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 : $$E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$$]
Answer (integer)
147
Solution
Zn | Zn<sup>2+</sup>(0.1 M) || Ag<sup>+</sup> (0.01 M) | Ag
<br><br>Zn(s) + 2Ag<sup>+</sup> $\rightleftharpoons$ 2Ag(s) + Zn<sup>+2</sup>
<br><br>$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$<br><br>$= 0.80 - ( - 0.76)$<br><br>$= 1.56V$<br><br>$${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$<br><br>$= 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$<br><br>$= 1.56 - {{0.059} \over 2} \times 3$<br><br>$= 1.56 - 0.0885$<br><br>$= 1.4715$<br><br>$= 147.15 \times {10^{ - 2}}$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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