JEE Main 2019 — Electrochemistry Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
Given the equilibrium constant:
KC of the reaction :
Cu(s) + 2Ag+ (aq) Cu2+ (aq) + 2Ag(s) is
10 1015, calculate the E of this reaciton at 298 K
[2.303 at 298 K = 0.059V]
KC of the reaction :
Cu(s) + 2Ag+ (aq) Cu2+ (aq) + 2Ag(s) is
10 1015, calculate the E of this reaciton at 298 K
[2.303 at 298 K = 0.059V]
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.4736 V
Step-by-step explanation
We know,
Go = -RTln(KC) ....(1)
Also Go = -nF ....(2)
-nF = -RTln(KC)
=
=
=
( n = no of electron transferred = 2 )
= 0.059
= 0.059 8
= 0.472 V
Go = -RTln(KC) ....(1)
Also Go = -nF ....(2)
-nF = -RTln(KC)
=
=
=
( n = no of electron transferred = 2 )
= 0.059
= 0.059 8
= 0.472 V
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