What is the EMF of the cell: Pt, H2 (1 atm) \| CH3COOH (0.1 M) \|\| (0.01 M) NH4OH \| H2 (1 atm), Pt? Given: Ka for CH3COOH = 1.8 × 10^–5, Kb for NH4OH = 1.8 × 10^–5, 2.303RT/F = 0.06, log 1.8 = 0.25)

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Prashant Kotian · Accepted Answer

Correct answer: B. Anode [H+] = sqrt(Ka*C) = sqrt(1.8*10^-5 * 0.1) = 1.34*10^-3 M. Cathode [OH-] = sqrt(Kb*C) = sqrt(1.8*10^-5 * 0.01) = 4.24*10^-4 M, so [H+]_c = 10^-14 / 4.24*10^-4 = 2.35*10^-11 M. E = 0.06 * log([H+]_c / [H+]_a) = 0.06 * log(2.35*10^-11 / 1.34*10^-3) = 0.06 * (-7.75) = -0.465 V. Therefore, correct answer is B.

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