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: $\log_{10} 2 = 0.301$)

2.96 V

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}$