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The e.m.f. of a Daniell cell at 298 K is $E_1$.
$Zn|ZnSO_4(0.01\ M)\,||\,CuSO_4(1.0\ M)|Cu$
When the concentration of $ZnSO_4$ is 1.0 M and that of $CuSO_4$ is 0.01 M, the e.m.f. is changed to $E_2$. What is the relationship between $E_1$ and $E_2$ :
$Zn|ZnSO_4(0.01\ M)\,||\,CuSO_4(1.0\ M)|Cu$
When the concentration of $ZnSO_4$ is 1.0 M and that of $CuSO_4$ is 0.01 M, the e.m.f. is changed to $E_2$. What is the relationship between $E_1$ and $E_2$ :
A
$E_1 \gt E_2$
B
$E_1 \lt E_2$
C
$E_1 = E_2$
D
$E_2 = 0 \neq E_1$
Detailed Solution
Cell reaction: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$, with n = 2.
Nernst equation: $E = E^\circ - \dfrac{0.0591}{2}\log\dfrac{[Zn^{2+}]}{[Cu^{2+}]}$
Case 1: $E_1 = E^\circ - \dfrac{0.0591}{2}\log\dfrac{0.01}{1.0} = E^\circ - \dfrac{0.0591}{2}(-2) = E^\circ + 0.0591$ V
Case 2: $E_2 = E^\circ - \dfrac{0.0591}{2}\log\dfrac{1.0}{0.01} = E^\circ - \dfrac{0.0591}{2}(2) = E^\circ - 0.0591$ V
With $E^\circ = 1.10$ V: $E_1 = 1.159$ V and $E_2 = 1.041$ V.
So $E_1 \gt E_2$.
Nernst equation: $E = E^\circ - \dfrac{0.0591}{2}\log\dfrac{[Zn^{2+}]}{[Cu^{2+}]}$
Case 1: $E_1 = E^\circ - \dfrac{0.0591}{2}\log\dfrac{0.01}{1.0} = E^\circ - \dfrac{0.0591}{2}(-2) = E^\circ + 0.0591$ V
Case 2: $E_2 = E^\circ - \dfrac{0.0591}{2}\log\dfrac{1.0}{0.01} = E^\circ - \dfrac{0.0591}{2}(2) = E^\circ - 0.0591$ V
With $E^\circ = 1.10$ V: $E_1 = 1.159$ V and $E_2 = 1.041$ V.
So $E_1 \gt E_2$.