Emf of the following cell at 298   K in V is x × 10 - 2 Zn Zn 2 + 0 . 1 M ‖ Ag + 0 . 01 M Ag…

Emf of the following cell at 298 K in V is x×10-2

ZnZn2+0.1MAg+0.01MAg

The value of x is ________________ (Rounded off to the nearest integer)

Given: EZn2+/Znθ=-0.76 V; EAg+/Agθ=+0.80 V;  2.303RTF=0.059

Solution

Ecell0=EAg+/Ag0-EZn2+/Zn0

=0.80--0.76

=1.56 V

Ecell=1.56-0.0592logZn2+Ag+2

=1.56-0.0592log0.10.012

=1.56-0.0592×3

=1.56-0.0885

=1.4715

=147.15×10-2

Asked in: JEE Main 2021 (26 Feb Shift 2)

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