1 Faraday electricity was passed through (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 29

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Solution

Q.
1 Faraday electricity was passed through ${Cu}^{2 +}$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through ${Ag}^{+}$ (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is:

$Given: E_{{Cu}^{2 +} / Cu}^{\circ} = 0.34 V$

$E_{{Ag}^{+} / Ag}^{\circ} = 0.80 V$

$\frac{2.303 R T}{F} = 0.06 V$ [2025]

Ans.
(400)

Cell reaction of the electrolytic cell:

$Cu (s) + \underset{0.1 M}{2 {Ag}^{+} (a q)} \to \underset{1 M}{{Cu}^{2 +} (a q)} + 2 Ag$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = E_{{Ag}^{+} / Ag}^{\circ} - E_{{Cu}^{2 +} / Cu}^{\circ}$

$= (0.8 - 0.34) V = 0.46 V$

Reaction quotient (Q):
$= \frac{[{Cu}^{2 +}]}{{[Ag^{+}]}^{2}} = \frac{1}{{(0.1)}^{2}} = 100$ , $\log Q = \log 100 = 2$

By Nernst equation:

$E_{cell} = 0.46 - \frac{0.0591}{2} \times 2 = 0.4 V = 400 mV$

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