Find the emf of the cell in which the following reaction takes place at 298 K Ni ( s ) + 2 Ag + ( 0 . 001 M ) ? Ni 2 + ( 0 . 001 M ) + 2 Ag ( s ) Given that E cell = 10 . 5 V , 2 . 303 R T F = 0 . 059 ? at 298 K [20

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Solution

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

${Ni}_{(s)} + 2 {Ag}^{+} (0.001 M) ⟶ {Ni}^{2 +} (0.001 M) + 2 {Ag}_{(s)}$

Given that $E_{cell}^{\circ} = 10.5$ $V, \frac{2.303 R T}{F} = 0.059 at 298 K$ [2022]

1 1.0385 V

2 1.385 V

3 0.9615 V

4 1.05 V

Ans.
(3)

According to Nernst equation,

$E = E_{cell}^{\circ} - \frac{0.059}{n} \log \frac{[{Ni}^{2 +}]}{{[{Ag}^{+}]}^{2}}$

$E_{cell}^{\circ} = 1.05 (Given)$

Note : Please read 10.5 as 1.05 in question.

$E = 1.05 - \frac{0.059}{2} \log \frac{0.001}{{(0.001)}^{2}}$

$= 1.05 - \frac{0.059}{2} \log 10^{3} = 1.05 - \frac{0.059 \times 3}{2}$

$= 1.05 - 0.0885 = 0.9615$ $V$

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