Question 17.8: Consider a voltaic cell in which the following reaction occu...
Consider a voltaic cell in which the following reaction occurs:
O_2(g, 0.98\> atm) + 4H^+(aq, pH = 1.24) + 4Br^-(aq, 0.15\> M) \longrightarrow 2H_2O + 2Br_2(l)a Calculate E for the cell at 25°C.
ANALYSIS
Information given: reaction: (O_2(g) + 4H^+(aq) + 4Br^-(aq) \longrightarrow 2Br_2(l) + 2H_2O)
P_{O_2}(0.98 atm); [H^+] (pH = 1.24); [Br^-] (0.15 M)
temperature (25°C)
Information implied: Table 17.1 (standard reduction potentials)
Asked for: E
STRATEGY
1. Change pH to [H^+] and find Q.
2. Assign oxidation numbers, write oxidation and reduction half-reactions, and cancel electrons to find n.
3. Find E°. (E^{\circ}_{red} + E^{\circ}_{ox})
4. Substitute into the Nernst equation (Equation 17.4) for T = 25°C
E = E^{\circ} – \frac{0.0257}{n} \ln Q
b When the voltaic cell is at 35°C, E is measured to be 0.039 V. What is E° at 35°C?
ANALYSIS
Information given: E (0.039 V) at T (35°C) From part (a): Q (1.8 \times 10^{8}); n (4 moles)
Information implied: R and F values in joules
Asked for: E° at 35°C
STRATEGY
Substitute into the Nernst equation.
E = E^{\circ} – \frac{RT}{nF} \ln Q
| Table 17.1 Standard Potentials in Water Solution at 25°C | ||||
| Acidic Solution, [H^+] = 1 M | ||||
| {E^{\circ}}_{red} (V) | ||||
 |
Li^+(aq) + e^- | \longrightarrow Li(s)\blacktriangleleft ⓘ |  |
-3.04 |
| K^+(aq) + e^- | \longrightarrow K(s) | -2.936 | ||
| Ba^{2+}(aq) + 2e^{-} | \longrightarrow Ba(s) | -2.906 | ||
| Ca^{2+}(aq) + 2e^- | \longrightarrow Ca(s) | -2.869 | ||
| Na^+(aq) + e^- | \longrightarrow Na(s) | -2.714 | ||
| Mg^{2+}(aq) + 2e^- | \longrightarrow Mg(s) | -2.357 | ||
| Al^{3+}(aq) + 3e^- | \longrightarrow Al(s) | -1.68 | ||
| Mn^{2+}(aq) + 2e^- | \longrightarrow Mn(s) | -1.182 | ||
| Zn^{2+}(aq) + 2e^- | \longrightarrow Zn(s) | -0.762 | ||
| Cr^{3+}(aq) + 3e^- | \longrightarrow Cr(s) | -0.744 | ||
| Fe^{2+}(aq) + 2e^- | \longrightarrow Fe(s) | -0.409 | ||
| Cr^{3+}(aq) + e^- | \longrightarrow Cr^{2+}(aq) | -0.408 | ||
| Cd^{2+}(aq) + 2e^- | \longrightarrow Cd(s) | -0.402 | ||
| PbSO_4(s) + 2e^- | \longrightarrow Pb(s) + {SO_4}^{2-}(aq) | -0.356 | ||
| Tl^+(aq) + e^- | \longrightarrow Tl(s) | -0.336 | ||
| Co^{2+}(aq) + 2e^- | \longrightarrow Co(s) | -0.282 | ||
| Ni^{2+}(aq) + 2e^- | \longrightarrow Ni(s) | -0.236 | ||
| AgI(s) + e^- | \longrightarrow Ag(s) + I^-(aq) | -0.152 | ||
| Sn^{2+}(aq) + 2e^- | \longrightarrow Sn(s) | -0.141 | ||
| Pb^{2+}(aq) + 2e^- | \longrightarrow Pb(s) | -0.127 | ||
| 2H^+(aq) + 2e^- | \longrightarrow H_2(g) | 0 | ||
| AgBr(s) + e^- | \longrightarrow Ag(s) + Br^-(aq) | 0.073 | ||
| S(s) + 2H^+(aq) + 2e^- | \longrightarrow H_2S(aq) | 0.144 | ||
| Sn^{4+}(aq) + 2e^- | \longrightarrow Sn^{2+}(aq) | 0.154 | ||
| {SO_4}^{2-}(aq) + 4H^+(aq) + 2e^- | \longrightarrow SO_2(g)+ 2H_2O | 0.155 | ||
| Cu^{2+}(aq) + e^- | \longrightarrow Cu^+(aq) | 0.161 | ||
| Cu^{2+}(aq) + 2e^- | \longrightarrow Cu(s) | 0.339 | ||
| Cu^+(aq) + e^- | \longrightarrow Cu(s) | 0.518 | ||
| I_2(s) + 2e^- | \longrightarrow 2I^-(aq) | 0.534 | ||
| Fe^{3+}(aq) + e^- | \longrightarrow Fe^{2+}(aq) | 0.769 | ||
| {Hg_2}^{2+}(aq) + 2e^- | \longrightarrow 2Hg(l) | 0.796 | ||
| Ag^+(aq) + e^- | \longrightarrow Ag(s) | 0.799 | ||
| 2Hg^{2+}(aq) + 2e^- | \longrightarrow {Hg_{2}}^{2+}(aq) | 0.908 | ||
| {NO_3}^-(aq) + 4H^+(aq) + 3e^- | \longrightarrow NO(g) + 2H_2O | 0.964 | ||
| {AuCl_4}^-(aq) + 3e^- | \longrightarrow Au(s) + 4Cl^-(aq) | 1.001 | ||
| Br_2(l) + 2e^- | \longrightarrow 2Br^-(aq) | 1.077 | ||
| O_2(g) + 4H^+(aq) + 4e^- | \longrightarrow 2H_2O | 1.229 | ||
| MnO_2(s) + 4H^+(aq) + 2e^- | \longrightarrow Mn^{2+}(aq) + 2H_2O | 1.229 | ||
| {Cr_2O_7}^{2-}(aq) + 14H^+(aq) + 6e^- | \longrightarrow 2Cr^{3+}(aq) + 7H_2O | 1.33 | ||
| Cl_2(g) + 2e^- | \longrightarrow 2Cl^-(aq) | 1.36 | ||
| {ClO_3}^{-}(aq) + 6H^+(aq) + 5e^- | \longrightarrow \frac{1}{2} Cl_2(g) + 3H_2O | 1.458 | ||
| Au^{3+}(aq) + 3e^- | \longrightarrow Au(s) | 1.498 | ||
| {MnO_4}^{-}(aq) + 8H^+(aq) + 5e^- | \longrightarrow Mn^{2+}(aq) + 4H_2O | 1.512 | ||
| PbO_2(s) + {SO_4}^{2-}(aq) + 4H^+(aq) + 2e^- | \longrightarrow PbSO_4(s) + 2H_2O | 1.687 | ||
| H_2O_2(aq) + 2H^+(aq) + 2e^- | \longrightarrow 2H_2O | 1.763 | ||
| Co^{3+}(aq) + e^- | \longrightarrow Co^{2+}(aq) | 1.953 | ||
| F_2(g) + 2e^- \blacktriangleleft ⓘ | \longrightarrow 2F^-(aq) | 2.889 | ||
| Basic Solution, [OH^-] = 1 M | ||||
| {E^{\circ}}_{red} (V) | ||||
| Fe(OH)_2(s) + 2e^- | \longrightarrow Fe(s) + 2OH^-(aq) | -0.891 | ||
| 2H_2O + 2e^- | \longrightarrow H_2(g) + 2OH^-(aq) | -0.828 | ||
| Fe(OH)_3(s) + e^- | \longrightarrow Fe(OH)_2(s) + OH^-(aq) | -0.547 | ||
| S(s) + 2e^- | \longrightarrow S^{2-}(aq) | -0.445 | ||
| {NO_3}^{-}(aq) + 2H_2O + 3e^- | \longrightarrow NO(g) + 4OH^-(aq) | -0.14 | ||
| {NO_3}^-(aq) + H_2O + 2e^- | \longrightarrow {NO_2}^-(aq) + 2OH^-(aq) | 0.004 | ||
| {ClO_4}^-(aq) + H_2O + 2e^- | \longrightarrow {ClO_3}^-(aq) + 2OH^-(aq) | 0.398 | ||
| O_2(g) + 2H_2O + 4e^- | \longrightarrow 4OH^-(aq) | 0.401 | ||
| {ClO_3}^-(aq) + 3H_2O + 6e^- | \longrightarrow Cl^-(aq) + 6OH^-(aq) | 0.614 | ||
| ClO^-(aq) + H_2O + 2e^- | \longrightarrow Cl^-(aq) + 2OH^-(aq) | 0.89 | ||
ⓘ Lithium is the strongest reducing agent.
ⓘ Lithium and fluorine are very dangerous materials to work with.
ⓘ 
O= strongest oxidizing agent;
R = strongest reducing agent.
ⓘ Fluorine is the strongest oxidizing agent.
