MHT CET · Chemistry · Electrochemistry
EMF of hydrogen electrode in term of \(\mathrm{pH}\) is (at 1 atm pressure).
- A \(E_{\mathrm{H}_{2}}=\frac{R T}{F} \times \mathrm{pH}\)
- B \(E_{\mathrm{H}_{2}}=\frac{R T}{F} \cdot \frac{1}{\mathrm{pH}}\)
- C \(E_{\mathrm{H}_{2}}=\frac{2.303 \mathrm{RT}}{F} \mathrm{pH}\)
- D \(E_{\mathrm{H}_{2}}=-0.591 \mathrm{pH}\)
Answer & Solution
Correct Answer
(D) \(E_{\mathrm{H}_{2}}=-0.591 \mathrm{pH}\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_{2}\)
According to Nernst equation,
\(
\begin{aligned}
E &=E^{\circ}+\frac{0.0591}{n} \log \frac{1}{\left[\mathrm{H}^{+}\right]^{2}} \\
E &=0-\frac{0.0591}{2} \log \left[\mathrm{H}^{+}\right]^{2} \\
&=-0.0591 \mathrm{pH}
\end{aligned}
\)
According to Nernst equation,
\(
\begin{aligned}
E &=E^{\circ}+\frac{0.0591}{n} \log \frac{1}{\left[\mathrm{H}^{+}\right]^{2}} \\
E &=0-\frac{0.0591}{2} \log \left[\mathrm{H}^{+}\right]^{2} \\
&=-0.0591 \mathrm{pH}
\end{aligned}
\)
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