1 FE Chemistry Electrochem F12
FE_Chemistry-electrochem_F12
Electrochemistry
An electrolyte is a substance that dissociate in solution to produce positive and negative ions. - It can be an aqueous solution of a soluble salt, or it can be an ionic substance in molten form.
Electrolysis is the passage of an electric current through an electrolyte, driven by an external voltage source. o Electrolysis occurs when the positive terminal and negative terminal of a voltage source are placed in an electrolyte. o Negative ions (anions) will be attracted to the anode, where they are oxidized (AnOx). o Positive ions (cations) will be attracted to the cathode, where they will be reduced (RedCat). o The passage of ions constitutes the current.
Example - Daniel cell Voltim eter
A zinc electrode in a zinc chloride solution connected to a copper electrode in a cupric chloride solution.
e-
e-
LEO is a lion, “GER”
K+ Cl-
Zn
LEO – Losing Electron is Oxidation GER – Gaining Electron is Reduction
Cu
Salt Bridge Zn2+
Anode is where oxidation occurs – An Ox Cathode is where reduction occurs – Red Cat
Zn
Cl-
Zn2+ + 2eAnode (negative)
Cu2+
Cl-
Cu2+ + 2eCathode (positive)
Electrolytic (electrochemical) reactions o The reactions (which do not proceed spontaneously) that are forced to proceed by supplying electrical energy.
Faraday’s Laws of Electrolysis - Faraday’s Laws of Electrolysis can be used to predict the duration and magnitude of a direct current needed to complete an electrolytic reaction. Law 1: The mass of a substance generated by electrolysis is proportional to the amount of electricity used. Law 2: For any constant amount of electricity, the mass of substance generated is proportional to its equivalent weight. Law 3: One faraday of electricity (96,485 C or 96,485 A·s) will produce one gram equivalent weight.
1
Cu
FE_Chemistry-electrochem_F12
The number of gram of a substance produced at an electrode in an electrolytic reaction can be found from the equation:
I t MW no. of faradays GEW 96,485 change in oxidation state
m
where m = mass of a substance produced, g I = current, A t = time, s MW = molecular weight or AW = atomic weight The number of gram-moles produced (n) is
m I t no. of faradays MW 96,485 change in oxidation state change in oxidation state
n
Problem: A current of 0.075 A passes through a solution of silver nitrate for 10 minutes. How much silver is deposited? (A) 0.030 g
(B) 0.035 g
(C) 0.040 g
(D) 0.050 g
(Solution) The number of gram of a substance produced at an electrode in an electrolytic reaction can be found from the equation (Faraday’s law):
m
I t MW no. of faradays GEW 96, 485 A s change in oxidation state
where m = mass of a substance produced, g I = current, A t = time, s MW = molecular weight or AW = atomic weight Given: I = 0.075 A t = (10 min)(60 s/min) = 600 s Note:
AgNO3 Ag+ + NO3– Ag+
-
+ e-
Ag
AgNO3 is soluble. Ag+ is readily reduced to Ag (metallic)
Change in oxidation state, z = 1/mol Silver exists as single atoms, so the molecular weight is the atomic weight, 107.87 g/mol. 2
FE_Chemistry-electrochem_F12
change in oxidation state = 1/mol MW or AW of Ag = 107.87 g/mol. m
0.075 A 600 s 107.87 g / mol 0.050 g 96,485 A s 1 / mol
Answer is (D).
Problem: How many grams of copper will be deposited at an electrode if a current of 1.5 A is supplied for 2 hours to a CuSO4 solution? (A) 2.4 g
(B) 3.6 g
(C) 7.1 g
(D) 48 g
Given:
I = 1.5 A t = (2 hrs)(3600 s/hr) = 7200 s AW of Cu = 63.5 g/mol. Note: CuSO4 Cu2+ + SO4 2-
The dissociation reaction for copper sulfate is: The electrolytic reaction equation is:
Cu2+ + 2e-
Cu
Since the change in oxidation number is 2 per atom of copper deposited, the change in oxidation state, z = 2/mol Use Faraday’s law
m
I t MW no. of faradays GEW 96,485 change in oxidation state
where m = mass of a substance produced, g I = current, A t = time, s MW = molecular weight or AW = atomic weight = 63.5 g/mol
m
1.5 A 7200 s 63.5g / mol 3.55 g 96,485 A s 2 / mol Answer is (B).
3
Electrochemistry
An electrolyte is a substance that dissociate in solution to produce positive and negative ions. - It can be an aqueous solution of a soluble salt, or it can be an ionic substance in molten form.
Electrolysis is the passage of an electric current through an electrolyte, driven by an external voltage source. o Electrolysis occurs when the positive terminal and negative terminal of a voltage source are placed in an electrolyte. o Negative ions (anions) will be attracted to the anode, where they are oxidized (AnOx). o Positive ions (cations) will be attracted to the cathode, where they will be reduced (RedCat). o The passage of ions constitutes the current.
Example - Daniel cell Voltim eter
A zinc electrode in a zinc chloride solution connected to a copper electrode in a cupric chloride solution.
e-
e-
LEO is a lion, “GER”
K+ Cl-
Zn
LEO – Losing Electron is Oxidation GER – Gaining Electron is Reduction
Cu
Salt Bridge Zn2+
Anode is where oxidation occurs – An Ox Cathode is where reduction occurs – Red Cat
Zn
Cl-
Zn2+ + 2eAnode (negative)
Cu2+
Cl-
Cu2+ + 2eCathode (positive)
Electrolytic (electrochemical) reactions o The reactions (which do not proceed spontaneously) that are forced to proceed by supplying electrical energy.
Faraday’s Laws of Electrolysis - Faraday’s Laws of Electrolysis can be used to predict the duration and magnitude of a direct current needed to complete an electrolytic reaction. Law 1: The mass of a substance generated by electrolysis is proportional to the amount of electricity used. Law 2: For any constant amount of electricity, the mass of substance generated is proportional to its equivalent weight. Law 3: One faraday of electricity (96,485 C or 96,485 A·s) will produce one gram equivalent weight.
1
Cu
FE_Chemistry-electrochem_F12
The number of gram of a substance produced at an electrode in an electrolytic reaction can be found from the equation:
I t MW no. of faradays GEW 96,485 change in oxidation state
m
where m = mass of a substance produced, g I = current, A t = time, s MW = molecular weight or AW = atomic weight The number of gram-moles produced (n) is
m I t no. of faradays MW 96,485 change in oxidation state change in oxidation state
n
Problem: A current of 0.075 A passes through a solution of silver nitrate for 10 minutes. How much silver is deposited? (A) 0.030 g
(B) 0.035 g
(C) 0.040 g
(D) 0.050 g
(Solution) The number of gram of a substance produced at an electrode in an electrolytic reaction can be found from the equation (Faraday’s law):
m
I t MW no. of faradays GEW 96, 485 A s change in oxidation state
where m = mass of a substance produced, g I = current, A t = time, s MW = molecular weight or AW = atomic weight Given: I = 0.075 A t = (10 min)(60 s/min) = 600 s Note:
AgNO3 Ag+ + NO3– Ag+
-
+ e-
Ag
AgNO3 is soluble. Ag+ is readily reduced to Ag (metallic)
Change in oxidation state, z = 1/mol Silver exists as single atoms, so the molecular weight is the atomic weight, 107.87 g/mol. 2
FE_Chemistry-electrochem_F12
change in oxidation state = 1/mol MW or AW of Ag = 107.87 g/mol. m
0.075 A 600 s 107.87 g / mol 0.050 g 96,485 A s 1 / mol
Answer is (D).
Problem: How many grams of copper will be deposited at an electrode if a current of 1.5 A is supplied for 2 hours to a CuSO4 solution? (A) 2.4 g
(B) 3.6 g
(C) 7.1 g
(D) 48 g
Given:
I = 1.5 A t = (2 hrs)(3600 s/hr) = 7200 s AW of Cu = 63.5 g/mol. Note: CuSO4 Cu2+ + SO4 2-
The dissociation reaction for copper sulfate is: The electrolytic reaction equation is:
Cu2+ + 2e-
Cu
Since the change in oxidation number is 2 per atom of copper deposited, the change in oxidation state, z = 2/mol Use Faraday’s law
m
I t MW no. of faradays GEW 96,485 change in oxidation state
where m = mass of a substance produced, g I = current, A t = time, s MW = molecular weight or AW = atomic weight = 63.5 g/mol
m
1.5 A 7200 s 63.5g / mol 3.55 g 96,485 A s 2 / mol Answer is (B).
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