JEE Main 2024 Chemistry question
The potential for the given half cell at 298 K is (–) × V. 2H(aq) + 2e → H(g), [H] = 1M, P = 2 atm (Given: 2.303RT/F = 0.06 V, log 2 = 0.3)
Answer: 1
Solution
E = −(0.06/2) log 2 = −0.009 V ≈ −1 × 10⁻² V (nearest integer).