JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Redox Reactions and Electrochemistry

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}^{o} = 0.34 V

E_{Ag^{+} / Ag}^{o} = 0.8 V

\frac{2.303 RT}{F} = 0.06 V

Answer: 400

Solution

Passing 1 F through the copper half-cell deposits 0.5 mol Cu, leaving

1.5 - 0.5 = 1.0 mol

C u^{2 +}

in 1 L, that is 1 M. Passing 0.1 F through the silver half-cell deposits 0.1 mol Ag, leaving

0.2 - 0.1 = 0.1 M

A g^{+}

. With copper as anode and silver as cathode the cell reaction is

C u + 2 A g^{+} \to C u^{2 +} + 2 A g

and

E^{\circ} = 0.80 - 0.34 = 0.46 V

. The Nernst equation gives

E = 0.46 - \frac{0.06}{2} lo g \frac{[ C u ^{2 +} ]}{[ A g ^{+} ] ^{2}} = 0.46 - 0.03 lo g \frac{1}{( 0.1 ) ^{2}} = 0.46 - 0.06 = 0.40 V

. The cell emf is 400 mV.

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Redox Reactions and Electrochemistry

Solution

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