Question 17.6: USING THE NERNST EQUATION TO CALCULATE THE CELL POTENTIAL UN...
USING THE NERNST EQUATION TO CALCULATE THE CELL POTENTIAL UNDER NONSTANDARD-STATE CONDITIONS
Consider a galvanic cell that uses the reaction
Zn(s) + 2 H^{+}(aq) → Zn^{2+}(aq) + H_{2}(g)
Calculate the cell potential at 25 °C when [H^{+}] = 1.0 M, [Zn^{2+}] = 0.0010 M, and P_{H_{2}} = 0.10 atm.
STRATEGY
We can calculate the standard cell potential E °from the standard reduction potentials in Table 17.1 and then use the Nernst equation to find the cell potential E under the cited conditions.
TABLE 17.1 Standard Reduction Potentials at 25 °C
| Reduction Half-Reaction | E° (V) | |||
| Stronger oxidizing agent Weaker |
F_{2}(g) + 2 e^{–} H_{2}O_{2}(aq) + 2 H^{+}(aq) + 2 e^{–} MnO_{4}^{–}(aq) + 8 H^{+}(aq) + 5 e^{–} Cl_{2}(g) + 2 e^{–} Cr_{2}O_{7}^{2–}(aq) + 14 H^{+}(aq) + 6 e^{–} O_{2}(g) + 4 H^{+}(aq) + 4 e^{–} Br_{2}(aq) + 2 e^{–} Ag^{+}(aq) + e^{–} Fe^{3+}(aq) + e^{–} O_{2}(g) + 2 H^{+}(aq) + 2 e^{–} I_{2}(s) + 2 e^{–} O_{2}(g) + 2 H_{2}O(l) + 4 e^{–} Cu^{2+}(aq) + 2 e^{–} Sn^{4+}(aq) + 2 e^{–} 2 H^{+}(aq) + 2 e^{–} Pb^{2+}(aq) + 2 e^{–} Ni^{2+}(aq) + 2 e^{–} Cd^{2+}(aq) + 2 e^{–} Fe^{2+}(aq) + 2 e^{–} Zn^{2+}(aq) + 2 e^{–} 2 H_{2}O(l) + 2 e^{–} Al^{3+}(aq) + 3 e^{–} Mg^{2+}(aq) + 2 e^{–} Na^{+}(aq) + e^{–} Li^{+}(aq) + e^{-} |
→ 2 F^{–}(aq)
→ 2 H_{2}O(l) → Mn^{2+}(aq) + 4 H_{2}O(l) → 2 Cl^{–}(aq) → 2 Cr^{3+}(aq) + 7 H_{2}O(l) → 2 H_{2}O(l) → 2 Br^{–}(aq) → Ag(s) → Fe^{2+}(aq) → H_{2}O_{2}(aq) → 2 I^{–}(aq) → 4 OH^{–}(aq) → Cu(s) → Sn^{2+}(aq) → H_{2}(g) →Pb(s) →Ni(s) →Cd(s) →Fe(s) →Zn(s) → H_{2} (g) + 2 OH^{-}(aq) → Al(s) → Mg(s) → Na(s) → Li(s) |
2.87
1.78 1.51 1.36 1.36 1.23 1.09 0.80 0.77 0.70 0.54 0.40 0.34 0.15 0 – 0.13 – 0.26 – 0.40 – 0.45 – 0.76 – 0.83 – 1.66 – 2.37 – 2.71 – 3.04 |
Weaker oxidizing agent Stronger |
The standard cell potential is
E° = E°_{Zn → Zn^{2+}} + E°_{H^{+} → H_{2}} = -(-0.76 V) + 0 V = 0.76 V
The cell potential at 25 °C under nonstandard-state conditions is given by the Nernst equation:
E = E° – \frac{0.0592 V}{n} log Q
=E° – \left(\frac{0.0592 V}{n}\right)\left(log\frac{[Zn^{2+}](P_{H_{2}})}{[H^{+}]^{2}} \right)
where the reaction quotient contains both molar concentrations of solutes and the partial pressure of a gas (in atm). As usual, zinc has been omitted from the reaction quotient because it is a pure solid. For this reaction, 2 mol of electrons are transferred, so n = 2. Substituting into the Nernst equation the appropriate values of E°, n, [H^{+}], [Zn^{2+}], and P_{H_{2}} gives
E = (0.76 V) – \left(\frac{0.0592 V}{2}\right)\left(log \frac{(0.0010)(0.10)}{(1.0)^{2}}\right) = (0.76 V) – \left(\frac{0.0592 V}{2}\right)(-4.0)
= 0.76 V + 0.12 V
0.88 V at 25 °C
BALLPARK CHECK
We expect that the reaction will have a greater tendency to occur under the cited conditions than under standard-state conditions because the product concentrations are lower than standard-state values. We therefore predict that the cell potential E will be greater than the standard cell potential E°, in agreement with the solution.