The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298 K is : Cu(s) — JEE Mains Chemistry Past Papers Chemistry Question
Question
The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298 K is : Cu(s) + Sn2+(aq.) Cu2+ (aq.) + Sn(s); ( Sn | Sn2 E = –0.16 V, Cu | Cu2 E = 0.34 V, Take F = 96500 C mol–1)
💡 Solution & Explanation
0 Cu / Cu Sn / Sn cell 2 E – E E = – 0.16 –0.34 = – 0.50V G0 = –nF cell E = – 2 × 96500 × (–0.5) = 96500J = 96.5 KJ = 96500 J | JEE MAIN-2020 | DATE : 02-09-2020 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 9
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