Q96 / 180 · NEET 2026 (Cancelled) · Chemistry

Calculate emf of the half cell given below :
$\text{Pt (s)} | \text{H}_2 \text{(g, 2 atm)} | \text{HCl (aq, 0.02 M)}$
$E^{\circ}_{\text{H}_2/\text{H}^{+}} = 0$ V
$\left(\text{Given}: \dfrac{2.303\ RT}{F} = 0.059, \log 2 = 0.3010\right)$

  • A
    $-0.109$ V
  • B
    $0.035$ V
  • C
    $-0.035$ V
  • Correct answer
    D
    $0.109$ V

Solution

The given half-cell is represented as an oxidation electrode: $\text{Pt (s)} | \text{H}_2 \text{(g)} | \text{H}^{+} \text{(aq)}$.

The corresponding oxidation half-cell reaction is:
$\text{H}_2 \text{(g)} \rightarrow 2\text{H}^{+} \text{(aq)} + 2\text{e}^{-}$

Using the Nernst equation for the oxidation potential:
$E = E^{\circ}_{\text{H}_2/\text{H}^{+}} - \dfrac{0.059}{n} \log \dfrac{[\text{H}^{+}]^2}{P_{\text{H}_2}}$

Given values:
$E^{\circ}_{\text{H}_2/\text{H}^{+}} = 0$ V
$n = 2$
$[\text{H}^{+}] = 0.02$ M (since HCl is a strong monoprotic acid)
$P_{\text{H}_2} = 2$ atm

Substituting the values into the Nernst equation:
$E = 0 - \dfrac{0.059}{2} \log \dfrac{(0.02)^2}{2}$

$E = -0.0295 \log \dfrac{4 \times 10^{-4}}{2}$

$E = -0.0295 \log (2 \times 10^{-4})$

$E = -0.0295 (\log 2 + \log 10^{-4})$

$E = -0.0295 (0.3010 - 4)$

$E = -0.0295 \times (-3.699)$

$E = 0.10912$ V $\approx 0.109$ V

Answer: $0.109$ V

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

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