At 298 , K temperature,a hydrogen gas electro | vedclass

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Question

(A) The reaction at the hydrogen electrode is: $H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-}$.Given $pH = 10$,so $[H^{+}] = 10^{-10} \, M$.Pressure of $H_

Vedclass · Accepted Answer

$0.59$

Vedclass · Answer

(A) The reaction at the hydrogen electrode is: $H_{2}(g) ightarrow 2H^{+}(aq) + 2e^{-}$.Given $pH = 10$,so $[H^{+}] = 10^{-10} \, M$.Pressure of $H_{2}$ gas,$P_{H_{2}} = 1 \, atm$.Using the Nernst equation for the oxidation potential:$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log \frac{[H^{+}]^{2}}{P_{H_{2}}}$Since $E^{\circ}_{ox}$ for $S.H.E$ is $0 \, V$ and $n = 2$:$E_{ox} = 0 - \frac{0.0591}{2} \log \frac{(10^{-10})^{2}}{1}$$E_{ox} = -0.02955 	imes \log(10^{-20})$$E_{ox} = -0.02955 	imes (-20) = 0.591 \, V$.Rounding to two decimal places,the potential is $0.59 \, V$.

At $298 \, K$ temperature,a hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at $1 \, atm$ pressure. The potential of the electrode would be ......... $V$.

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