The pressure of H_{2} required to make the po | vedclass

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(A) The electrode reaction is: $2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}$Using the Nernst equation: $E = E^{0} - \frac{0.0591}{2} \log \frac{P_{H_{

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$10^{-14} \ atm$

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(A) The electrode reaction is: $2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}$Using the Nernst equation: $E = E^{0} - \frac{0.0591}{2} \log \frac{P_{H_{2}}}{[H^{+}]^{2}}$Given $E = 0$,$E^{0} = 0$,and in pure water $[H^{+}] = 10^{-7} \ M$:$0 = 0 - 0.0295 \log \frac{P_{H_{2}}}{(10^{-7})^{2}}$$0 = \log \frac{P_{H_{2}}}{10^{-14}}$Taking the antilog: $1 = \frac{P_{H_{2}}}{10^{-14}}$Therefore,$P_{H_{2}} = 10^{-14} \ atm$.

The pressure of $H_{2}$ required to make the potential of $H_{2}$ electrode zero in pure water at $298 \ K$ is

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