potential for the zinc electrode is determined to be -0.76V.
This may seem a little strange because the cell reads +0.76V and yet the determined reduction potential for the zinc electrode is -0.76V.
The key here is that the original definition of the standard cell potential is set up such that each cell is referenced as a reduction reaction. In other words what we have actually determined is the half-cell potential (in reference to the SHE) for the reduction of zinc:
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Zn2+(aq) + 2e- Zn(s) E0red = -0.76V Whenever we assign a potential to a half-reaction using the equation for the standard cell potential, it is always done in reference to a reduction reaction
Thus, in reference to the SHE, the oxidation of Zn(s) is associated with a half-cell potential of +0.76V (this is the spontaneous direction for the reaction; the reduction of Zn2+ and the oxidation of H2(g) is not spontaneous)
The standard cell potential (Voltage) describes the potential per unit charge that the reaction produces
This value of the potential per unit charge (i.e. electron) is a constant that is a property of the two half-reactions that are reacting in the cell. It is independent of the stoichiometry or amount of the reactants in the cell
The more positive the value of E0red the greater the driving force for reduction In reference to H+/H2(g) , Ag+ will preferentially be reduced to Ag(s) and H2(g) will be oxidized to H+(aq). In reference to H+/H2(g) , Zn(s) will preferentially be oxidized and H+ will be reduced to H2(g). The cathode is the electrode where reduction occurs, therefore, in order for a cell to operate spontaneously, the half-reaction at the cathode must have a more positive value of E0red than the half-reaction at the anode o The greater driving force for reduction at the cathode can force the reaction at the anode to operate "in reverse" - i.e. to undergo the oxidation reaction The difference between the standard reduction potentials of the two reactions, E0red o (cathode) - E0red (anode), is the excess potential that can be used to drive electrons through the cell, E0cell
Oxidizing and Reducing Agents In the above discussion of reduction potentials and their relationship to one another in half-reactions, we can predict which direction current will flow in a voltaic cell. In other words, we can predict which reactant will be oxidized and which will be reduced
The same information can be used for redox reactions where the compounds are in direct contact, and not separated into half-cells. For example, from the example above, we would predict that silver ion will oxidize lithium metal as a spontaneous redox reaction. Alternatively, we would predict that the reduction of lithium ion by silver metal is not a spontaneous reaction (and would require the input of energy to drive)
The more positive the E0red value of a half-reaction, the greater the tendency of the compound to be reduced - and therefore to oxidize another compound (i.e. the greater the tendency to act as an oxidizing agent)
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Substances that are strong oxidizing agents are subsequently weak reducing agents. Likewise, substances that are strong reducing agents are weak oxidizing agents. Spontaneity of Redox Reactions
A positive voltage that forms across the electrodes of a voltaic cell indicates that the oxidation-reduction reaction is a spontaneous reaction for reduction at the cathode and oxidation at the anode.
Conversely, if the potentials of the half-cells are known, then it is possible to predict whether a given redox reaction will be spontaneous (i.e. result in a positive voltage) A negative voltage indicates that the reverse reaction is spontaneous (i.e. oxidation at the cathode, and reduction at the anode; by convention you would need to swap the labels on the electrodes) Basically, any current flow indicates that there is a spontaneous redox reaction occurring in a voltaic cell. The sign of the voltage indicates at which electrode the reduction or oxidation is occurring. (electrons flow towards the half-cell where reduction is occurring - by convention, the cathode)
As long as we can identify the actual reduction and oxidation processes that will occur in a redox reaction, the general description of the standard reduction potential for any redox reaction (and not just one occurring in a voltaic cell) would be: E0 = E0red (reduction process) - E0red (oxidation process)
Thus, E0 will be positive for the case where the reaction is spontaneous E0 will be zero for a redox reaction at equilibrium E0 will be negative for the case where the reaction is spontaneous in the reverse direction Note that there is no reference here to what is the cathode and what is the anode
Activity Series of Metals The activity series of metals lists the metals in decreasing order of their relative ease of oxidation:
The stronger the reduction potential, the more difficult it is to oxidize the compound. The weaker the reduction potential, the easier it is to oxidize the compound
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EMF and Free Energy Change Free energy change, ΔG, is a measure of the spontaneity of a chemical reaction or process. Likewise, the standard reduction potential (E0cell) for a redox reaction is also a measure of the spontaneity of a redox reaction. What is the relationship between these two values?
ΔG has units of J/mole Reduction potential (electromotive force, E) has units of Volts. 1 Volt describes the potential difference necessary to impart 1J of energy to a charge of 1 coulomb (6.24 x 10 18 electrons) o We need some way to relate charge in Coulombs (a collection of electrons) to moles of electrons (another collection of electrons) ΔG = -nFE
n = number of moles of electrons transferred in the reaction F = the quantity of electrical charge (in coulombs) that is contained in 1 mole of electrons (this is the Faraday constant). One Faraday is equal to 96,500 coulombs/mole of e-. Since 1 volt = 1Joule/coulomb, one Faraday also equals 96,500 J/volt*mole eJoules/mol = -(electrons)* (coulombs/mol of electrons) * (Joules/coulomb)
F, and n, are positive values. Therefore, a positive value of E (which indicates spontaneity) is a negative value for ΔG (which also indicates spontaneity)
If everything is in the standard state: ΔG0 = -nFE0
Effect of Concentration on Cell EMF The EMF of a redox reaction in a voltaic cell is determined not only by the type of redox reaction, but also the concentrations of the reactants and products (i.e. the reducing agent and oxidizing agent)
The EMF of the cell will fall as the reactants are used up and products increase in concentration At equilibrium concentrations of reactants and products, the EMF = 0. Electrons flow spontaneously in a redox reaction because the system is attempting to achieve equilibrium. When equilibrium is achieved, net electron flow is zero.
The Nernst Equation Walther Hermann Nernst (1864 - 1941) was a German chemist who came up with an equation that related the EMF of a redox reaction on the concentration of reactants and products
Free energy and the equilibrium constant (this is from the section on thermodynamics) ΔG = ΔG0 + RT lnQ where Q is the reaction quotient
Relationship between ΔG and the electromotive force (from that genius Faraday) ΔG = -nFE where n = moles of electrons transferred, F = 96,500 J/V mole e -, E = EMF in volts
Substituting into the free energy equation:
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-nFE = -nFE0 + RT lnQ solving for E yields:
A common form of the equation does away with the natural log and puts the equation in the form of log10:
Yet another simplification occurs if the temperature of the reaction is known. For example, 298K is a common reference temperature for redox reactions (this is room temperature). At 298K the value of 2.303*RT/F = 0.0592 V mole:
This is the Nernst Equation. It allows us to do the following for redox reactions: 1. 2.
If we know the value of E0 and we measure the EMF of the cell (under non-standard conditions) we can determine the value of Q (and therefore the concentrations of reactants and products) If we know the value of E0 and the concentrations of reactants and products, then we can determine the resulting EMF of the cell
Equilibrium Constants for Redox Reactions What happens when the redox reaction achieves equilibrium concentrations of reactants and products?
From thermodynamics, ΔG = 0 If ΔG = 0, then from Faraday's analysis of the EMF, E = 0 (because ΔG = -nFE, and n and F are positive and non-zero, so E must be zero) In other words, if the redox reaction is at equilibrium, there is no longer any net reduction or oxidation reaction that drives an electromotive force (the cell is "dead")
or, rearranging to solve for K:
What does this equation tell us?
This says that the equilibrium constant is directly related to the standard EMF for the redox reaction (conversely, the standard EMF for the reaction is directly related to the value of the equilibrium constant)
A redox reaction where the equilibrium lies far to the right, will have a large value for K, and a large value for the standard EMF
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If we can calculate the standard EMF for a reaction (by comparing the relative difference between the two half-reactions) we can derive the equilibrium constant for the reaction
Commercial Voltaic Cells Voltaic cells provide a convenient, safe and portable supply of electrical energy.
The Industrial Revolution marked the development of heat engines, and other devices, that utilized the energy released from combustion reactions in the form of heat The Digital Age (Electronic Revolution?) revolves around devices that require energy in the form of electrical energy. Such devices require voltaic cells and redox chemistry (as opposed to gasoline and combustion chemistry)
Technology in transition
The automobile industry has relied upon combustion reactions (the internal combustion engine). But these pollute, causing acid rain and global warming. Also, fossil fuels are a limited resource. Electric cars promise lower pollution, and are essentially silent (an important feature in a population-dense world). A few electric cars have been developed, but they have problems: o Range is limited o The electrical systems are heavy o They are expensive All of the above drawbacks are related primarily to the voltaic cells
Hybrid combustion/voltaic systems have been developed and can be purchased o These designs combine a small internal combustion (IC) engine with an electric motor and voltaic cells o The IC engine may power an electric generator for the electric subsystem. The IC engine can therefore run at its most efficient rpm at all times (even when the car is not moving). Alternatively, the IC engine can drive the wheels directly during situations when extra power is needed (starting, passing, going uphill, etc.). All other times, the car is powered by the electric motor. o These cars still have problems related to the size/weight/storage capacity of voltaic cells
Lead-Storage Battery The redox half-reactions in a lead-storage (lead acid) battery are as follows:
Cathode: PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- PbSO4(s) + 2H2O(l)
Anode: Pb(s) + SO42-(aq) PbSO4(s) + 2e-
Overall redox reaction:
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PbO2(s) + Pb(s) + 4H+(aq) + 2SO42-(aq) 2PbSO4(s) + 2H2O(l)
Sulfuric acid provides the protons and sulfate ions: 2H 2SO4(aq) 2SO42-(aq) + 4H+(aq) The solid electrodes (Pb and PbO2) do not react in their respective redox reactions to produce soluble ions. In both cases, PbSO4(s) is formed and remains attached as a solid to the electrode(s). Thus, ions do not diffuse from one half-cell to the other. Therefore, the two electrodes can be placed in the same container of acid. Water is produced and sulfuric acid is consumed during the reaction. The EMF per "cell" under standard conditions is: E0cell = E0red (cathode) - E0red (anode) = (+1.685 V) - (-0.356 V) = 2.041 V o
6 cells can be combined end-to-end (i.e. in series) to produce about 12 V (what you find in a typical car) This is a reversible reaction. If electrical current is applied in the opposite direction (this is the job of a generator or alternator in your car) the electrodes are regenerated 2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 4H+(aq) + 2SO42-(aq)
Dry Cell These are your basic (not alkaline) type battery. The 6V battery in your emergency flashlight is most likely a dry cell type. It was invented in 1866. The reactions are curiously rather complex. A simple version of the half-reactions is as follows:
Cathode: 2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O(l)
Anode: Zn(s) Zn2+(aq) + 2e-
The construction consists of a zinc electrode (for the anode). The cathode is a bit weird - it is an inert support of graphite is immersed in a paste of ammonium chloride and manganese dioxide. The dimanganese trioxide solid precipitates out on the surface of the inert graphite This is not reversible, so the battery cannot be recharged (the electrode reaction products diffuse throughout the cell) In an "alkaline" type battery the ammonium chloride is replace by potassium hydroxide (KOH). This provides more useable voltage and greater capacity than the typical dry cell
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Nickel Cadmium (Nicad, or Cadnica cell)
Cathode: NiO2(s) + 2H2O(l) + 2e- Ni(OH)2(s) + 2OH-(aq)
Anode:
Cd(s) + 2OH-(aq) Cd(OH)2(s) + 2eThis type of cell uses a cadmium anode and a nickel dioxide cathode
The solid products of the respective electrode reactions adhere to the electrodes and do not diffuse throughout the cell. Thus, the redox reaction is reversible (i.e. like the lead acid cell, the nickel cadmium cell is reversible) No gases are produced, so the cell can be sealed
Fuel Cells Combustion reactions produce heat that, in turn, can be used to produce electricity. However, typically less than 40% of the heat energy is converted to electrical energy - the rest is "wasted" as heat. Combustion reactions are actually redox reactions: diatomic oxygen (0 oxidation number) is reduced to carbon dioxide (-2 oxidation number) or water (-2 oxidation number). Direct production of electricity from redox chemistry, instead of combustion, for these reactions could result in higher efficiency of production of electrical energy. Voltaic cells that perform this type of redox reaction for conventional fuels (such as hydrogen or methane) are called fuel cells
A common reaction being utilized in fuel cells is the reduction of oxygen by hydrogen o Cathode: O2(g) + 2H2O(l) + 4e- 4OH-(aq)
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Anode: 2H2(g) + 4OH-(aq) 4H2O(l) + 4e-
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Overall reaction:
2H2(g) + O2(g) 2H2O(l) This is currently an very expensive way to generate energy, but is extremely efficient and compact. Its broadest application to date has been to provide electricity (and drinking water) for spacecraft. Electrolysis
Voltaic cells are based upon the spontaneity of a particular redox reaction
This in turn is based upon the relative reduction potentials for the two half-reactions The half-reaction with the greater reduction potential will drive the other half-reaction in the direction of oxidation
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The relative difference between the two reduction potentials provides a quantitative description of the electromotive force of the voltaic cell
However, we have seen that some commercial voltaic cells can be "recharged"
This is achieved by the application of current in the opposite direction to that of the spontaneous direction of current flow in the voltaic cell This application of current in the opposite direction will cause the redox reaction to proceed in the opposite direction o The half cell where oxidation occurred will now undergo a reduction reaction o The half cell where reduction occurred will now undergo an oxidation reaction
Sodium metal and chlorine gas spontaneously react to form the ionic compound Sodium Chloride: 2Na(s) + Cl2(g) 2NaCl(s)
This reaction is a redox reaction: Anode: 2Na(s) 2Na+(s) + 2eCathode: Cl2(g) + 2e- 2Cl-(s)
The reverse reaction, the decomposition of ionic NaCl into Na(s) and Cl 2(g), can be achieved by the application of an EMF (i.e. external current)
2NaCl(l) 2Na(l) + Cl2(g) Redox reactions that are driven by an external EMF are called electrolysis reactions and take place in electrolytic cells The electrolytic cell contains two electrodes and either a molten salt solution (as with NaCl) or some other type of solution (e.g. aqueous)
The external EMF is provided by a source of direct current (sometimes another voltaic cell, or battery) This external current acts as an electron pump to push electrons onto one electrode (and drive a reduction reaction) and to withdraw electrons from the other electrode (and drive an oxidation reaction) o The electrode where oxidation occurs is still called the anode. However, for an electrolytic cell it is labeled as "+" to indicate that this is the electrode where electrons are being withdrawn by the external EMF o The electrode where reduction occurs is stall called the cathode. However, for an electrolytic cell it is labeled as "-" to indicate that this is the electrode where electrons are being pumped into by the external EMF
Electrolysis of Aqueous Solutions With aqueous solutions of salts it must be considered that the electrolysis might end up driving the reduction or oxidation of H2O(l)
H2O(l) can be oxidized to produce O2(g) H2O(l) can be reduced to produce H2(g) If a salt is present, the cation of the salt may or may not be preferentially reduced compared to H2O(l). Likewise, the anion may or may not be preferentially oxidized compared to H 2O(l)
Here are the reduction potential data for the reduction of H2O(l) and Na+(aq): 2H2O(l) + 2e- H2(g) + 2OH-(aq) E0red = -0.83 V 0 Na+(aq) + e- red = -2.71 V
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When comparing any two reduction reactions the reaction that is favored is the one with the more positive reduction potential. In this case, the reduction of H2O(l) is the favored reduction reaction
Here are the reduction potential data for the oxidation of H2O(l) and Cl-(aq). Remember that the reduction potential provided is referencing the reduction (i.e. reverse) reaction + 2H2O(l) (aq) + O2(g) + 4e- E0red = 1.23 V 2Cl (aq) Cl2(g) + 2e- E0red = 1.36 V
When comparing the reduction potentials, the reduction of Cl- is favored. However, we are interested in the reverse reactions, i.e. the oxidation reactions. In this case, remember that a substance that is readily reduced is difficult to oxidize, and vice versa. Thus, if the Cl - is favored for reduction, then the water reaction is favored for oxidation What this analysis suggests is that electrolysis of an aqueous solution of Na +(aq) and Cl-(aq) ions will not result in the reduction of Na+ or oxidation of Cl-, but rather, will result in the reduction of H2O to produce hydrogen, and the oxidation of H 2O to produce oxygen.
Note: although this analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E0cell. This may be due to kinetic rather than thermodynamic considerations. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame we must increase the voltage of the external source (hence, overvoltage). In the case of the oxidation of H2O versus Cl-(aq), the oxidation of H2O to produce O2(g) is thermodynamically favored, but kinetically quite slow. Thus we could try to crank up the voltage. However, the oxidation of chlorine occurs at a voltage not much higher than that for water. Thus, in practical terms, oxidation of chlorine will preferentially take place at the anode.
The redox reactions for the electrolysis of an aqueous solution of NaCl are therefore: (reduction) 2H2O(l) + 2e- H2(g) + 2OH-(aq) (oxidation) 2Cl-(aq) 2(g) + 2e
And the overall reaction is:
2H2O(l) + 2Cl-(aq) 2(g) + H2(g) + 2OH (aq) Thus, while electrolysis of molten NaCl produces chlorine gas and sodium metal, the electrolysis of an aqueous solution of NaCl produces chlorine gas, hydrogen gas and hydroxide ion
What external EMF is required do drive this particular electrolysis?
(reduction; cathode) 2H2O(l) + 2e- H2(g) + 2OH-(aq) E0red = -0.83 V - 0 (oxidation; anode) 2Cl-(aq) 2(g) + 2e E red = 1.36 V 0 0 0 E cell = E (cathode) - E (anode) = -0.83 V - (1.36 V) = -2.19 V This value is negative, indicating that the reaction is not spontaneous as written and must be driven by an external EMF Electrolysis with Active Electrodes
Inert electrodes do not enter into the chemical reaction, and just serve as a surface upon which electron transfers can occur (and be conducted to the other half-cell) Active electrodes chemically participate in the redox reaction
If an aqueous solution contains solid metal electrodes, the electrodes may be oxidized to the ionic form if the metal is easier to oxidize than water. Since we think in terms of reduction potentials and not ease of
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oxidation, this means that the metal electrodes will be oxidized to ionic form if the reduction potential of the metal is less than that of water Ni(s) Ni2+(aq) + 2e- E0red = -0.28 V 2H2O(l) 4H+(aq) + O2(g) + 4e- E0red = 1.23 V
In this case the reduction potential of H2O(l) is greater than that for Nickel metal. Thus, H2O is easier to reduce, and therefore more difficult to oxidize, than Nickel. Thus, at the anode, Nickel will preferentially be oxidized compared to H2O(l) if we provide an external EMF that "sucks" electrons out at this electrode
What about the cathode (reduction) reaction involving Nickel ion and water? Ni2+(aq) + 2e- Ni(s) E0red = -0.28 V 2H2O(l) + 2e- H2(g) + 2OH-(aq) E0red = -0.83 V
In this case, the half-reaction with the greater reduction potential is the reduction of Nickel. Therefore, at the cathode, Nickel will preferentially be reduced if we provide an external EMF that forces electrons onto this electrode
Thus, a metallic cathode may be coated with solid nickel metal in a process known as "electroplating" Quantitative Aspects of Electrolysis
Reduction half-reactions provide information about the stoichiometric requirements for the production of the elemental form of the metal. In particular, the number of electrons needed per molar basis of metal reduced Na+ + e- Na Cu2+ + 2e- Cu Al3+ + 3e- Al
From the above information, 1 mole of electrons will allow the production (i.e. reduction) of 1 mole of sodium metal, 0.5 mole of copper metal, and 0.333 mole of aluminum metal
For any half-reaction the amount of substance reduced is proportional to the number of electrons that pass into the cell
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We can quantitate the number of electrons by keeping track of the charge that passes into a cell
Charge is measured in Coulombs 1 mole of electrons has a charge of 96,500 Coulombs (C). This number of Coulombic charge is equal to 1 Faraday (F)
Charge of 1 mole of electrons = 96,500 C = 1F The word, current, refers to the rate of flow of electricity
One ampere (amp) is equal to a rate of flow of charge of 1 Coulomb every second amps = Coulombs/second and Coulombs = amps * seconds
In summary… Amps = charge / unit of time Charge the number of electrons The number of electrons the number of moles of a compound reduced
Electrical Work Relationship between E and ΔG:
A positive value for E (electromotive force) is associated with a spontaneous redox reaction Likewise, a negative value for ΔG is associated with a spontaneous reaction or process +E or -ΔG means a spontaneous reaction
The relationship between ΔG and E: ΔG = -nFE
Useful work
For any spontaneous process, ΔG is a measure of the maximum useful work (wmax)that can be done by the process. Therefore, -nFE also is a measure of the maximum useful work that can be done by a redox reaction:
wmax = -nFE wmax = moles * (Coulombs/moles) * (Joules/Coulomb) = Joules Note: work done by the system on the surroundings is indicated by a negative sign for w Non-spontaneous redox reactions
Electrolysis describes the application of an external EMF to drive a non-spontaneous process For a non-spontaneous process, ΔG is positive and E will be negative To force the redox reaction to proceed, the external potential must be larger in magnitude than the EMF of the cell (i.e. Ecell) Eexternal - Ecell
When an external potential is applied, the surroundings do work on the system. Therefore, we expect the sign of the w term to be positive
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wmax = +nFEexternal Units of power
A watt is a unit of energy expenditure, i.e. the amount of energy consumed or produced per unit of time. It has units of Joules/second: 1 watt = 1 Joule/second and 1 Joule = 1 watt * second
How does the Power Company charge you for your electric usage? It charges you for the watt*hours that you consume, or kilowatt*hours 1 k watt hour = (1000 watt)*(1 hour) (360 sec/hour) ((1 J/s)/watt) = 3.6 x 10 6 Joules
Corrosion Corrosion reactions are redox reactions in which the elemental form of a metal is oxidized to produce an unwanted compound or a water soluble ion
Oxidized forms of a metal may not have the strength and conductive properties of the metal in the elemental form
Diatomic oxygen in the atmosphere has a relatively high reduction potential - higher than many metals
For metals exposed to air, oxygen (in the presence of H +) is reduced to H2O, and the metal is oxidized, in a spontaneous redox reaction
For some metals, however, the initial oxidation on the surface protects the underlying metal against further oxidation. This is because the metal oxide layer is impervious to both oxygen and H 2O (a source of H+)
This is the case for aluminum, which although quite reactive upon exposure to oxygen and water, develops a metal oxide surface layer that protects the metal underneath
Corrosion of Iron The corrosion of iron (i.e. rusting) costs the U.S. economy about $70 billion annually
Formation of rust requires both O2 and H2O Anode reaction: Fe(s) Fe2+(aq) + 2e- E0red = -0.44 V
Cathode reaction: O2(g) + 4H+(aq) + 4e- 2H2O(l) E0red = 1.23 V Thus, the reduction of oxygen is favored over the reduction of iron. In other words, the iron will preferentially oxidize.
The Fe2+ that is formed eventually is oxidized further to Fe3+ which forms a hydrated iron (III) oxide (i.e. rust)
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Prevention of Corrosion If iron is painted, then oxygen and water are prevented from contacting the metal, and corrosion is avoided Sometimes the iron is coated with a thin layer of another metal, such as Tin (the layer of Tin keeps oxygen and water away from the iron) 2000 Dr. Michael Blaber
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Oxidation - refers to the loss of electrons by a molecule, atom or ion Reduction - refers to the gain of electrons by an molecule, atom or ion Oxidation number - the oxidation number of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom; it is the charge an atom would possess if the bonding were ionic
In oxidation numbers, the charge is written before the number (to distinguish them from actual electronic charges - where the number is placed in front of the charge)
HCl Hydrogen is less electronegative than Chlorine, therefore, Chlorine gets all the shared electrons in the HCl bond. In this compound, the H oxidation number is therefore +1, and the oxidation number of Cl is -1. Chemical reactions in which the oxidation state of one or more substances changes are called oxidationreduction reactions (or redox reactions)
Keeping track of the oxidation numbers of the elements involved in a reaction will identify whether the reaction is a redox reaction or not
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) The oxidation numbers for the above elements and ions are:
Zn = 0, Zn2+ = +2 H+ = +1, H2 = 0 Thus, the oxidation number of both the Zn(s) and H +(aq) change during the course of the reaction, and so, this must be a redox reaction
In this reaction an unambiguous transfer of electrons occurs:
The elemental form of zinc (i.e. zinc metal) loses two electrons (and becomes soluble zinc ion) Two H+ ions in solution gain two electrons and form a hydrogen (H2) molecule
In other types of reactions, we can identify changes in oxidation number, but it is not so clear that a particular atom has gained or lost an electron:
Hydrogen has been oxidized and oxygen has been reduced in the above reaction involving the production of water from elemental hydrogen and oxygen However, the H-O bonds are not ionic, they are polar covalent, so there is not a complete transfer of shared electrons
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Using oxidation numbers is just a convenient bookkeeping method to understand reduction/oxidation types of reactions. It does not necessarily mean that complete electron transfer has occurred.
In a redox reaction, both oxidation and reduction must occur. In other words, something must be oxidized for something else to be reduced
The compound that contributes the electrons is called the reducing agent. The reducing agent gives up electrons, causing another compound to be reduced (and is therefore oxidized in the process) The compound that accepts electrons is called the oxidizing agent. The oxidizing agent accepts electrons from another compound, causing the other compound to be oxidized (in the process, the oxidizing agent is reduced)
In the above reaction of hydrogen and oxygen:
Hydrogen is the reducing agent (it contributes electrons to oxygen, and is itself oxidized) Oxygen is the oxidizing agent (it accepts electrons from hydrogen - oxidizing it, and is itself reduced) Balancing Oxidation-Reduction Reactions
In balancing redox reactions, gains and losses of electrons must be balanced
In many cases the equation can be balanced without explicitly considering the transfer of electrons. For example: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
However, many redox reactions are more complicated and cannot be balanced without taking into account the number of electrons lost and gained during the reaction
Half-Reactions Although oxidation and reduction must take place simultaneously (i.e. for something to be reduced, something else must be oxidized) it is often convenient to consider them as separate processes
The following is the oxidation of Tin(II) by Iron(III) Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
This reaction can be considered as two separate processes: o The oxidation of Sn2+ o The reduction of Fe3+ Oxidation: Sn2+(aq) Sn4+(aq) + 2eReduction: 2Fe3+(aq) + 2e- 2Fe2+(aq)
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In the oxidation reaction the electrons are shown as products (i.e. the tin is oxidized and releases electrons) In the reduction reaction the electrons are shown as reactants (i.e. the iron ion combines with electrons as it is reduced) Equations that show either oxidation, or reduction, alone are called half-reactions
The overall electrons produced in the oxidation half-reaction must equal the number of electrons consumed in the reduction half-reaction o When the equations have been balanced in this way, they are combined to produce the balanced overall redox reaction
Balancing Equations By The Method Of Half-Reactions The use of half-reactions provides a general way to balance redox reactions The reaction between permanganate ion (MnO4-) and oxalate ion (C2O42-) in acidic solution:
When these ions combine in an acidic solution, they react to produce manganese ion (Mn2+) and CO2 gas Here is the unbalanced equation that describes this reaction:
MnO4- + C2O42- Mn2+(aq) + CO2(g) To balance this redox reaction using the method of half-reactions, begin by writing the incomplete oxidation and reduction half-reactions: MnO4- Mn2+(aq) C2O42- CO2(g) Which compound is being reduced and which one is being oxidized?
Now, balance the atoms undergoing oxidation or reduction by the appropriate addition of coefficients.
In the above half-reactions, the Mn appears to be balanced on both sides of its half-reaction However, the carbon is not balanced in the second half-reaction - we need another carbon in the products
MnO4- Mn2+(aq) C2O42- 2CO2(g) If the reaction is done under acidic aqueous solution, H+ and H2O can be added to reactants or products to balance H and O atoms. Likewise, if the reaction is done under basic aqueous solution, OH - and H2O can be added to balance H and O atoms
In the manganese half-reaction we have 4 O atoms on the left and none on the right. We can add 4 H2O molecules on the right to take care of this: MnO4- Mn2+(aq) + 4H2O(l)
We now have 8 H atoms on the right that need to be balanced. This reaction is done under acidic conditions, so we can balance the H atoms by adding H+(aq) ions on the left: MnO4- + 8H+ Mn2+(aq) + 4H2O(l)
The atoms are now balanced, but the charges are not. We have a net charge of 7+ on the left, and 2+ on the right. We want to balance the charge without altering the number of atoms (they are balanced already). We balance the charge by adding the appropriate number of electrons:
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MnO4- + 8H+ + 5e- Mn2+(aq) + 4H2O(l) (balanced atoms, balanced charge)
How about the other half-reaction? C2O42- 2CO2(g)
The atoms are balanced, but the charges are not. We have a net charge of 2- on the left and 0 on the right. We can balance this by adding the appropriate number of electrons: C2O42- 2CO2(g) + 2e(balanced atoms, balanced charge)
Are we done? No. By the original definition above, in a redox reaction the overall electrons produced in the oxidation half-reaction must equal the number of electrons consumed in the reduction half-reaction. We have not yet identified which is the oxidation half-reaction and which is the reduction half-reaction, but it is clear that the electrons are not balancing (we have one halfreaction with 5e- and the other with 2e-) o The half-reactions must be multiplied by some factor, such that the electrons present in the half-reactions balance each other o For the above half-reactions, the lowest common multiple is 10. Thus, the half-reaction with 5e- should be multiplied by 2, and the half-reaction with 2e- should be multiplied by 5: 2MnO4- + 16H+ + 10e- 2Mn2+(aq) + 8H2O(l) 5C2O42- 10CO2(g) + 10e-
We can now describe the redox reaction accurately: o The oxidation half-reaction is: 5C2O42- 10CO2(g) + 10eo
The reduction half-reaction is:
2MnO4- + 16H+ + 10e- 2Mn2+(aq) + 8H2O(l) In other words, the oxalate ion is oxidized by the permanganate ion (and the permanganate ion is reduced) o
The overall balanced redox reaction is the combination of the two balanced halfreactions: 2MnO4- + 16H+ + 10e- 2Mn2+(aq) + 8H2O(l) 5C2O42- 10CO2(g) + 10e+ 2MnO4 + 16H + 5C2O42- 10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Summary of balancing half-reactions in acidic solutions: 1. Divide reaction into two incomplete half-reactions 2. Balance each half-reaction by doing the following: a. Balance all elements, except O and H b. Balance O by adding H2O c. Balance H by adding H+ d. Balance charge by adding e-as needed
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3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to get a common multiple 4. The overall reaction is the sum of the half-reactions 5. The oxidation half-reaction is the one that produces electrons as products, and the reduction half-reaction is the one that uses electrons as reactants Balancing Redox Reactions That Occur in Basic Solution In a basic aqueous solution the half-reactions must be balanced using OH- and H2O (instead of H+ and H2O)
The procedure for balancing in this case is actually the same as for balancing in an acidic solution, with one small twist: o Since we are under basic conditions, we have to neutralize any H + ions in solution o We achieve this by adding the appropriate number of OH- ions o We keep things balanced by adding the same number of OH - ions to both sides of the half-reaction
Voltaic Cells A spontaneous redox reaction involves the transfer of electrons
In principle, this transfer or movement of electrons can be used to do a type of work - specifically, electrical work A type of device that makes use of redox reactions to produce electron flow, and to allow electrical work, is known as a Voltaic Cell (named after Count Alessandro Volta, 1745-1827, an Italian physicist)
The development of the voltaic cell begins as follows... A spontaneous redox reaction that we have previously discussed involves the metals Zinc (Zn) and Copper (Cu)
If you recall the Activity Series for metals and hydrogen ion:
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When comparing Zinc with Copper in the activity series, we expect that Zinc should be easier to oxidize (i.e. should give up electrons easier than Copper) The spontaneity of the reaction between zinc and copper ion reflects the Activity Series relationship: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) o o
The reaction as shown is spontaneous. The Zinc metal is oxidized, and the Copper ion is reduced The reverse reaction, the oxidation of Copper and the reduction of Zinc ion, is not spontaneous
The redox half-reactions for the above reaction would be: Oxidation: Zn(s) Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- Cu(s)
The oxidation of a Zinc atom releases 2 electrons The reduction of a Copper ion is achieved by the acceptance of 2 electrons
Thus, there would appear to be a movement, or flow, of electrons from the Zinc metal to the Copper ions According to the above equation, if Zinc metal is placed in an aqueous solution containing Cu2+ ions (e.g. a solution of copper sulfate salt), the following will occur:
At the surface of the Zinc metal, the oxidation of Zinc atoms will be coupled to the reduction of Cu2+ ions Electrons from the oxidation of the zinc will reduce the copper ions to elemental copper on the surface of the zinc Over time, the zinc metal will dissolve (as solid zinc is oxidized to zinc ion) and zinc ion will build up in the solution. Also over time, the solution will no longer contain copper ions, an the copper will be present in elemental form
In the above reaction, the zinc and copper metal/ions are in direct contact with each other and that is why the reduced form of the copper (i.e. the metal) builds up on the surface of the zinc
What if we separated the Cu(s)/Cu2+ from the Zn(s)/Zn2+?
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The reaction will not occur because there is no way for the electrons released by the oxidation of Zinc metal to get over to the Copper ions and reduce them to Copper metal
What if we provide a path of conductance for the electrons released by the oxidation of the zinc to get over to the copper?
Now that the electrons have a path to the Cu/Cu2+ side it would appear that the reaction could proceed However, we now have another problem ...
At the beginning of the redox reaction we have a neutral aqueous salt solution, e.g. ZnSO 4 (i.e. the concentration of cation equals the concentration of anion) As the redox reaction proceeds we build up Zn2+ ions in the solution where the Zinc is being oxidized. Conversely, we remove Cu2+ ions from solution where the Copper ion is being reduced. The anion concentration (sulfate ion in this case) does not change. Thus, we are building up a net positive charge in the zinc solution, and a net negative charge in the copper solution These charges will oppose the flow of electrons. The positive charge in the zinc solution will make it harder for the negative electrons to leave. Likewise, the negative charge in the copper solution will repel the electrons that are trying to come over from the zinc side
We need a way to neutralize the charge build-up in the solutions due to the change in soluble ion concentration What if we had a tube filled with aqueous solution that connected the two redox reactions?
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This would allow anion to move from the copper reduction side to the zinc oxidation side and keep the charges in solution neutral In turn, this would now allow the electrons to flow However, as the reaction proceeds, we have yet another problem...
As the reaction proceeds, although movement of anion allows the overall charges in solution to remain neutral, the net movement of anion produces a concentration gradient across the two solutions. In other words, after a while the net concentration of ions in the zinc oxidation side will be greater than in the copper reduction side. This concentration gradient will oppose movement of anion Actually, this is not really a problem that we will see. What it means is that we have to realize that charge neutralization can occur by either anions moving to the left or cations moving to the right
In consideration of keeping the overall concentration of ions in balance between the two sides, cations will also be moving to the right:
Thus, in the connecting tube of solution we have net movement of both types of ions: o anions are going into the oxidation side o cations going into the reduction side
The connecting tube of solution is called a Salt Bridge Summary of the movement of ions, electrons and the redox half-reactions in a voltaic cell:
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The two solid metals in the different half-reactions are called electrodes
The metal in the half-reaction where oxidation is occurring is called the Anode The metal in the half-reaction where reduction is occurring is called the Cathode The cathode is often labeled with a "+"; "this electrode attracts electrons" The anode is often labeled with a "-"; "this electrode repels electrons"
Cell EMF Here is a summary of a voltaic cell:
A key question is: what causes the electrons to flow from the anode to the cathode?
The electrons flow from the anode to the cathode because of a difference in the potential energy of the electrons at the anode compared to the cathode In particular, the potential energy of the electrons is higher at the anode than at the cathode Like water molecules in a waterfall, the electrons move in a direction from high potential energy to low potential energy (i.e. "downhill" in an energy landscape)
The potential difference between two electrodes is measured in Volts (V)
One volt is the potential difference, between an anode and a cathode, that would be required in order to impart 1 Joule of energy to a charge of 1 coulomb (note: 1 coulomb is the charge of a collection of 6.24 x 1018 electrons) 1 V = 1 J/C (i.e. energy per electron)
The potential difference between two electrodes is called the electromotive force, or emf The emf of a voltaic cell is called the cell potential, or the cell voltage
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The cell voltage of a voltaic cell will be positive value
The cell voltage will depend upon the specific half-reactions, concentrations of ions, and temperature
Common tabulations of cell voltages are performed at 25°C and standard conditions of concentrations of liquid (i.e. 1M) and gas components (i.e. 1 atm) Under standard conditions, the emf of a voltaic cell is called the standard emf or the standard cell potential (E0cell) By convention, the potential associate with an electrode is the potential for reduction The standard cell potential is given by the standard reduction potential of the cathode, minus the standard reduction potential of the anode
E0red(cathode)
E0cell = E0red(cathode) - E0red(anode) 0 E red(anode) then the electron flow from anode
If > to cathode is spontaneous Standard Reduction Potentials The cell potential of a voltaic cell depends upon the two redox half-reactions that comprise the functioning cell
There are many different combinations of redox half-reactions. Do we need to tabulate all possible unique combinations? As you might have guessed, the answer is no, we don't. In principle, we could assign a unique standard potential to each half-reaction and by comparing these values determine the cell potential for any combination of half-reactions
There is one problem, however: we have seen that an isolated half-reaction doesn't result in reduction or oxidation or electron flow. Therefore, how are we going to assign a standard potential to a half-reaction?
We are going to use a generally agreed upon reduction reaction as the reference half-reaction against which to compare and tabulate all possible half-reactions The reference reaction is the reduction of H+(aq) ions to produce H2(g): 2H+(aq, 1M) + 2e- H2(g, 1 atm) E0red = 0.0V This is a Standard Hydrogen Electrode (SHE)
How do we use this SHE to determine the standard reduction potential, E 0red, for some redox half-reaction?
Let's say we are interested in determining the standard reduction potential for Zn2+ We would set up a voltaic cell using a SHE as one half-cell, and a Zinc electrode as the other halfcell
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The half-cell reactions are: Zn(s) Zn2+(aq) + 2e- (Zinc electrode) 2H+(aq) + 2e- H2(g) (SHE)
The overall cell reaction: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)
Note: this is performed under Standard Conditions, where concentrations = 1.0M, and p = 1 atm
For the reaction as shown, the zinc electrode is the anode (oxidation), and the SHE is the cathode (reduction) o The voltage for the overall cell is measured at 0.76 Volts (i.e. this is the standard cell potential) o From the previous definition of the standard cell potential: E0cell = E0red(cathode) - E0red(anode) o
We know the value for the standard cell potential, it is +0.76V. We also know the value for the reduction potential for the SHE, it is 0V. Thus, 0.76V = E0red(cathode) - E0red(anode) 0.76V = 0 - E0red(anode) E0red(anode) = -0.76V
In other words, in reference to the SHE, the standard reduction potential for the zinc electrode is determined to be -0.76V.
This may seem a little strange because the cell reads +0.76V and yet the determined reduction potential for the zinc electrode is -0.76V.
The key here is that the original definition of the standard cell potential is set up such that each cell is referenced as a reduction reaction. In other words what we have actually determined is the half-cell potential (in reference to the SHE) for the reduction of zinc:
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Zn2+(aq) + 2e- Zn(s) E0red = -0.76V Whenever we assign a potential to a half-reaction using the equation for the standard cell potential, it is always done in reference to a reduction reaction
Thus, in reference to the SHE, the oxidation of Zn(s) is associated with a half-cell potential of +0.76V (this is the spontaneous direction for the reaction; the reduction of Zn2+ and the oxidation of H2(g) is not spontaneous)
The standard cell potential (Voltage) describes the potential per unit charge that the reaction produces
This value of the potential per unit charge (i.e. electron) is a constant that is a property of the two half-reactions that are reacting in the cell. It is independent of the stoichiometry or amount of the reactants in the cell
The more positive the value of E0red the greater the driving force for reduction In reference to H+/H2(g) , Ag+ will preferentially be reduced to Ag(s) and H2(g) will be oxidized to H+(aq). In reference to H+/H2(g) , Zn(s) will preferentially be oxidized and H+ will be reduced to H2(g). The cathode is the electrode where reduction occurs, therefore, in order for a cell to operate spontaneously, the half-reaction at the cathode must have a more positive value of E0red than the half-reaction at the anode o The greater driving force for reduction at the cathode can force the reaction at the anode to operate "in reverse" - i.e. to undergo the oxidation reaction The difference between the standard reduction potentials of the two reactions, E0red o (cathode) - E0red (anode), is the excess potential that can be used to drive electrons through the cell, E0cell
Oxidizing and Reducing Agents In the above discussion of reduction potentials and their relationship to one another in half-reactions, we can predict which direction current will flow in a voltaic cell. In other words, we can predict which reactant will be oxidized and which will be reduced
The same information can be used for redox reactions where the compounds are in direct contact, and not separated into half-cells. For example, from the example above, we would predict that silver ion will oxidize lithium metal as a spontaneous redox reaction. Alternatively, we would predict that the reduction of lithium ion by silver metal is not a spontaneous reaction (and would require the input of energy to drive)
The more positive the E0red value of a half-reaction, the greater the tendency of the compound to be reduced - and therefore to oxidize another compound (i.e. the greater the tendency to act as an oxidizing agent)
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Substances that are strong oxidizing agents are subsequently weak reducing agents. Likewise, substances that are strong reducing agents are weak oxidizing agents. Spontaneity of Redox Reactions
A positive voltage that forms across the electrodes of a voltaic cell indicates that the oxidation-reduction reaction is a spontaneous reaction for reduction at the cathode and oxidation at the anode.
Conversely, if the potentials of the half-cells are known, then it is possible to predict whether a given redox reaction will be spontaneous (i.e. result in a positive voltage) A negative voltage indicates that the reverse reaction is spontaneous (i.e. oxidation at the cathode, and reduction at the anode; by convention you would need to swap the labels on the electrodes) Basically, any current flow indicates that there is a spontaneous redox reaction occurring in a voltaic cell. The sign of the voltage indicates at which electrode the reduction or oxidation is occurring. (electrons flow towards the half-cell where reduction is occurring - by convention, the cathode)
As long as we can identify the actual reduction and oxidation processes that will occur in a redox reaction, the general description of the standard reduction potential for any redox reaction (and not just one occurring in a voltaic cell) would be: E0 = E0red (reduction process) - E0red (oxidation process)
Thus, E0 will be positive for the case where the reaction is spontaneous E0 will be zero for a redox reaction at equilibrium E0 will be negative for the case where the reaction is spontaneous in the reverse direction Note that there is no reference here to what is the cathode and what is the anode
Activity Series of Metals The activity series of metals lists the metals in decreasing order of their relative ease of oxidation:
The stronger the reduction potential, the more difficult it is to oxidize the compound. The weaker the reduction potential, the easier it is to oxidize the compound
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EMF and Free Energy Change Free energy change, ΔG, is a measure of the spontaneity of a chemical reaction or process. Likewise, the standard reduction potential (E0cell) for a redox reaction is also a measure of the spontaneity of a redox reaction. What is the relationship between these two values?
ΔG has units of J/mole Reduction potential (electromotive force, E) has units of Volts. 1 Volt describes the potential difference necessary to impart 1J of energy to a charge of 1 coulomb (6.24 x 10 18 electrons) o We need some way to relate charge in Coulombs (a collection of electrons) to moles of electrons (another collection of electrons) ΔG = -nFE
n = number of moles of electrons transferred in the reaction F = the quantity of electrical charge (in coulombs) that is contained in 1 mole of electrons (this is the Faraday constant). One Faraday is equal to 96,500 coulombs/mole of e-. Since 1 volt = 1Joule/coulomb, one Faraday also equals 96,500 J/volt*mole eJoules/mol = -(electrons)* (coulombs/mol of electrons) * (Joules/coulomb)
F, and n, are positive values. Therefore, a positive value of E (which indicates spontaneity) is a negative value for ΔG (which also indicates spontaneity)
If everything is in the standard state: ΔG0 = -nFE0
Effect of Concentration on Cell EMF The EMF of a redox reaction in a voltaic cell is determined not only by the type of redox reaction, but also the concentrations of the reactants and products (i.e. the reducing agent and oxidizing agent)
The EMF of the cell will fall as the reactants are used up and products increase in concentration At equilibrium concentrations of reactants and products, the EMF = 0. Electrons flow spontaneously in a redox reaction because the system is attempting to achieve equilibrium. When equilibrium is achieved, net electron flow is zero.
The Nernst Equation Walther Hermann Nernst (1864 - 1941) was a German chemist who came up with an equation that related the EMF of a redox reaction on the concentration of reactants and products
Free energy and the equilibrium constant (this is from the section on thermodynamics) ΔG = ΔG0 + RT lnQ where Q is the reaction quotient
Relationship between ΔG and the electromotive force (from that genius Faraday) ΔG = -nFE where n = moles of electrons transferred, F = 96,500 J/V mole e -, E = EMF in volts
Substituting into the free energy equation:
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-nFE = -nFE0 + RT lnQ solving for E yields:
A common form of the equation does away with the natural log and puts the equation in the form of log10:
Yet another simplification occurs if the temperature of the reaction is known. For example, 298K is a common reference temperature for redox reactions (this is room temperature). At 298K the value of 2.303*RT/F = 0.0592 V mole:
This is the Nernst Equation. It allows us to do the following for redox reactions: 1. 2.
If we know the value of E0 and we measure the EMF of the cell (under non-standard conditions) we can determine the value of Q (and therefore the concentrations of reactants and products) If we know the value of E0 and the concentrations of reactants and products, then we can determine the resulting EMF of the cell
Equilibrium Constants for Redox Reactions What happens when the redox reaction achieves equilibrium concentrations of reactants and products?
From thermodynamics, ΔG = 0 If ΔG = 0, then from Faraday's analysis of the EMF, E = 0 (because ΔG = -nFE, and n and F are positive and non-zero, so E must be zero) In other words, if the redox reaction is at equilibrium, there is no longer any net reduction or oxidation reaction that drives an electromotive force (the cell is "dead")
or, rearranging to solve for K:
What does this equation tell us?
This says that the equilibrium constant is directly related to the standard EMF for the redox reaction (conversely, the standard EMF for the reaction is directly related to the value of the equilibrium constant)
A redox reaction where the equilibrium lies far to the right, will have a large value for K, and a large value for the standard EMF
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If we can calculate the standard EMF for a reaction (by comparing the relative difference between the two half-reactions) we can derive the equilibrium constant for the reaction
Commercial Voltaic Cells Voltaic cells provide a convenient, safe and portable supply of electrical energy.
The Industrial Revolution marked the development of heat engines, and other devices, that utilized the energy released from combustion reactions in the form of heat The Digital Age (Electronic Revolution?) revolves around devices that require energy in the form of electrical energy. Such devices require voltaic cells and redox chemistry (as opposed to gasoline and combustion chemistry)
Technology in transition
The automobile industry has relied upon combustion reactions (the internal combustion engine). But these pollute, causing acid rain and global warming. Also, fossil fuels are a limited resource. Electric cars promise lower pollution, and are essentially silent (an important feature in a population-dense world). A few electric cars have been developed, but they have problems: o Range is limited o The electrical systems are heavy o They are expensive All of the above drawbacks are related primarily to the voltaic cells
Hybrid combustion/voltaic systems have been developed and can be purchased o These designs combine a small internal combustion (IC) engine with an electric motor and voltaic cells o The IC engine may power an electric generator for the electric subsystem. The IC engine can therefore run at its most efficient rpm at all times (even when the car is not moving). Alternatively, the IC engine can drive the wheels directly during situations when extra power is needed (starting, passing, going uphill, etc.). All other times, the car is powered by the electric motor. o These cars still have problems related to the size/weight/storage capacity of voltaic cells
Lead-Storage Battery The redox half-reactions in a lead-storage (lead acid) battery are as follows:
Cathode: PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- PbSO4(s) + 2H2O(l)
Anode: Pb(s) + SO42-(aq) PbSO4(s) + 2e-
Overall redox reaction:
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PbO2(s) + Pb(s) + 4H+(aq) + 2SO42-(aq) 2PbSO4(s) + 2H2O(l)
Sulfuric acid provides the protons and sulfate ions: 2H 2SO4(aq) 2SO42-(aq) + 4H+(aq) The solid electrodes (Pb and PbO2) do not react in their respective redox reactions to produce soluble ions. In both cases, PbSO4(s) is formed and remains attached as a solid to the electrode(s). Thus, ions do not diffuse from one half-cell to the other. Therefore, the two electrodes can be placed in the same container of acid. Water is produced and sulfuric acid is consumed during the reaction. The EMF per "cell" under standard conditions is: E0cell = E0red (cathode) - E0red (anode) = (+1.685 V) - (-0.356 V) = 2.041 V o
6 cells can be combined end-to-end (i.e. in series) to produce about 12 V (what you find in a typical car) This is a reversible reaction. If electrical current is applied in the opposite direction (this is the job of a generator or alternator in your car) the electrodes are regenerated 2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 4H+(aq) + 2SO42-(aq)
Dry Cell These are your basic (not alkaline) type battery. The 6V battery in your emergency flashlight is most likely a dry cell type. It was invented in 1866. The reactions are curiously rather complex. A simple version of the half-reactions is as follows:
Cathode: 2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O(l)
Anode: Zn(s) Zn2+(aq) + 2e-
The construction consists of a zinc electrode (for the anode). The cathode is a bit weird - it is an inert support of graphite is immersed in a paste of ammonium chloride and manganese dioxide. The dimanganese trioxide solid precipitates out on the surface of the inert graphite This is not reversible, so the battery cannot be recharged (the electrode reaction products diffuse throughout the cell) In an "alkaline" type battery the ammonium chloride is replace by potassium hydroxide (KOH). This provides more useable voltage and greater capacity than the typical dry cell
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Nickel Cadmium (Nicad, or Cadnica cell)
Cathode: NiO2(s) + 2H2O(l) + 2e- Ni(OH)2(s) + 2OH-(aq)
Anode:
Cd(s) + 2OH-(aq) Cd(OH)2(s) + 2eThis type of cell uses a cadmium anode and a nickel dioxide cathode
The solid products of the respective electrode reactions adhere to the electrodes and do not diffuse throughout the cell. Thus, the redox reaction is reversible (i.e. like the lead acid cell, the nickel cadmium cell is reversible) No gases are produced, so the cell can be sealed
Fuel Cells Combustion reactions produce heat that, in turn, can be used to produce electricity. However, typically less than 40% of the heat energy is converted to electrical energy - the rest is "wasted" as heat. Combustion reactions are actually redox reactions: diatomic oxygen (0 oxidation number) is reduced to carbon dioxide (-2 oxidation number) or water (-2 oxidation number). Direct production of electricity from redox chemistry, instead of combustion, for these reactions could result in higher efficiency of production of electrical energy. Voltaic cells that perform this type of redox reaction for conventional fuels (such as hydrogen or methane) are called fuel cells
A common reaction being utilized in fuel cells is the reduction of oxygen by hydrogen o Cathode: O2(g) + 2H2O(l) + 4e- 4OH-(aq)
o
Anode: 2H2(g) + 4OH-(aq) 4H2O(l) + 4e-
o
Overall reaction:
2H2(g) + O2(g) 2H2O(l) This is currently an very expensive way to generate energy, but is extremely efficient and compact. Its broadest application to date has been to provide electricity (and drinking water) for spacecraft. Electrolysis
Voltaic cells are based upon the spontaneity of a particular redox reaction
This in turn is based upon the relative reduction potentials for the two half-reactions The half-reaction with the greater reduction potential will drive the other half-reaction in the direction of oxidation
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The relative difference between the two reduction potentials provides a quantitative description of the electromotive force of the voltaic cell
However, we have seen that some commercial voltaic cells can be "recharged"
This is achieved by the application of current in the opposite direction to that of the spontaneous direction of current flow in the voltaic cell This application of current in the opposite direction will cause the redox reaction to proceed in the opposite direction o The half cell where oxidation occurred will now undergo a reduction reaction o The half cell where reduction occurred will now undergo an oxidation reaction
Sodium metal and chlorine gas spontaneously react to form the ionic compound Sodium Chloride: 2Na(s) + Cl2(g) 2NaCl(s)
This reaction is a redox reaction: Anode: 2Na(s) 2Na+(s) + 2eCathode: Cl2(g) + 2e- 2Cl-(s)
The reverse reaction, the decomposition of ionic NaCl into Na(s) and Cl 2(g), can be achieved by the application of an EMF (i.e. external current)
2NaCl(l) 2Na(l) + Cl2(g) Redox reactions that are driven by an external EMF are called electrolysis reactions and take place in electrolytic cells The electrolytic cell contains two electrodes and either a molten salt solution (as with NaCl) or some other type of solution (e.g. aqueous)
The external EMF is provided by a source of direct current (sometimes another voltaic cell, or battery) This external current acts as an electron pump to push electrons onto one electrode (and drive a reduction reaction) and to withdraw electrons from the other electrode (and drive an oxidation reaction) o The electrode where oxidation occurs is still called the anode. However, for an electrolytic cell it is labeled as "+" to indicate that this is the electrode where electrons are being withdrawn by the external EMF o The electrode where reduction occurs is stall called the cathode. However, for an electrolytic cell it is labeled as "-" to indicate that this is the electrode where electrons are being pumped into by the external EMF
Electrolysis of Aqueous Solutions With aqueous solutions of salts it must be considered that the electrolysis might end up driving the reduction or oxidation of H2O(l)
H2O(l) can be oxidized to produce O2(g) H2O(l) can be reduced to produce H2(g) If a salt is present, the cation of the salt may or may not be preferentially reduced compared to H2O(l). Likewise, the anion may or may not be preferentially oxidized compared to H 2O(l)
Here are the reduction potential data for the reduction of H2O(l) and Na+(aq): 2H2O(l) + 2e- H2(g) + 2OH-(aq) E0red = -0.83 V 0 Na+(aq) + e- red = -2.71 V
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When comparing any two reduction reactions the reaction that is favored is the one with the more positive reduction potential. In this case, the reduction of H2O(l) is the favored reduction reaction
Here are the reduction potential data for the oxidation of H2O(l) and Cl-(aq). Remember that the reduction potential provided is referencing the reduction (i.e. reverse) reaction + 2H2O(l) (aq) + O2(g) + 4e- E0red = 1.23 V 2Cl (aq) Cl2(g) + 2e- E0red = 1.36 V
When comparing the reduction potentials, the reduction of Cl- is favored. However, we are interested in the reverse reactions, i.e. the oxidation reactions. In this case, remember that a substance that is readily reduced is difficult to oxidize, and vice versa. Thus, if the Cl - is favored for reduction, then the water reaction is favored for oxidation What this analysis suggests is that electrolysis of an aqueous solution of Na +(aq) and Cl-(aq) ions will not result in the reduction of Na+ or oxidation of Cl-, but rather, will result in the reduction of H2O to produce hydrogen, and the oxidation of H 2O to produce oxygen.
Note: although this analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E0cell. This may be due to kinetic rather than thermodynamic considerations. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame we must increase the voltage of the external source (hence, overvoltage). In the case of the oxidation of H2O versus Cl-(aq), the oxidation of H2O to produce O2(g) is thermodynamically favored, but kinetically quite slow. Thus we could try to crank up the voltage. However, the oxidation of chlorine occurs at a voltage not much higher than that for water. Thus, in practical terms, oxidation of chlorine will preferentially take place at the anode.
The redox reactions for the electrolysis of an aqueous solution of NaCl are therefore: (reduction) 2H2O(l) + 2e- H2(g) + 2OH-(aq) (oxidation) 2Cl-(aq) 2(g) + 2e
And the overall reaction is:
2H2O(l) + 2Cl-(aq) 2(g) + H2(g) + 2OH (aq) Thus, while electrolysis of molten NaCl produces chlorine gas and sodium metal, the electrolysis of an aqueous solution of NaCl produces chlorine gas, hydrogen gas and hydroxide ion
What external EMF is required do drive this particular electrolysis?
(reduction; cathode) 2H2O(l) + 2e- H2(g) + 2OH-(aq) E0red = -0.83 V - 0 (oxidation; anode) 2Cl-(aq) 2(g) + 2e E red = 1.36 V 0 0 0 E cell = E (cathode) - E (anode) = -0.83 V - (1.36 V) = -2.19 V This value is negative, indicating that the reaction is not spontaneous as written and must be driven by an external EMF Electrolysis with Active Electrodes
Inert electrodes do not enter into the chemical reaction, and just serve as a surface upon which electron transfers can occur (and be conducted to the other half-cell) Active electrodes chemically participate in the redox reaction
If an aqueous solution contains solid metal electrodes, the electrodes may be oxidized to the ionic form if the metal is easier to oxidize than water. Since we think in terms of reduction potentials and not ease of
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oxidation, this means that the metal electrodes will be oxidized to ionic form if the reduction potential of the metal is less than that of water Ni(s) Ni2+(aq) + 2e- E0red = -0.28 V 2H2O(l) 4H+(aq) + O2(g) + 4e- E0red = 1.23 V
In this case the reduction potential of H2O(l) is greater than that for Nickel metal. Thus, H2O is easier to reduce, and therefore more difficult to oxidize, than Nickel. Thus, at the anode, Nickel will preferentially be oxidized compared to H2O(l) if we provide an external EMF that "sucks" electrons out at this electrode
What about the cathode (reduction) reaction involving Nickel ion and water? Ni2+(aq) + 2e- Ni(s) E0red = -0.28 V 2H2O(l) + 2e- H2(g) + 2OH-(aq) E0red = -0.83 V
In this case, the half-reaction with the greater reduction potential is the reduction of Nickel. Therefore, at the cathode, Nickel will preferentially be reduced if we provide an external EMF that forces electrons onto this electrode
Thus, a metallic cathode may be coated with solid nickel metal in a process known as "electroplating" Quantitative Aspects of Electrolysis
Reduction half-reactions provide information about the stoichiometric requirements for the production of the elemental form of the metal. In particular, the number of electrons needed per molar basis of metal reduced Na+ + e- Na Cu2+ + 2e- Cu Al3+ + 3e- Al
From the above information, 1 mole of electrons will allow the production (i.e. reduction) of 1 mole of sodium metal, 0.5 mole of copper metal, and 0.333 mole of aluminum metal
For any half-reaction the amount of substance reduced is proportional to the number of electrons that pass into the cell
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We can quantitate the number of electrons by keeping track of the charge that passes into a cell
Charge is measured in Coulombs 1 mole of electrons has a charge of 96,500 Coulombs (C). This number of Coulombic charge is equal to 1 Faraday (F)
Charge of 1 mole of electrons = 96,500 C = 1F The word, current, refers to the rate of flow of electricity
One ampere (amp) is equal to a rate of flow of charge of 1 Coulomb every second amps = Coulombs/second and Coulombs = amps * seconds
In summary… Amps = charge / unit of time Charge the number of electrons The number of electrons the number of moles of a compound reduced
Electrical Work Relationship between E and ΔG:
A positive value for E (electromotive force) is associated with a spontaneous redox reaction Likewise, a negative value for ΔG is associated with a spontaneous reaction or process +E or -ΔG means a spontaneous reaction
The relationship between ΔG and E: ΔG = -nFE
Useful work
For any spontaneous process, ΔG is a measure of the maximum useful work (wmax)that can be done by the process. Therefore, -nFE also is a measure of the maximum useful work that can be done by a redox reaction:
wmax = -nFE wmax = moles * (Coulombs/moles) * (Joules/Coulomb) = Joules Note: work done by the system on the surroundings is indicated by a negative sign for w Non-spontaneous redox reactions
Electrolysis describes the application of an external EMF to drive a non-spontaneous process For a non-spontaneous process, ΔG is positive and E will be negative To force the redox reaction to proceed, the external potential must be larger in magnitude than the EMF of the cell (i.e. Ecell) Eexternal - Ecell
When an external potential is applied, the surroundings do work on the system. Therefore, we expect the sign of the w term to be positive
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wmax = +nFEexternal Units of power
A watt is a unit of energy expenditure, i.e. the amount of energy consumed or produced per unit of time. It has units of Joules/second: 1 watt = 1 Joule/second and 1 Joule = 1 watt * second
How does the Power Company charge you for your electric usage? It charges you for the watt*hours that you consume, or kilowatt*hours 1 k watt hour = (1000 watt)*(1 hour) (360 sec/hour) ((1 J/s)/watt) = 3.6 x 10 6 Joules
Corrosion Corrosion reactions are redox reactions in which the elemental form of a metal is oxidized to produce an unwanted compound or a water soluble ion
Oxidized forms of a metal may not have the strength and conductive properties of the metal in the elemental form
Diatomic oxygen in the atmosphere has a relatively high reduction potential - higher than many metals
For metals exposed to air, oxygen (in the presence of H +) is reduced to H2O, and the metal is oxidized, in a spontaneous redox reaction
For some metals, however, the initial oxidation on the surface protects the underlying metal against further oxidation. This is because the metal oxide layer is impervious to both oxygen and H 2O (a source of H+)
This is the case for aluminum, which although quite reactive upon exposure to oxygen and water, develops a metal oxide surface layer that protects the metal underneath
Corrosion of Iron The corrosion of iron (i.e. rusting) costs the U.S. economy about $70 billion annually
Formation of rust requires both O2 and H2O Anode reaction: Fe(s) Fe2+(aq) + 2e- E0red = -0.44 V
Cathode reaction: O2(g) + 4H+(aq) + 4e- 2H2O(l) E0red = 1.23 V Thus, the reduction of oxygen is favored over the reduction of iron. In other words, the iron will preferentially oxidize.
The Fe2+ that is formed eventually is oxidized further to Fe3+ which forms a hydrated iron (III) oxide (i.e. rust)
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Prevention of Corrosion If iron is painted, then oxygen and water are prevented from contacting the metal, and corrosion is avoided Sometimes the iron is coated with a thin layer of another metal, such as Tin (the layer of Tin keeps oxygen and water away from the iron) 2000 Dr. Michael Blaber
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