For the cell represented by the following reaction
$Mg(s) + 2Ag^+ (0.0001M) → Mg^{2+} (0.2M) + 2Ag(s)$
If $E^°_{(cell)} = 3.17 V$, its $E_{(cell)}$ at $298\, K$ will be:
(Given: $\log_{10} 2 = 0.301$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.96 V
Use Nernst Equation at $298 \, \text{K}$:
$E_{\text{cell}} = E^o_{\text{cell}} - \left( \frac{0.0591}{n} \right) \log Q$
Where:
$n = \text{electrons transferred}$
$Q = \text{reaction quotient}$
Reaction:
$Mg \rightarrow Mg^{2+} + 2e^-$
$2Ag^+ + 2e^- \rightarrow 2Ag$
So, $n = 2$.
Reaction quotient ($Q$):
$Q = \frac{[Mg^{2+}]}{[Ag^+]^2}$
$Q = \frac{0.2}{(0.0001)^2}$
$Q = \frac{0.2}{10^{-8}}$
$Q = 2 \times 10^7$
$\log Q$ calculation:
$\log(2 \times 10^7) = \log 2 + \log 10^7$
$ = 0.301 + 7$
$ = 7.30$
Apply Nernst equation:
$E_{\text{cell}} = 3.17 - \left(\frac{0.0591}{2}\right)(7.301)$
$E_{\text{cell}} = 3.17 - (0.02955 \times 7.301)$
$ = 3.17 - 0.2156 \approx 2.96 \text{ V}$