If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$
(N/A) The cell reaction is: $H_2(g) + 2H^{+}(x \, M) \rightarrow 2H^{+}(10^{-6} \, M) + H_2(g)$.
Using the Nernst equation at $298 \, K$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.
Since $E^{\circ}_{cell} = 0 \, V$ and $n = 2$,we have $0.118 = 0 - \frac{0.0591}{2} \log \frac{(10^{-6})^2}{x^2}$.
$0.118 = -0.02955 \times \log (10^{-6}/x)^2 = -0.02955 \times 2 \times \log (10^{-6}/x) = -0.0591 \times (-6 - \log x)$.
$0.118 / 0.0591 = -(-6 - \log x) \Rightarrow 2 = 6 + \log x$.
$\log x = -4$,so $x = [H^{+}] = 10^{-4} \, M$.
$pH = -\log [H^{+}] = -\log (10^{-4}) = 4.0$.
Using the Nernst equation at $298 \, K$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.
Since $E^{\circ}_{cell} = 0 \, V$ and $n = 2$,we have $0.118 = 0 - \frac{0.0591}{2} \log \frac{(10^{-6})^2}{x^2}$.
$0.118 = -0.02955 \times \log (10^{-6}/x)^2 = -0.02955 \times 2 \times \log (10^{-6}/x) = -0.0591 \times (-6 - \log x)$.
$0.118 / 0.0591 = -(-6 - \log x) \Rightarrow 2 = 6 + \log x$.
$\log x = -4$,so $x = [H^{+}] = 10^{-4} \, M$.
$pH = -\log [H^{+}] = -\log (10^{-4}) = 4.0$.
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