NEET 2026 — Chemistry solution

NEET2026Chemistry

Calculate emf of the half cell given below :

$P t (s) ∣ H_{2} (g, 2 atm) ∣ H C l (a q, 0.02 M)$

$E_{H_{2} / H^{+}}^{\circ} = 0$ V

Given : $\frac{2.303 R T}{F} = 0.059$ , log 2 = 0.3010

Solution

The oxidation half reaction is $H_{2} (g) \to 2 H^{+} (a q) + 2 e^{-}$ .

Using the Nernst equation for the oxidation potential:

$E_{O P} = 0 - \frac{0.059}{2} lo g \frac{[ H ^{+} ] ^{2}}{P _{H_{2}}}$

$= - \frac{0.059}{2} lo g \frac{( 2 \times 1 0 ^{- 2} ) ^{2}}{2}$

$= - \frac{0.059}{2} lo g (2 \times 1 0^{- 4})$

$= - \frac{0.059}{2} (0.3010 - 4) = - \frac{0.059}{2} (- 3.7) \approx + 0.109$ V.

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