Choose the problem corresponding to the eighth digit of your Student ID Number.
For the electrochemical cells described below, find the standard cell potential.
0) Pt(s) | MnO2(s) | MnO4–(aq) | pH=0 || Co2+(aq) | Co3+(aq) | Pt(s)
1) Pt(s) | Mn2+(aq) | MnO4–(aq) | pH = 0 || Au3+(aq) | Au(s)
2) Pt(s) | Au3+(aq) | Au+(aq) || ClO3–(aq) | Cl–(aq) | pH = 0 | Pt(s)
3) Pt(s) | NO2(g) | HNO2(aq) | pH = 0 || MnO2(s) | Mn2+(aq) | pH = 0 | Pt(s)
4) Hg(l) | Hg22+(aq) || NO(g) | NO3–(aq) | pH = 0 | Pt(s)
5) Pt(s) | MnO4–(aq) | MnO42–(aq) || O2(g) | H2O2(aq) | pH = 0 | Pt(s)
6) Cu(s) | Cu2+(aq) || H2SO3(aq) | S(s) | pH = 0 | Pt(s)
7) Hg(l) | HgCl2(s) | Cl–(aq) || VO2+(aq) | V3+(aq) | pH = 0 | Pt(s)
8) Fe(s) | Fe3+(aq) || H2S(g) | S(s) | pH = 0 | Pt(s)
9) Pb(s) | PbSO4(s) | SO42–(aq) || H3PO4(aq) | H3PO3(aq) | pH = 0 | Pt(s)
0) Pt(s) | MnO2(s) | MnO4–(aq) | pH = 0 || Co2+(aq) | Co3+(aq) | Pt(s)
Anode: MnO2(s) + 2 H2O(l) → MnO4–(aq) + 4 H+(aq) + 3 e–
E°ox = –E°red = –1.70 V
Cathode: Co3+(aq) + e– → Co2+(aq)
E°red = 1.82 V
Net: MnO2(s) + 3 Co3+(aq) + 2 H2O(l) → MnO4–(aq) + 3 Co2+(aq) + 4 H+(aq)
E° = E°red + E°ox = 1.82 – 1.70 = +0.12 V
1) Pt(s) | Mn2+(aq) | MnO4–(aq) | pH = 0 || Au3+(aq) | Au(s)
Anode: Mn2+(aq) + 4 H2O(l) → MnO4–(aq) + 8 H+(aq) + 5 e–
E°ox = –E°red = –1.51 V
Cathode: Au3+(aq) + 3 e– → Au(s)
E°red = 1.52 V
Net: 3 Mn2+(aq) + 5 Au3+(aq) + 12 H2O(l) → 3 MnO4–(aq) + 5 Au(s) + 24 H+(aq)
E° = E°red + E°ox = 1.52 – 1.51 = +0.01 V
2) Pt(s) | Au3+(aq) | Au+(aq) || ClO3–(aq) | Cl–(aq) | pH = 0 | Pt(s)
Anode: Au+(aq) → Au3+(aq) + 2 e–
E°ox = –E°red = –1.36 V
Cathode: ClO3–(aq) + 6 H+(aq) + 6 e– → Cl–(aq) + 3 H2O(l)
E°red = 1.450 V
Net: 3 Au+(aq) + ClO3–(aq) + 6 H+(aq) → 3 Au3+(aq) + Cl–(aq) + 3 H2O(l)
E° = E°red + E°ox = 1.450 – 1.36 = +0.09 V
3) Pt(s) | NO2(g) | HNO2(aq) | pH = 0 || MnO2(s) | Mn2+(aq) | pH = 0 | Pt(s)
Anode: HNO2(aq) → NO2(g) + H+(aq) + e–
E°ox = –E°red = –1.07 V
Cathode: MnO2(s) + 4 H+(aq) + 2 e– → Mn2+(aq) + 2 H2O(l)
E°red = 1.23 V
Net: 2 HNO2(aq) + MnO2(s) + 2 H+(aq) → 2 NO2(g) + Mn2+(aq) + 2 H2O(l)
E° = E°red + E°ox = 1.23 – 1.07 = +0.16 V
4) Hg(l) | Hg22+(aq) || NO(g) | NO3–(aq) | pH = 0 | Pt(s)
Anode: 2 Hg(l) → Hg22+(aq) + 2 e–
E°ox = –E°red = –0.90 V
Cathode: NO3–(aq) + 4 H+(aq) + 3 e– → NO(g) + 2 H2O(l)
E°red = 0.956 V
Net: 6 Hg(l) + 2 NO3–(aq) + 8 H+(aq) → 3 Hg22+(aq) + 2 NO(g) + 4 H2O(l)
E° = E°red + E°ox = 0.956 – 0.90 = +0.06 V
5) Pt(s) | MnO4–(aq) | MnO42–(aq) || O2(g) | H2O2(aq) | pH = 0 | Pt(s)
Anode: MnO42–(aq) → MnO4–(aq) + e–
E°ox = –E°red = –0.56
Cathode: O2(g) + 2 H+(aq) + 2 e– → H2O2(aq)
E°red = 0.695 V
Net: 2 MnO42–(aq) + O2(g) + 2 H+(aq) → 2 MnO4–(aq) + H2O2(aq)
E° = E°red + E°ox = 0.695 – 0.56 = +0.14 V
6) Cu(s) | Cu2+(aq) || H2SO3(aq) | S(s) | pH = 0 | Pt(s)
Anode: Cu(s) → Cu2+(aq) + 2 e–
E°ox = –E°red = –0.340 V
Cathode: H2SO3(aq) + 4 H+(aq) + 4 e– → S(s) + 3 H2O(l)
E°red = 0.449 V
Net: 2 Cu(s) + H2SO3(aq) + 4 H+(aq) → 2 Cu2+(aq) + S(s) + 3 H2O(l)
E° = E°red + E°ox = 0.449 – 0.340 = +0.109
7) Hg(l) | HgCl2(s) | Cl–(aq) || VO2+(aq) | V3+(aq) | pH = 0 | Pt(s)
Anode: Hg(l) + 2 Cl–(aq) → HgCl2(s) + 2 e–
E°ox = –E°red = –0.2676
Cathode: VO2+(aq) + 2 H+(aq) + e– → V3+(aq) + H2O(l)
E°red = 0.337
Net: Hg(l) + 2 Cl–(aq) + 2 VO2+(aq) + 4 H+(aq) → HgCl2(s) + 2 V3+(aq) + 2 H2O(l)
E° = E°red + E°ox = 0.337 –0.2676 = +0.069 V
8) Fe(s) | Fe3+(aq) || H2S(g) | S(s) | pH = 0 | Pt(s)
Anode: Fe(s) → Fe3+(aq) + 3 e–
E°ox = –E°red = 0.04 V
Cathode: S(s) + 2 H+(aq) + 2 e– → H2S(g)
E°red = 0.14 V
Net: 2 Fe(s) + 3 S(s) + 6 H+(aq) → 2 Fe3+(aq) + 3 H2S(g)
E° = E°red + E°ox = 0.14 + 0.04 = +0.18 V
9) Pb(s) | PbSO4(s) | SO42–(aq) || H3PO4(aq) | H3PO3(aq) | pH = 0 | Pt(s)
Anode: Pb(s) + SO42–(aq) → PbSO4(s) + 2 e–
E°ox = –E°red = 0.356
Cathode: H3PO4(aq) + 2 H+(aq) + 2 e– → H3PO3(aq) + H2O(l)
E°red = –0.276
Net: Pb(s) + SO42–(aq) + H3PO4(aq) + 2 H+(aq) → PbSO4(s) + H3PO3(aq) + H2O(l)
E° = E°red + E°ox = –0.276 + 0.356 = +0.080 V