Electrochemistry Homework Set Worked Out Solutions

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Unformatted text preview: metalg metal The solution contains Pt2+, thus the plating reaction is: Pt2+(aq) + 2e‐ Pt(s) 0.500 A=0.500 C * 55.0 s= 27.5 C 1 s 27.5 C * 1 mol e‐ = 0.000285 mol e‐ 96,486 C 0.000285 mol e‐ * 1 mol Pt = 0.000143 mol Pt ‐ 2 mol e 0.000143 mol Pt * 195.08 g Pt = 0.0278 g = 27.8 mg 1 mol Pt 28. The solution contains Pt2+, thus the plating reaction is: 3+(aq) + 3e‐ Cr(s) Cr 5.00 A=5.00 C * (30.0 min* 60)= 9,000 C 1 s 9.000 C * 1 mol e‐ = 0.0933 mol e‐ 96,486 C 0.0933 mol e‐ * 1 mol Cr = 0.0311 mol Cr ‐ 3 mol e 0.0311 mol Cr * 52.00 g Cr = 1.62 g 1 mol Cr 29. For this problem, we need to work backwards to find the moles of vanadium oxidized and the moles of electrons transferred: 114 mg V=0.114 g V * 1 mol V = 0.00224 mol V 50.94 g V 650 C * 1 mol e‐ = 0.00674 mol e‐ 96,486 C Thus, 0.00674 moles of electrons were transferred to oxidize 0.00224 moles of vanadium, or: ‐=2.8 3 electrons transferred per vanadium atom oxidized 0.00674 mol e 0.00224 mol V oxidation state = +3 30. In order to liberate the most metal, we need the least amount of electrons required to reduce a mole of metal, or the smallest change in oxidation state. This corresponds to a lesser amount of electricity. All of the given choices involve a +10 change, except C, which involves a +20 change. Thus C can be eliminated. Next, we need the greatest molecular weight, so that 1.0 g of metal can be liberated with the least amount of electricity. Of the remaining choices, Ag has the greatest molecular weight, so answer choice D is correct. Dr. Fus Electrochemistry Problem Set Solutions Chemistry 123 31. CrO42‐(aq) + 8H+ + 6e‐ Cr(s) + 4H2O(l) 1.00 g Cr * 1 mol Cr = 0.0192 mol Cr 52.00 g Cr 0.0192 mol Cr * 6 mol e‐ = 0.115 mol e‐ 1 mol Cr 0.115 mol e‐ * 96,486 C = 11,133 C 1 mol e‐ 11,133 C * 1 s = 1,855.5 s = 30.9 min 6 C 32. Use the equation: Ecell=E°cell­ (0.0592/n)*logQ (*at 25 °C; n=mol e‐) For Sn2+ + 2Fe3+ Sn4+ + 2Fe2+, find Q: 4+][Fe2+]2= (0.10)(0.10)2= 0.008 Q=[Sn [Sn2+][Fe3+]2 (0.50)(0.50)2 Ecell= 0.617 – (0.0592/2)*log(0.008)= +0....
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This homework help was uploaded on 03/28/2014 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

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