CVEN 3454/5404 Water Chemistry Exam 3 Solutions Redox and Adsorption 1. (5 points) Steel drums containing plutonium cuttings and carbon tetrachloride (CCl4) were stored for about two decades at the Rocky Flats Plant south of Boulder. When the drums were inspected, workers found that severe corrosion had resulted in the release of the wastes to the underlying soil. Should the Rocky Flats engineers have expected that CCl4 would cause the corrosion? Here is the reaction of carbon tetrachloride to chloroform (CHCl3): CCl4 + H+ + 2 e- = CHCl3 + Cl- EH0 = 0.79 V (a) What is the pe0 of the oxidation of zero-valent iron (Fe(s), or Fe(0)) to ferric hydroxide (Fe(OH3)(s)) by the reduction of CCl4 to chloroform? (b) Will the oxidation of zero-valent iron by the reduction of CCl4 proceed spontaneously (i.e., should the engineers have expected the corrosion to occur)? The reaction of Fe(s) to Fe(OH)3(s) can be assembled from a series of half-cell and precipitation reactions, like this: Fe(s) = Fe2+ + 2 eFe2+ = Fe3+ + eFe3+ + 3 OH- = Fe(OH)3(s) Combined, these give an overall reaction of Fe(s) + 3 OH- = Fe(OH)3(s) + 3 eWe have the pe0 for the Fe2+/Fe3+ half-cell in Table 9.3 and the Ks0 for the precipitation of the ferric hydroxide from Table 8.7, but we don’t have a pe0 for the Fe(s)/Fe2+ half-cell. We could calculate the free energy of that half-cell and convert it into a K or pe0, or we can do the same for the full reaction, which appears to be a quicker calculation. So, the standard free energy of the Fe(s)/Fe(OH)3(s) half-cell reaction is ΔGr0 = ΔG 0f , Fe ( OH ) ( s ) + 3ΔG 0f , e− − ΔG 0f , Fe ( s ) − 3ΔG 0f ,OH − 3
ΔG = ( −712 ) + 2 ( 0 ) − ( 0 ) − 3 ( −157.3) 0 r
ΔGr0 = −240 kJ mol-1 The equilibrium constant for this reaction is ⎛ ΔGr0 ⎞ K eq = exp ⎜ − ⎟ ⎝ RT ⎠ ⎛ ⎞ −240, 000 J mol -1 K eq = exp ⎜ − ⎟ -1 -1 ⎜ ( 8.314 J mol K ) ( 298 K ) ⎟ ⎝ ⎠ 42 K eq = 1.17 × 10 K eq = 10 42.07
The pe0 value of the Fe(s)/Fe(OH)3(s) half-cell reaction is 1 log K ne 1 pe0 = ( 42.07 ) 3 0 pe = 14.02 The pe0 of the CCl4/CHCl3 half-cell reaction can be determined using its redox potential, EH0: 0.79 V EH0 pe 0 = = 0.059 V 0.059 V pe 0 =
pe 0 = 13.39 The logarithm of the equilibrium constant for this half-cell reaction is log K = ne pe 0 = 2 ( 13.39 ) log K = 26.78
Now we can combine the two half-cell reactions, Fe(s) + 3 OH- = Fe(OH)3(s) + 3 elog K = 42.07 pe0 = 14.02 + CCl4 + H + 2 e = CHCl3 + Cl log K = 26.78 pe0 = 13.39 We need to normalize to a common number of electrons, say 6, which does not change the pe0 values,
2 Fe(s) + 6 OH- = 2 Fe(OH)3(s) + 6 elog K = 84.14 pe0 = 14.02 + 3 CCl4 + 3 H + 6 e = 3 CHCl3 + 3 Cl log K = 80.34 pe0 = 13.39 And the overall is 2 Fe(s) + 3 CCl4 + 3 H+ + 6 OH- = 2 Fe(OH)3(s) + 3 CHCl3 + 3 Cllog K = 164.48 pe0 = 27.41 Just to clean up this final reaction, let’s get rid of the protons and hydroxides by adding some water: 2 Fe(s) + 3 CCl4 + 3 H+ + 6 OH- = 2 Fe(OH)3(s) + 3 CHCl3 + 3 Cllog K = 164.48 pe0 = 27.41 + 6 H2O = 6 H + 6 OH 6(log K) = 6(-14.0) = -84.0 And the final overall reaction is 2 Fe(s) + 3 CCl4 + 6 H2O = 2 Fe(OH)3(s) + 3 CHCl3 + 3 Cl- + 3 H+ log K = 80.48 Because there six electrons transferred in this final reaction, the pe0 value is 1 pe 0 = ( 80.48 ) 6 pe 0 = 13.41
(b) Will the oxidation of zero-valent iron by the reduction of CCl4 proceed spontaneously (i.e., should the engineers have expected the corrosion to occur)? The calculated pe0 value is positive, so the reaction should proceed spontaneously if at standard state. To determine whether or not the reaction would proceed under the actual conditions in the drum would have been challenging. The activities of the solids would have been 1, but the activities of the liquids (carbon tetrachloride, chloroform, water) and the ions (chloride, H+) would have been challenging. Initially, at least, the activity of the carbon tetrachloride would have been 1 for a pure liquid.
2. Millions of people in southern Bangladesh are drinking groundwater contaminated with arsenic (see the figure below; McArthur et al., 2004, Arsenic in groundwater: Testing pollution mechanisms for sedimentary aquifers in Bangladesh, Water Resources Research 37, 109-117). The installation of thousands of tube wells, which were intended to prevent ingestion of pathogens, caused the mobilization of arsenate by reduction of ferric oxyhydroxide in the aquifer sediments. As the ferric oxyhydroxide is microbially reduced, adsorbed arsenic is released. The mobility of the released arsenic depends on its oxidation state -- arsenate (AsO43-) is relatively mobile and arsenite (AsO33-) is not. The arsenic is released from the ferric oxyhydroxide as arsenate, but it may subsequently be reduced to arsenite by ferrous iron. The reduction of ferric oxyhydroxide releases ferrous iron. Understanding the risk of arsenic exposure and the removal of arsenic from the groundwater depends on knowing the arsenic oxidation state. (a) What is the equilibrium pe of an iron-arsenic-water system composed of TOTFe(II) = 0.15 mM, TOTFe(III) = 0.05 mM, and TOTAs(V) = 0.1 mM at pH 7.0? Use the log C-pe graph below to solve the problem. (b) What is the ratio of As(III):As(V) at the equilibrium pe?
(a) What is the equilibrium pe of an iron-arsenic-water system composed of TOTFe(II) = 1.5×10-4 M, TOTFe(III) = 0.5×10-4 M, and TOTAs(V) = 1.0×10-4 M at pH 7.0? Use the log C-pe graph below to solve the problem. The iron is added to the system as both Fe(II) and Fe(III), so there is no clear choice of reference oxidation state for iron in the electron condition. Most of the iron is added as Fe(II), so let’s select Fe(II) as the reference oxidation state. The Fe(III) species will be in the -1 column of the electron condition because Fe(III) has one fewer electron than Fe(II). At pH 7.0, the Fe(II) species on the graph are Fe2+ and FeOH+. The Fe(III) species are Fe(OH)2+ and Fe(OH)30 All of the arsenic is added as As(V), so let’s select As(V) as the reference oxidation state for arsenic. The As(III) species will be in the +2 column for the electron condition because As(III) has two more electrons than As(V). The electron condition table is -1
0
Fe(OH)2+, Fe(OH)30
Fe2+, FeOH+ HAsO42-, H2AsO4Fe(II)
Fe(III)
+2
total
H3AsO3, H2AsO31.5x10-4 0.5x10-4
As(V)
1x10-4
TOTeeq = -{Fe(OH)2+} - {Fe(OH)30} + 2{H3AsO3} + 2{H2AsO3-} TOTein = -1 TOTFe(III) = -0.5×10-4 -{Fe(OH)2+} - {Fe(OH)30} + 2{H3AsO3} + 2{H2AsO3-} = -0.5x10-4 2{H3AsO3} + 2{H2AsO3-} + 0.5x10-4 = {Fe(OH)2+} + {Fe(OH)30} 2{H3AsO3} + 2{H2AsO3-} + 10-4.3 = {Fe(OH)2+} + {Fe(OH)30} On the graph prepared by Benjamin (2002) and supplied with the exam, the LHS and RHS solution lines intersect at a pe of about 4.05. Acceptable values of the solution pe are 4.0 to 4.1.
-4.3
(b) What is the ratio of As(III):As(V) at the equilibrium pe? At the equilibrium pe of 4.05, the activities of the arsenic species are {HAsO42-} ≈ 10-4.15 = 7.1×10-5 {H2AsO4-} ≈ 10-4.5 = 3.2×10-5 TOTAs(V) = {HAsO42-} + {H2AsO4-} = 1.03×10-4 Now, we know that the total amount of As(V) added was 1.0×10-4, so we can’t have exceeded the total. And if at equilibrium, TOTAs(V) is essentially the same as the TOTAs(V) added, then there can’t have been any significant amount of As(III) produced by reduction of the As(V) – the Fe(II) was not a sufficiently strong reductant to reduce As(V) to As(III). Sure enough, both {H3AsO3} and {H2AsO3-} are somewhere well below the bottom of the graph. Therefore, the ratio of As(III):As(V) at equilibrium is about 0.
3. (5 points) At the 100N Area of the U.S. Department of Energy’s Hanford Site in Hanford, Washington, cesium-137 (the radioactive isotope of Cs) is leaking from the 1301-N liquid waste disposal facility. The 137Cs has migrated to the Columbia River, which is adjacent to the site. To prevent the further release of 137Cs to the river, engineers plan to install a "permeable reactive barrier" (shown as the dark bar labeled just "reactive barrier" in the figure below) that will sorb Cs+ from the groundwater. The barrier will be composed of clinoptilolite, a natural zeolite mineral. The barrier dimensions are 550 m in length and 100 m depth. The porosity of the barrier is 0.40 and the density of the clinoptilolite is 2.4 g cm-3. The amount of water projected to flow through the barrier is 2.0x108 m3 yr-1. The average 137Cs concentration in the plume is 3.0 μM, and the cesium is present predominantly as Cs+ at the groundwater pH of 6.0. The width of the barrier (the dimension of the barrier parallel to the flow path shown by the arrow) is still being determined by the engineers. They need to make the barrier wide enough to adsorb cesium for a design life of thirty years. To do this, they needed to determine the maximum adsorption capacity of the clinoptilolite. The adsorption of Cs+ onto clinoptilolite was tested over a wide range of Cs+ concentrations and the adsorption isotherm data is presented in the table below. (a) Determine the maximum adsorption capacity of the clinoptilolite for cesium in units of mol/kg. (b) Using the maximum adsorption capacity and the information provided above, determine the width of the barrier (units of m) required to remove Cs+ from the contaminated groundwater for 30 years. [Cs+] (M)
qCs+ (mol/kg)
1.5x10-6
1.6x10-3
6.5x10-6
6.0x10-3
2.4x10-5
2.1x10-2
1.1x10-4
5.0x10-2
4.8x10-4
9.2x10-2
2.5x10-3
1.7x10-1
5.0x10-3
1.9x10-1
1.1x10-2
2.3x10-1
To determine the width of the barrier, we need to determine the adsorption capacity of the clinoptilolite. The Langmuir adsorption isotherm has a parameter that represents the adsorption capacity – qmax (Eqn. 10.10, Benjamin, 2001). If the data fits a Langmuir isotherm, we can use the qmax parameter to solve the problem. To determine the Langmuir adsorption parameters, we plot [Cs+]/qCs+ versus [Cs+]. The goodness of fit of the linear regression of this data is R2 = 0.991, which indicates that the adsorption isotherm fits the Langmuir model. The slope of the linear regression is 4.3. The slope is equal to the reciprocal of qmax: 0.06 1 = 4.3 qmax
This means that the clinoptilolite in the barrier can adsorb 0.23 moles of Cs+ per kilogram of clinoptilolite. Note that the Langmuir equation does not require adsorption to be measured per area of the adsorbent. To last for t = 30 years, how much Cs+ needs to be adsorbed? The flow rate through the barrier is Q = 2.0x108 m3 yr-1. The cesium concentration in the water is 3.0x10-6 mol L-1, or C = 3.0x10-3 mol m-3. The total amount of Cs+ moving into the barrier is
0.05
[Cs+]/qCs+ (kg L-1)
qmax = 0.23 mol kg -1
0.04 0.03 0.02 [Cs +]/qCs+ = 4.3[Cs +] + 0.0021 R2 = 0.991
0.01 0.00 0.000
0.005
0.010 +
[Cs ] (M)
0.015
MCs+ = CQt = ( 3.0 × 10 −3 mol m -3 )( 2.0 × 10 −8 m 3 yr −1 ) ( 30 yr ) MCs+ = 1.8 × 107 mol
How much clinoptilolite do we need to adsorb all this Cs+? The required mass of clinoptilolite is M + 1.8 × 107 mol Mclino = Cs = 0.23 mol kg -1 qmax Mclino = 7.8 × 107 kg What is the volume of this mass of clinoptilolite? The density of the clinoptilolite is ρclino = 2.4 g cm-3 = 2,400 kg m-3, so the clinoptilolite volume is 7.8 × 107 kg M Vclino = clino = ρclino 2, 400 kg m -3 Vclino = 3.3 × 10 4 m 3 The porosity of the clinoptilolite barrier is θ = 0.40, so the clinoptilolite occupies (1 – θ) = 0.6, or 60% of the barrier volume. That makes the total volume of the barrier 3.3 × 10 4 m 3 V Vbarrier = clino = 1−θ 0.6 4 3 Vbarrier = 5.5 × 10 m
The width of the barrier (the length of the flow path) can be determined using the length (l = 550 m) and depth (d = 100 m) of the barrier: 5.5 × 10 4 m 3 Vbarrier wbarrier = = lbarrier dbarrier ( 550 m )( 100 m ) wbarrier = 1.0 m