If the standard electrode potential, $E_{cell}^o = 1.10 V$, then the $E_{cell}$ for the following reaction at 298 K is

$Zn(s) + Cu^{2+} (0.1 M) →Zn^{2+} (0.01M) + Cu(s)$

1.1295 V

The correct answer is Option (1) → 1.1295 V

Use the Nernst equation at 298 K:

$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$

For the reaction:

$\text{Zn(s)} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu(s)}$

• n = 2
• Reaction quotient:

$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.01}{0.1} = 0.1$

Now substitute:

$E_{\text{cell}} = 1.10 - \frac{0.0591}{2}\log(0.1)$

$\log(0.1) = -1$

$E_{\text{cell}} = 1.10 - \frac{0.0591}{2}(-1)$

$E_{\text{cell}} = 1.10 + 0.02955 = 1.1295 \text{ V}$