Question
Explain the reaction between pH of the solution and the hydrogen electrode potential.
✓
Answer
$
\begin{array}{l}
E_{H^{+} / H_2}=0.059 \log \left[H^{+}\right] \\
E_{H^{+} / H_2}=-0.059 pH
\end{array}
$
\begin{array}{l}
E_{H^{+} / H_2}=0.059 \log \left[H^{+}\right] \\
E_{H^{+} / H_2}=-0.059 pH
\end{array}
$
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