Question 17.6: GRADED Consider a voltaic cell in which the following reacti...
GRADED
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) → 2H_{2}O + 2Br_{2}(l)
ⓐ Calculate E for the cell at 25°C.
ⓑ When the voltaic cell is at 35°C, E is measured to be 0.039 V. What is E° at 35°C?
ⓐ
| ANALYSIS | |
| reaction: (O_{2}(g) + 4H^{+}(aq) + 4Br^{-}(aq) → 2Br_{2}(l) + 2H_{2}O)
P_{O_{2}} (0.98 atm); [H^{+}] (pH = 1.24); [Br^{-}] (0.15 M) temperature (25°C) |
Information given: |
| Table 17.1 (standard reduction potentials) | Information implied: |
| E | Asked for: |
Table 17.1 Standard Potentials in Water Solution at 25°C
| Lithium is the strongest reducing agent.
Fluorine is the strongest oxidizing agent.
Lithium and fluorine are very dangerous materials to work with.
|
AcidicSolution,[H^{+}] = 1 M | ||||
| E°_{red}(V) | |||||
| -3.040 -2.936 -2.906 -2.869 -2.714 -2.357 -1.680 -1.182 -0.762 -0.744 -0.409 -0.408 -0.402 -0.356 -0.336 -0.282 -0.236 -0.152 -0.141 -0.127 0.000 0.073 0.144 0.154 0.155 0.161 0.339 0.518 0.534 0.769 0.796 0.799 0.908 0.964 1.001 1.077 1.229 1.229 1.330 1.360 1.458 1.498 1.512 1.687 1.763 1.953 2.889 |
 |
→Li(s) → K(s) → Ba(s) → Ca(s) →Na(s) → Mg(s) → Al(s) → Mn(s) → Zn(s) → Cr(s) → Fe(s) → Cr^{2+}(aq) → Cd(s) → Pb(s) + SO_{4}^{2-}(aq) → Tl(s) → Co(s) →Ni(s) → Ag(s) + I^{-}(aq) → Sn(s) → Pb(s) →H_{2}(g) → Ag(s) + Br^{-}(aq) →H_{2}S(aq) → Sn^{2+}(aq) → SO_{2}(g) + 2H_{2}O → Cu^{+}(aq) → Cu(s) → Cu(s) → 2I^{-}(aq) → Fe^{2+}(aq) → 2Hg(l) → Ag(s) →Hg_{2}^{2+}(aq) →NO(g) + 2H_{2}O → Au(s) + 4Cl^{-}(aq) → 2Br^{-}(aq) → 2H_{2}O → Mn^{2+}(aq) + 2H_{2}O → 2Cr^{3+}(aq) + 7H_{2}O → 2Cl^{-}(aq) →\frac{1}{2} Cl_{2}(g) + 3H_{2}O → Au(s) → Mn^{2+}(aq) + 4H_{2}O → PbSO_{4}(s) + 2H_{2}O → 2H_{2}O → Co^{2+}(aq) → 2F^{-}(aq) |
Li^{+}(aq) + e^{-} K^{+}(aq) + e^{-} Ba^{2+}(aq) + 2e^{-} Ca^{2+}(aq) + 2e^{-} Na^{+}(aq) + e^{-} Mg^{2+}(aq) + 2e^{-} Al^{3+}(aq) + 3e^{-} Mn^{2+}(aq) + 2e^{-} Zn^{2+}(aq) + 2e^{-} Cr^{3+}(aq) + 3e^{-} Fe^{2+}(aq) + 2e^{-} Cr^{3+}(aq) + e^{-} Cd^{2+}(aq) + 2e^{-} PbSO_{4}(s) + 2e^{-} Tl^{+}(aq) + e^{-} Co^{2+}(aq) + 2e^{-} Ni^{2+}(aq) + 2e^{-} AgI(s) + e^{-} Sn^{2+}(aq) + 2e^{-} Pb^{2+}(aq) + 2e^{-} 2H^{+}(aq) + 2e^{-} AgBr(s) + e^{-} S(s) + 2H^{+}(aq) + 2e^{-} Sn^{4+}(aq) + 2e^{-} SO_{4}^{2-}(aq) + 4H^{+}(aq) + 2e^{-} Cu^{2+}(aq) + e^{-} Cu^{2+}(aq) + 2e^{-} Cu^{+}(aq) + e^{-} I_{2}(s) + 2e^{-} Fe^{3+}(aq) + e^{-} Hg_{2}^{2+}(aq) + 2e^{-} Ag^{+}(aq) + e^{-} 2Hg^{2+}(aq) + 2e^{-} NO_{3}^{-}(aq) + 4H^{+}(aq) + 3e^{-} AuCl_{4}^{-}(aq) + 3e^{-} Br_{2}(l) + 2e^{-} O_{2}(g) + 4H^{+}(aq) + 4e^{-} MnO_{2}(s) + 4H^{+}(aq) + 2e^{-} Cr_{2}O_{7}^{2-}(aq) + 14H^{+}(aq) + 6e^{-} Cl_{2}(g) + 2e^{-} ClO_{3}^{-}(aq) + 6H^{+}(aq) + 5e^{-} Au^{3+}(aq) + 3e^{-} MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-} PbO_{2}(s) + SO_{4}^{2-}(aq) + 4H^{+}(aq) + 2e^{-} H_{2}O_{2}(aq) + 2H^{+}(aq) + 2e^{-} Co^{3+}(aq) + e^{-} F_{2}(g) + 2e^{-} |
 |
|
| Basic Solution, [OH^{-}] = 1 M | |||||
| E°_{red}(V) | |||||
| -0.891 -0.828 -0.547 -0.445 -0.140 0.004 0.398 0.401 0.614 0.890 |
→ Fe(s) + 2 OH^{-}(aq) →H_{2}(g) + 2 OH^{-}(aq) → Fe(OH)_{2}(s) + OH^{-}(aq) → S^{2-}(aq) →NO(g) + 4 OH^{-}(aq) →NO_{2}^{-}(aq) + 2 OH^{-}(aq) → ClO_{3}^{-}(aq) + 2 OH^{-}(aq) → 4 OH^{-}(aq) → Cl^{-}(aq) + 6 OH^{-}(aq) → Cl^{-}(aq) + 2 OH^{-}(aq) |
Fe(OH)_{2}(s) + 2e^{-} 2H_{2}O + 2e^{-} Fe(OH)_{3}(s) + e^{-} S(s) + 2e^{-} NO_{3}^{-}(aq) + 2H_{2}O + 3e^{-} NO_{3}^{-}(aq) + H_{2}O + 2e^{-} ClO_{4}^{-}(aq) + H_{2}O + 2e^{-} O_{2}(g) + 2H_{2}O + 4e^{-} ClO_{3}^{-}(aq) + 3H_{2}O + 6e^{-} ClO^{-}(aq) + H_{2}O + 2e^{-} |
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O = strongest oxidizing agent;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°_{red} + E°_{ox})
4. Substitute into the Nernst equation (Equation 17.4) for T = 25°C
E = E° – \frac{(0.0257 V)}{n}ln Q at 25°C (17.4)
E = E° – \frac{0.0257}{n}ln Q
ⓑ
| ANALYSIS | |
| E (0.039 V) at T (35°C) From part (a): Q (1.8 × 10^{8}); n (4 moles) |
Information given: |
| R and F values in joules | Information implied: |
| E° at 35°C | Asked for: |
STRATEGY
Substitute into the Nernst equation for any T.
E = E° – \frac{RT}{nF}ln Q
ⓐ
| 1.24 = -log_{10}[H^{+}]; [H^{+}] = 0.058 M
Q = \frac{1}{(P_{O_{2}}) [H^{+}]^{4} [Br^{-}]^{4}} = \frac{1}{(0.98)(0.058)^{4}(0.15)^{4}} = 1.8 × 10^{8} |
1. [H^{+}]
Q |
| O: 0 → -2 (reduction); Br: -1 → 0 (oxidation)
O_{2}(g) + 4H^{+}(aq) + 4e^{-}(aq) → 2H_{2}O
2Br^{-}(aq) → Br_{2}(l) + 2e^{-}
The oxidation half-reaction must be multiplied by 2 to cancel out the four electrons in the reduction half-reaction. n = 4 |
2. Oxidation numbers
Half-reactions
n |
| E°_{red} for O_{2} = 1.299 V; E°_{red} for Br^{-} = -1.077 V E° = 1.229 V + (-1.077 V) = 0.152 V |
3. E° |
| E = 0.152 V – \frac{0.0257}{4} ln (1.8 × 10^{8}) = 0.030 V | 4. E |
ⓑ
| 0.039 V = E° – \frac{(8.31 J/mol · K)(308K)}{4 (9.648 × 10^{4} J/mol · V)} ln (1.8 × 10^{8})
E° = 0.039 V + 0.126 V = 0.165 V |
E° |