E=E0−nF2.303RTlogQ
Here,
E=+0.801VE0=ECathode∘−EAnode∘(BothareSRPvalues)E0=0.008−(−0.763)
=+0.771V
∴0.801=+0.771−n0.06log10−2
⇒n=4
JEE Main 2022 — Chemistry Physical Chemistry
The cell potential for Zn∣Zn2+(aq)∥Snx+∣Sn is 0.801V at 298K. The reaction quotient for the above reaction is 10−2. The number of electrons involved in the given electrochemical cell reaction is _____ .
(Given EZn2+∣Zn0=−0.763V,ESnx+∣Sn0=+0.008V and F2.303RT=0.06V)
Held on 25 Jul 2022 · Verified 6 Jul 2026.
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