The emf of the cell M|Mn+ (0.02 M) || H+ (1 M) | H2 (g) (1 atm), Pt a…

The correct answer is option 2. 2.

\(Anode : M −ne \rightarrow \underset{(0.02M)}{M^{n+}}\)  \(E^0 = +0.76\)

\(Cathode: \underset{(1m)}{nH^+}  +  ne  \rightarrow \frac{n}{2}H_2 (g)\) \(E^0 = 0\)

\(M  +  nH^+ \rightarrow \frac{n}{2}H_2 (g)\) \(E^0 = +0.76\)

\(E_{cell}= 0.81 = 0.76 − \frac{0.0591}{n}log\frac{[M^{n+}]}{[H^+]^n}\)

or \(E_{cell} =0.76 − \frac{0.0591}{n}log\frac{0.02}{1}\)

or, \(E_{cell} = 0.76 + \frac{0.0591}{n}log\frac{1}{0.02}\)

or, \(0.81= 0.76 + \frac{0.0591}{n}× 1.7\)  \([Since, E_{cell}= 0.81]\)

or, \(\frac{0.591}{n} × 1.7 = 0.81 − 0.76\)

or, \(n = \frac{0.0591 × 1.7}{0.05}\)

or, \(n = 2\)