Verified Solution • 2025–26 Session moderate

QJEEM21Chbdd4: JEE Main 2021 Chemistry - Class 12 JEEM21Chbdd4 Solution: Emf of the following cell at 298K in V is x 10 2 . Zn

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

Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag

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

[Given : EZn2+/Znθ=−0.76V;EAg2+/Agθ=+0.80V;2.303RTF=0.059E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059 ]

Step-by-Step Verified Solution

Zn | Zn2+(0.1 M) || Ag+ (0.01 M) | Ag

Zn(s) + 2Ag+ ⇌\rightleftharpoons 2Ag(s) + Zn+2

Ecell0=EAg+/Ag0−EZn2+/Zn0E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0

=0.80−(−0.76)= 0.80 - ( - 0.76)

=1.56V= 1.56V

Ecell=1.56−0.0592log⁡[Zn2+][Ag+]2{E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}

=1.56−0.0592log⁡0.1(0.01)2= 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}

=1.56−0.0592×3= 1.56 - {{0.059} \over 2} \times 3

=1.56−0.0885= 1.56 - 0.0885

=1.4715= 1.4715

=147.15×10−2= 147.15 \times {10^{ - 2}}

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