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(A) The oxidation half-reaction for the hydrogen electrode is: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$Using the Nernst equation for oxidation potent

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$0.177$

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(A) The oxidation half-reaction for the hydrogen electrode is: $H_{2(g)} ightarrow 2H^+_{(aq)} + 2e^-$Using the Nernst equation for oxidation potential:$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log Q$For the standard hydrogen electrode,$E^{\circ}_{ox} = 0.00 \ V$ and $n = 2$.The reaction quotient $Q$ is given by $[H^+]^2 / P_{H_2}$.Given $pH = 3$,so $[H^+] = 10^{-3} \ M$.$Q = \frac{(10^{-3})^2}{1} = 10^{-6}$.$E_{ox} = 0 - \frac{0.0591}{2} \log(10^{-6})$$E_{ox} = -0.02955 	imes (-6) = 0.1773 \ V$.Thus,the oxidation potential is approximately $0.177 \ V$.

What will be the oxidation potential for the following hydrogen half cell at $1 \ bar$ pressure and $25^{\circ} C$ temperature (in $V$)?
$Pt \mid H_{2(g)} (1 \ bar) \mid HCl_{(aq)} \ pH = 3$

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What will be the oxidation potential for the following hydrogen half cell at $1 \ bar$ pressure and $25^{\circ} C$ temperature (in $V$)?$Pt \mid H_{2(g)} (1 \ bar) \mid HCl_{(aq)} \ pH = 3$