Question 17.7: Consider a voltaic cell at 25°C in which the reaction is Zn(...
Consider a voltaic cell at 25°C in which the reaction is
Zn(s) + 2H^{+}(aq) → Zn^{2+}(aq) + H_{2}(g)
It is found that the voltage is +0.560 V when [Zn^{2+}] = 0.85 M and P_{H_{2}} = 0.988 atm. What is the pH in the H_{2}-H half-cell?
ANALYSIS | |
reaction: (Zn(s) + 2H^{+}(aq) → Zn^{2+}(aq) + H_{2}(g)) E (0.560 V); T (25°C) P_{H_{2}}(0.988 atm); [Zn^{2+}](0.85 M) temperature (25°C) |
Information given: |
Table 17.1 (standard reduction potentials) | Information implied |
pH | Asked for |
Table 17.1 Standard Potentials in Water Solution at 25°C
Lithium is the strongest reducing agent.
O = strongest oxidizing agent;
Fluorine is the strongest oxidizing agent.
Lithium and fluorine are very dangerous materials to work with.
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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^{-} |
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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^{-} |
STRATEGY
1. Assign oxidation numbers, write oxidation and reduction half-reactions, and cancel electrons to find n.
2. Find E°. (E°_{red} + E°_{ox})
3. Substitute into the Nernst equation (Equation 17.4) for T = 25°C and find Q.
E = E° – \frac{(0.0257 V)}{n}ln Q at 25°C (17.4)
4. Write the Q expression and substitute given concentrations and pressures to find [H^{+}]. Change [H^{+}] to pH.
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