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NEET Chemistry Electrochemistry 2026 NEET 2026 (Cancelled)

NEET Chemistry Question (2026) — Solution

Question

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

Options

  1. A. -0.109 V
  2. B. 0.035 V
  3. C. -0.035 V
  4. D. 0.109 V

Answer

D. 0.109 V

Step-by-step solution

The given half-cell is represented as an oxidation electrode: Pt (s) | H _2 (g) | H ^ + (aq) . The corresponding oxidation half-cell reaction is: H _2 (g) 2 H ^ + (aq) + 2 e ^ - Using the Nernst equation for the oxidation potential: E = E^ _ H _2/ H ^ + - 0.059 n [ H ^ + ]^2 P_ H _2 Given values: E^ _ H _2/ H ^ + = 0 V n = 2 [ H ^ + ] = 0.02 M (since HCl is a strong monoprotic acid) P_ H _2 = 2 atm Substituting the values into the Nernst equation: E = 0 - 0.059 2 (0.02)^2 2 E = -0.0295 4 10^ -4 2 E = -0.0295 (2 10^ -4 ) E = -0.0295 ( 2 + 10^ -4 ) E = -0.0295 (0.3010 - 4) E = -0.0295 (-3.699) E = 0.10912 V 0.109 V Answer: 0.109 V

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