Calculate the emf of the cell

\(Pt, H_2 (1.0 atm) | CH_3COOH (0.1 M) || NH_3 (aq, 0.01M) | H_2 (1.0 atm), Pt,\)

\(K_a (CH_3COOH) =1.8 × 10^{–5}, K_b (NH_3) = 1.8 × 10^{–5}\)

–0.46 V

The correct answer is option 2. –0.46 V.

\(Pt, H_2 (1.0 atm) | CH_3COOH (0.1 M) || NH_3 (aq, 0.01M) | H_2 (1.0 atm), Pt,\)

\(K_a (CH_3COOH) =1.8 × 10^{–5}\)

\(K_b (NH_3) = 1.8 × 10^{–5}\)

At Anode: \(\frac{1}{2}H_2 − e^− \rightarrow H^+\)  \(E^0 = 0\)

At Cathode: \(H^+ e^−  \rightarrow \frac{1}{2}H_2 (g)\)  \(E^0_{cell} = 0\)

\(E_{cell} = E^0_{cell} − \frac{0.0591}{1}log\frac{[H^+]_A}{[H^+]_C}\) -----(1)

\([OH^−]^2 = 0.01 × 1.8 × 10^{−5}\)

\([OH^−] = 4.2 × 10^{−4}\)

now,

\([H^+]_C = \frac{10^{−14}}{4.2 × 10^{−4}}\)

Similarly,

\([H^+]^2 = 1.8 × 10^{−5} − 0.1\)

\([H^+]_A = \sqrt{1.8 × 10^{−6}}\)

From equation (1), we get

\(E_{cell} = −0.0591 × 7.78\)

\(∴ E_{cell} = −0.46 V\)