Choose the problem corresponding to the seventh digit of your Student ID Number.
For the electrochemical cells described below, find the cell potential for the indicated concentration conditions. Unless otherwise given, assume the concentration of ions is 1.00 M and the partial pressure of gases is 1.00 atm. T = 25 °C.
0) Pt(s) | MnO2(s) | MnO4–(aq) | pH = 2.00 || Co2+(aq) | Co3+(aq) | Pt(s)
1) Pt(s) | Mn2+(aq) | MnO4–(aq) | pH = 2.00 || Au3+(aq) | Au(s)
2) Pt(s) | Au3+(aq) | Au+(aq) || ClO3–(aq) | Cl–(aq) | pH = 2.00 | Pt(s)
3) Pt(s) | NO2(g) | HNO2(aq) | pH = 2.00 || MnO2(s) | Mn2+(aq) | pH = 2.00 | Pt(s)
4) Hg(l) | Hg22+(aq) || NO(g) | NO3–(aq) | pH = 2.00 | Pt(s)
5) Pt(s) | MnO4–(aq) | MnO42–(aq) || O2(g) | H2O2(aq) | pH = 2.00 | Pt(s)
6) Cu(s) | Cu2+(aq) || H2SO3(aq) | S(s) | pH = 2.00 | Pt(s)
7) Hg(l) | HgCl2(s) | Cl–(aq) || VO2+(aq) | V3+(aq) | pH = 2.00 | Pt(s)
8) Fe(s) | Fe3+(aq) || H2S(g) | S(s) | pH = 2.00 | Pt(s)
9) Pb(s) | PbSO4(s) | SO42–(aq) || H3PO4(aq) | H3PO3(aq) | pH = 2.00 | Pt(s)
Use the Nernst Equation to find E from E° and the given pH.
0) Pt(s) | MnO2(s) | MnO4–(aq) | pH = 2.00 || Co2+(aq) | Co3+(aq) | Pt(s)
Net: MnO2(s) + 3 Co3+(aq) + 2 H2O(l) → MnO4–(aq) + 3 Co2+(aq) + 4 H+(aq)
E° = +0.12 V
E = E° – (RT/nF)ln([MnO4–] [Co3+]3[H+]4/[Co2+]3)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 3
[H+] = 10–2.00 = 1.0×10–2 M
[MnO4–] = [Co3+] = [Co2+] = 1.00 M
E = 0.12 – (8.314×298)/(3×96485)ln(1.0×10–2)4 = +0.28 V
1) Pt(s) | Mn2+(aq) | MnO4–(aq) | pH = 2.00 || Au3+(aq) | Au(s)
Net: 3 Mn2+(aq) + 5 Au3+(aq) + 12 H2O(l) → 3 MnO4–(aq) + 5 Au(s) + 24 H+(aq)
E° = +0.01 V
E = E° – (RT/nF)ln([MnO4–]3 [H+]24/[Mn2+]3[Au3+]5)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 15
[H+] = 10–2.00 = 1.0×10–2 M
[MnO4–] = [Au3+] = [Mn2+] = 1.00 M
E = 0.01 – (8.314×298)/(15×96485)ln(1.0×10–2)24 = +0.20 V
2) Pt(s) | Au3+(aq) | Au+(aq) || ClO3–(aq) | Cl–(aq) | pH = 2.00 | Pt(s)
Net: 3 Au+(aq) + ClO3–(aq) + 6 H+(aq) → 3 Au3+(aq) + Cl–(aq) + 3 H2O(l)
E° = +0.09 V
E = E° – (RT/nF)ln([Au3+]3[Cl–] /[Au+]3[ClO3–][H+]6)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 6
[H+] = 10–2.00 = 1.0×10–2 M
[ClO3–] = [Au3+] = [Au+] = [Cl–] = 1.00 M
E = 0.09 – (8.314×298)/(6×96485)ln(1/1.0×10–2)6 = –0.03 V
3) Pt(s) | NO2(g) | HNO2(aq) | pH = 2.00 || MnO2(s) | Mn2+(aq) | pH = 2.00 | Pt(s)
Net: 2 HNO2(aq) + MnO2(s) + 2 H+(aq) → 2 NO2(g) + Mn2+(aq) + 2 H2O(l)
E° = +0.16 V
E = E° – (RT/nF)ln([PNO2]2 [Mn2+]/[HNO2]2[H+]2)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 2
[H+] = 10–2.00 = 1.0×10–2 M
[HNO2] = [Mn2+] = 1.00 M, PNO2 = 1.00 atm
E = 0.16 – (8.314×298)/(2×96485)ln(1/1.0×10–2)2 = +0.04 V
4) Hg(l) | Hg22+(aq) || NO(g) | NO3–(aq) | pH = 2.00 | Pt(s)
Net: 6 Hg(l) + 2 NO3–(aq) + 8 H+(aq) → 3 Hg22+(aq) + 2 NO(g) + 4 H2O(l)
E° = +0.06 V
E = E° – (RT/nF)ln([Hg22+]3[PNO]2 /[NO3–]2[H+]8)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 6
[H+] = 10–2.00 = 1.0×10–2 M
[NO3–] = [Hg22+] = 1.00 M, PNO = 1.00 atm
E = 0.06 – (8.314×298)/(6×96485)ln(1/1.0×10–2)8 = –0.10 V
5) Pt(s) | MnO4–(aq) | MnO42–(aq) || O2(g) | H2O2(aq) | pH = 2.00 | Pt(s)
Net: 2 MnO42–(aq) + O2(g) + 2 H+(aq) → 2 MnO4–(aq) + H2O2(aq)
E° = +0.14 V
E = E° – (RT/nF)ln([MnO4–]2[H2O2] /PO2[MnO42–]2[H+]2)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 2
[H+] = 10–2.00 = 1.0×10–2 M
[MnO4–] = [MnO42–] = [H2O2] = 1.00 M, PO2 = 1.00 atm
E = 0.14 – (8.314×298)/(2×96485)ln(1/1.0×10–2)2 = +0.02 V
6) Cu(s) | Cu2+(aq) || H2SO3(aq) | S(s) | pH = 2.00 | Pt(s)
Net: 2 Cu(s) + H2SO3(aq) + 4 H+(aq) → 2 Cu2+(aq) + S(s) + 3 H2O(l)
E° = +0.109
E = E° – (RT/nF)ln([Cu2+]2/ [H2SO3][H+]4)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 4
[H+] = 10–2.00 = 1.0×10–2 M
[H2SO3] = [Cu2+] = 1.00 M
E = 0.109 – (8.314×298)/(4×96485)ln(1/1.0×10–2)4 = –0.009 V
7) Hg(l) | HgCl2(s) | Cl–(aq) || VO2+(aq) | V3+(aq) | pH = 2.00 | Pt(s)
Net: Hg(l) + 2 Cl–(aq) + 2 VO2+(aq) + 4 H+(aq) → HgCl2(s) + 2 V3+(aq) + 2 H2O(l)
E° = +0.069 V
E = E° – (RT/nF)ln([V3+]2 /[Cl–]2[VO2+]2[H+]4)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 2
[H+] = 10–2.00 = 1.0×10–2 M
[Cl–] = [V3+] = [VO2+] = 1.00 M
E = 0.069 – (8.314×298)/(2×96485)ln(1/1.0×10–2)4 = –0.168 V
8) Fe(s) | Fe3+(aq) || H2S(g) | S(s) | pH = 2.00 | Pt(s)
Net: 2 Fe(s) + 3 S(s) + 6 H+(aq) → 2 Fe3+(aq) + 3 H2S(g)
E° = +0.18 V
E = E° – (RT/nF)ln([Fe3+]2[PH2S]3 /[H+]6)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 6
[H+] = 10–2.00 = 1.0×10–2 M
[Fe3+] = 1.00 M, PH2S = 1.00 atm
E = 0.18 – (8.314×298)/(6×96485)ln(1/1.0×10–2)6 = +0.06 V
9) Pb(s) | PbSO4(s) | SO42–(aq) || H3PO4(aq) | H3PO3(aq) | pH = 2.00 | Pt(s)
Net: Pb(s) + SO42–(aq) + H3PO4(aq) + 2 H+(aq) → PbSO4(s) + H3PO3(aq) + H2O(l)
E° = +0.080 V
E = E° – (RT/nF)ln([H3PO3] /[SO42–][H3PO4][H+]2)
R = 8.314 J/mol·K
T = 25 °C = 298 K
F = 96485 C/mol
n = 2
[H+] = 10–2.00 = 1.0×10–2 M
[SO42–] = [H3PO3] = [H3PO4] = 1.00 M
E = 0.080 – (8.314×298)/(2×96485)ln(1/1.0×10–2)2 = –0.038 V