Question:
The reduction potential of hydrogen electrode at \(25^\circ C\) in a neutral solution is
\[
(p_{H_2}=1 \text{ bar})
\]
The reduction potential of hydrogen electrode at \(25^\circ C\) in a neutral solution is
\[ (p_{H_2}=1 \text{ bar}) \]
\[ (p_{H_2}=1 \text{ bar}) \]
Show Hint
For hydrogen electrode at \(25^\circ C\):
\[
E=-0.0591\times pH
\]
when hydrogen pressure is \(1\,bar\).
Updated On: Jun 15, 2026
- \(0.059\,V\)
- \(-0.059\,V\)
- \(-0.413\,V\)
- \(0.0\,V\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the hydrogen electrode reaction.
The standard hydrogen electrode reaction is
\[ 2H^+ + 2e^- \rightarrow H_2(g) \] For this reaction, standard electrode potential is
\[ E^\circ = 0\,V \]
Step 2: Use the Nernst equation.
The Nernst equation for hydrogen electrode is
\[ E=E^\circ-\frac{0.0591}{2}\log\frac{p_{H_2}}{[H^+]^2} \] Given,
\[ p_{H_2}=1\,\text{bar} \] So,
\[ E=0-\frac{0.0591}{2}\log\frac{1}{[H^+]^2} \] \[ E=-0.0591\,pH \]
Step 3: Use the pH of neutral solution.
At \(25^\circ C\), a neutral solution has
\[ pH=7 \] Therefore,
\[ E=-0.0591\times7 \] \[ E=-0.4137\,V \]
Step 4: Final conclusion.
Hence, the reduction potential of hydrogen electrode is approximately
\[ \boxed{-0.413\,V} \]
The standard hydrogen electrode reaction is
\[ 2H^+ + 2e^- \rightarrow H_2(g) \] For this reaction, standard electrode potential is
\[ E^\circ = 0\,V \]
Step 2: Use the Nernst equation.
The Nernst equation for hydrogen electrode is
\[ E=E^\circ-\frac{0.0591}{2}\log\frac{p_{H_2}}{[H^+]^2} \] Given,
\[ p_{H_2}=1\,\text{bar} \] So,
\[ E=0-\frac{0.0591}{2}\log\frac{1}{[H^+]^2} \] \[ E=-0.0591\,pH \]
Step 3: Use the pH of neutral solution.
At \(25^\circ C\), a neutral solution has
\[ pH=7 \] Therefore,
\[ E=-0.0591\times7 \] \[ E=-0.4137\,V \]
Step 4: Final conclusion.
Hence, the reduction potential of hydrogen electrode is approximately
\[ \boxed{-0.413\,V} \]
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