Question

Find the emf of the cell in which the following reaction takes place at 298 K.

\[\ce{Ni_{(s)} + 2Ag+ (0.001 M) -> Ni^{+2} (0.001 M) + 2Ag_{(s)}}\]

(Given that \[\ce{E{^{\circ}_{cell}}}\] = 1.05 V, \[\ce{\frac{2.303 RT}{F}}\] = 0.059 at 298 K)

विकल्प

• 1.385 V

• 0.9615 V

• 1.05 V

• 1.0385 V

Answer 1

0.9615 V

Explanation:

The given reaction is:

\[\ce{Ni_{(s)} + 2Ag+ (0.001 M) -> Ni^{+2} (0.001 M) + 2Ag_{(s)}}\]

\[\ce{E^{\circ}_{cell}}\] = 1.05 V,

\[\ce{\frac{2.303 RT}{F}}\] = 0.059

n = 2 (electrons exchanged)

Ni²⁺ = 0.001 M

Ag⁺ = 0.001 M

\[\ce{E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} log \frac{[Ni^{2+}]}{[Ag^{+2}]}}\]

\[\ce{E_{cell} = 1.05 - \frac{0.059}{2} log \frac{0.001}{(0.001)^2}}\]

= 1.05 − 0.0295 log(10³)

= 1.05 − 0.0295 × 3

= 1.05 − 0.0885

= 0.9615 V