Q9 / 150 · MHT CET 2007 · Chemistry

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
    $E_{\mathrm{H}_{2}}=\frac{2.303 \mathrm{RT}}{F} \mathrm{pH}$
  • Correct answer
    D
    $E_{\mathrm{H}_{2}}=-0.591 \mathrm{pH}$

Solution

$2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_{2}$ According to Nernst equation, $$ \begin{aligned} E &=E^{\circ}+\frac{0.0591}{n} \log \frac{1}{\left[\mathrm{H}^{+}\right]^{2}} \\ E &=0-\frac{0.0591}{2} \log \left[\mathrm{H}^{+}\right]^{2} \\ &=-0.0591 \mathrm{pH} \end{aligned} $$

Hence, the correct answer is option: 'D'.

Relevant Questions