Cell potential for the following reaction at $298 \ K$ is $0.096 \ V$. Calculate the $pH$ of the $HCl$ solution for the cell: $Sn_{(s)} | Sn^{2+} (0.05 \ M) || H^{+} (x \ M) | H_{2(g)} (1 \ bar) | Pt$ given $E_{Sn^{2+}|Sn}^{\circ} = -0.14 \ V$ and $E_{H^{+}|H_2}^{\circ} = 0.0 \ V$.

(1.395) The cell reaction is: $Sn_{(s)} + 2H^{+}_{(aq)} \rightarrow Sn^{2+}_{(aq)} + H_{2(g)}$.
Standard cell potential $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.0 - (-0.14) = 0.14 \ V$.
Using the Nernst equation at $298 \ K$: $E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log \frac{[Sn^{2+}]}{[H^{+}]^2}$.
Here $n = 2$,$[Sn^{2+}] = 0.05 \ M$,$E_{cell} = 0.096 \ V$.
$0.096 = 0.14 - \frac{0.0591}{2} \log \frac{0.05}{[H^{+}]^2}$.
$0.096 - 0.14 = -0.02955 \log \frac{0.05}{[H^{+}]^2}$.
$-0.044 = -0.02955 (\log 0.05 - 2 \log [H^{+}])$.
$1.489 = -1.301 - 2 \log [H^{+}]$.
$2.79 = -2 \log [H^{+}]$.
$pH = -\log [H^{+}] = \frac{2.79}{2} = 1.395$.