JEE MAIN - Chemistry (2021 - 26th February Evening Shift - No. 21)
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 : $$E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$$]
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
147
Explanation
Zn | Zn2+(0.1 M) || Ag+ (0.01 M) | Ag
Zn(s) + 2Ag+ $$ \rightleftharpoons $$ 2Ag(s) + Zn+2
$$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$$
$$ = 0.80 - ( - 0.76)$$
$$ = 1.56V$$
$${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$
$$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$$
$$ = 1.56 - {{0.059} \over 2} \times 3$$
$$ = 1.56 - 0.0885$$
$$ = 1.4715$$
$$ = 147.15 \times {10^{ - 2}}$$
Zn(s) + 2Ag+ $$ \rightleftharpoons $$ 2Ag(s) + Zn+2
$$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$$
$$ = 0.80 - ( - 0.76)$$
$$ = 1.56V$$
$${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$
$$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$$
$$ = 1.56 - {{0.059} \over 2} \times 3$$
$$ = 1.56 - 0.0885$$
$$ = 1.4715$$
$$ = 147.15 \times {10^{ - 2}}$$
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