JEE Main 2020 — Electrochemistry Question with Solution
From: JEE Main 2020 (Online) 2nd September Morning Slot
Question
The Gibbs change (in J) for the given reaction at
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) Cu2+(aq.) + Sn(s);
(,
)
Take F = 96500 C mol–1)
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) Cu2+(aq.) + Sn(s);
(,
)
Take F = 96500 C mol–1)
Enter your answer
Show full solutionCorrect answer: 96500
Correct answer
96500
Step-by-step explanation
G = Go + RTln
= –2 × 96500 [(–0.16) – 0.34] + RT
= 96500 J
= –2 × 96500 [(–0.16) – 0.34] + RT
= 96500 J
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