Question
How is the potential of hydrogen electrode obtained?
✓
Answer
Hydrogen gas electrode is represented as,
$H ^{+}( aq )\left| H _2\left( g , P _{ H _2}\right)\right| Pt$
Electrode reduction reaction is,
$2 H ^{+}{ }_{( aq )}+2 e ^{-} \rightarrow H _{2( g )}$
By Nernst equation, the reduction potential is,
$\begin{aligned}
& E_{ H ^{+} / H _2}=E_{ H ^{+} / H _2}^0-\frac{0.0592}{2} \log _{10} \frac{P_{ H _2}}{\left[ H ^{+}\right]^2} \\
& \because E_{ H ^{+} / H _2}^0=E_{ SHE }=0.0 V , \\
& E_{ H ^{+} / H _2}=-\frac{0.0592}{2} \log _{10} \frac{P_{ H _2}}{\left[ H ^{+}\right]^2}
\end{aligned}$
If $H _2$ gas is passed at $1 atm$, then $P _{ H _2}=1 atm$
$\begin{aligned}
E_{ H ^{+} / H _2} & =-\frac{0.0592}{2} \log _{10} \frac{1}{\left[ H ^{+}\right]^2} \\
& =-\frac{0.0592}{2} \log _{10}\left[ H ^{+}\right]^2 \\
& =-0.0592 \log _{10}\left[ H ^{+}\right] \\
& =0.0592\left[-\log _{10}\left( H ^{+}\right)\right] \\
& =0.0592 pH \\
\therefore pH & =\frac{E_{ H ^{+} / H _2}}{0.0592}\end{aligned}$
$H ^{+}( aq )\left| H _2\left( g , P _{ H _2}\right)\right| Pt$
Electrode reduction reaction is,
$2 H ^{+}{ }_{( aq )}+2 e ^{-} \rightarrow H _{2( g )}$
By Nernst equation, the reduction potential is,
$\begin{aligned}
& E_{ H ^{+} / H _2}=E_{ H ^{+} / H _2}^0-\frac{0.0592}{2} \log _{10} \frac{P_{ H _2}}{\left[ H ^{+}\right]^2} \\
& \because E_{ H ^{+} / H _2}^0=E_{ SHE }=0.0 V , \\
& E_{ H ^{+} / H _2}=-\frac{0.0592}{2} \log _{10} \frac{P_{ H _2}}{\left[ H ^{+}\right]^2}
\end{aligned}$
If $H _2$ gas is passed at $1 atm$, then $P _{ H _2}=1 atm$
$\begin{aligned}
E_{ H ^{+} / H _2} & =-\frac{0.0592}{2} \log _{10} \frac{1}{\left[ H ^{+}\right]^2} \\
& =-\frac{0.0592}{2} \log _{10}\left[ H ^{+}\right]^2 \\
& =-0.0592 \log _{10}\left[ H ^{+}\right] \\
& =0.0592\left[-\log _{10}\left( H ^{+}\right)\right] \\
& =0.0592 pH \\
\therefore pH & =\frac{E_{ H ^{+} / H _2}}{0.0592}\end{aligned}$
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