Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

(A) The cell reaction is: $Ag^{+}(0.75 \ M) + Ag_{(s)} \rightarrow Ag_{(s)} + Ag^{+}(0.25 \ M)$
For this concentration cell,the standard cell potential is $\Delta E_{\text{cell}}^{\ominus} = E_{\text{cathode}}^{\ominus} - E_{\text{anode}}^{\ominus} = 0.80 \ V - 0.80 \ V = 0.0 \ V$.
Using the Nernst equation at $298 \ K$:
$E_{\text{cell}} = \Delta E_{\text{cell}}^{\ominus} - \frac{0.0591}{n} \log \frac{[Ag^{+}]_{\text{anode}}}{[Ag^{+}]_{\text{cathode}}}$
Here,$n = 1$,$[Ag^{+}]_{\text{anode}} = 0.25 \ M$,and $[Ag^{+}]_{\text{cathode}} = 0.75 \ M$.
$E_{\text{cell}} = 0.0 - \frac{0.0591}{1} \log \left( \frac{0.25}{0.75} \right)$
$E_{\text{cell}} = -0.0591 \log \left( \frac{1}{3} \right)$
$E_{\text{cell}} = -0.0591 \times (-0.4771)$
$E_{\text{cell}} = +0.0282 \ V$