1 Faraday electricity was passed through Cu2+ (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

Question

1 Faraday electricity was passed through Cu2+ (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 _______ mV (nearest integer) Given : V . Eo Cu / Cu2 V . Eo Ag / Ag V . F RT .

Prashant Kotian · Accepted Answer

Correct answer: .. Cu+2 + 2e2–  Cu (1 farraday = charge on 1 mole electron) t = 0 1.5 1 mol t = 1 – 0.5 mol [Cu+2] = 1 M after electrolysis Ag+ + e–  Ag t = 0 0.2 0.1 mol t = 1 0.1 – – [Ag+] = 0.1 M after electrolysis Cell Cu(s) + 2Ag+(aq)  Cu+2(aq) + 2Ag(s) reaction E = Eº – n . log 2 ] Ag [ ] Cu [   E = (0.8 – 0.34) – 06 . log 2)1 . ( = 0.4 v correct answer = 400 mv | JEE(Main) 2025 | DATE : 07-04-2025 (SHIFT-1) | PAPER-1 | CHEMISTRY PAGE # 10

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