Question
Calculate emf of the half cell given below :
$$Pt(s) | H_2 (g, 2 \text{ atm}) | HCl (aq, 0.02 \text{ M})$$
$$E_{H_2 /H^+}^\circ = 0 \text{ V}$$
(Given : $\frac{2.303 RT}{F} = 0.059$, $\log 2 = 0.3010$)
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NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$0.109 \text{ V}$
Calculate Half-Cell EMF using Nernst Equation
The electromotive force (emf) of the half-cell can be calculated using the Nernst equation. The reaction at the electrode is:
$ H_2 (g) \rightarrow 2H^+ (aq) + 2e^- $
The Nernst equation for this half-reaction is:
$ E = E^\circ - \frac{2.303 RT}{nF} \log \frac{[H^+]^2}{P_{H_2}} $
Given values:
- Standard electrode potential, $E^\circ$ = 0 V
- Number of electrons, $n$ = 2
- $\frac{2.303 RT}{F}$ = 0.059
- Concentration of HCl, $[HCl]$ = 0.02 M, thus $[H^+]$ = 0.02 M
- Partial pressure of Hydrogen gas, $P_{H_2}$ = 2 atm
Nernst Equation Calculation Steps
- Substitute the given values into the Nernst equation:
$ E = 0 - \frac{0.059}{2} \log \frac{(0.02)^2}{2} $
- Simplify the equation:
$ E = -0.0295 \log \frac{0.0004}{2} $
$ E = -0.0295 \log (0.0002) $
$ E = -0.0295 \log (2 \times 10^{-4}) $
- Apply logarithm properties ($\log(a \times b) = \log a + \log b$ and $\log(10^x) = x$):
$ E = -0.0295 (\log 2 + \log 10^{-4}) $
$ E = -0.0295 (\log 2 - 4) $
- Substitute the value of $\log 2 = 0.3010$:
$ E = -0.0295 (0.3010 - 4) $
$ E = -0.0295 (-3.699) $
- Calculate the final emf:
$ E \approx 0.109 \text{ V} $
The calculated emf of the half-cell is approximately 0.109 V.
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