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.

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

\(K_a (CH_3COOH) =1.8 × 10^{–5}\)

\(K_b (NH_3) = 1.8 × 10^{–5}\)

\([H^+]_{CH_3COOH} = \sqrt{K_a \times C}\)

\(⇒ [H^+]_{CH_3COOH} = \sqrt{1.8 \times 10^{-5} \times 0.1}\)

\(⇒ [H^+]_{CH_3COOH} = 1.34 \times 10^{-3}\)

Also,

\([OH^-]= \sqrt{K_b\times C}\)

\(⇒ [OH^-]= \sqrt{1.8 \times 10^{-5}\times 0.01}\)

\(⇒ [OH^-]=  4.24 \times 10^{-4}\)

And

\([H^+]_{NH_4OH} = \frac{10^{-14}}{[OH^-]}\)

\(⇒ [H^+]_{NH_4OH} = \frac{10^{-14}}{4.24 \times 10^{-4}}\)

\(⇒ [H^+]_{NH_4OH} = 2.35 \times 10^{-11}\)

\(E_{cell} = E^0_{cell} - \frac{0.0591}{n}log\frac{[H^+]_{CH_3COOH}}{[H^+]_{NH_4OH}}\)

\(⇒E_{cell} = E^0_{cell} - \frac{0.0591}{1}log\frac{1.34 \times 10^{-3}}{2.35 \times 10^{-11}}\)

\(⇒E_{cell} \approx -0.46\)