Question 21.7: Using the Nernst Equation to Calculate Ecell Problem In a te......
Using the Nernst Equation to Calculate E_{\text{cell}}
Problem In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn^{2+} half-cell and an H_2/H^+ half-cell under the following conditions:
[Zn^{2+}] = 0.010 M [H^+] = 2.5 M P_{H_2} = 0.30 \text{ atm}Calculate E_{\text{cell}} at 298.15 K.
Plan To apply the Nernst equation and determine E_{\text{cell}}, we must know E^\circ_{\text{cell}} and Q. We write the spontaneous reaction and calculate E^\circ_{\text{cell}} from standard electrode potentials (Appendix D). Then we substitute into Equation 21.10.
E_{\text{cell}} = E^\circ_{\text{cell}} − \frac{0.0592 V}{n} \log Q (\text{at} 298.15 K) (21.10)
Solution Determining the cell reaction and E^\circ_{\text{cell}}:
2H^+(aq) + 2e^− ⟶ H_2(g) E^\circ = 0.00 V
Zn(s) ⟶ Zn^{2+}(aq) + 2e^− E^\circ = −0.76 V
Calculating Q:
Q = \frac{P_{H_2} × [Zn^{2+}]}{[H^+]^2} = \frac{0.30 × 0.010}{2.5^2} = 4.8×10^{−4}
Solving for E_{\text{cell}} at 25°C (298.15 K), with n = 2:
E_{\text{cell}} = E^\circ_{\text{cell}} − \frac{0.0592 V}{n} \log Q
= 0.76 V − \frac{0.0592 V}{2} \log (4.8×10^{-4}) = 0.76 V − ( −0.098 V) = 0.86 V
Check After you check the arithmetic, reason through the answer: E_{\text{cell}} > E^\circ_{\text{cell}} (0.86 > 0.76) because the log Q term was negative, which is consistent with Q < 1.