In two vessels each containing 500 mL water, 0.5 m mol of aniline (K_b = 10^–9) and 25 m mol of HCl are added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of cells made by connecting them appropriately.

0.395 V

The correct answer is option 1. 0.395 V.

We know from the Nernst equation,

\(E_{cell} = E^0_{cell} − \frac{0.0591}{n}log Q\)

For hydrogen electrode, \(E^0_{cell} = 0\)

At Anode: \(H_2  −  2e^−  \rightarrow  2H^+ (aq)\)

At Cathode: \(2H^+(aq)  +  2e^−  \rightarrow  H_2 (g)\)

\(E_{cell} = 0 − \frac{0.0591}{2} log\frac{[H^+]^2_A}{[H^+]^2_C}\)

At Cthode:

\([H^+] = \frac{25 × 10^{−3}}{500 × 10^{−3}} = 5 × 10^{−2}\)

At Anode:

\(PhNH_2  +  H_2O\) ⇌ \(PhNH_3^+  +  OH^−\)

Also,

\([OH^−] = \sqrt{\frac{K_b}{C}}\)

\(∴ [OH^−]^2 = \frac{10^{−9}}{\frac{0.5}{500}} = 10^{−6}\)

\(∴ [H^+]  =  10^{−8}\)

\(∴ E_{cell} = \frac{0.0591}{2}log\frac{(5 × 10^{−2})^2}{(10^{−8})^2}\)

or, \(E_{cell} = \frac{0.0591}{2} log[5 × 10^{−6}]\)

or, \(E_{cell} = 0.0591 × 6.7\)

∴ \(E_{cell} = 0.395 V\)