
Ecell0=EAg+/Ag0−EZn2+/Zn0
=0.80−(−0.76)
=1.56V
Ecell=1.56−20.059log[Ag+]2[Zn2+]
=1.56−20.059log(0.01)20.1
=1.56−20.059×3
=1.56−0.0885
=1.4715
=147.15×10−2
Emf of the following cell at 298K in V is x×10−2
Zn∣Zn2+(0.1M)∥Ag+(0.01M)∣Ag
The value of x is ________________ (Rounded off to the nearest integer)
[Given:EZn2+/Znθ=−0.76V;EAg+/Agθ=+0.80V;F2.303RT=0.059]
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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