In the electrochemical cell: $ ext{Zn}| ext{ZnSO}_4(0.01 ext{ M}) || ext{CuSO}_4(1.0 ext{ M})| ext{Cu

Question

In the electrochemical cell: $	ext{Zn}|	ext{ZnSO}_4(0.01 	ext{ M}) || 	ext{CuSO}_4(1.0 	ext{ M})|	ext{Cu}$, the emf of this Daniel cell is $E_1$. When the concentration of $	ext{ZnSO}_4$ is changed to $1.0 	ext{ M}$ and that of $	ext{CuSO}_4$ is changed to $0.01 	ext{ M}$, the emf changes to $E_2$. The relationship between $E_1$ and $E_2$ is : (Given, $\frac{RT}{F} = 0.059$)

Accepted Answer

According to the Nernst equation for the Daniell cell:
$E_{	ext{cell}} = E^\circ_{	ext{cell}} - \frac{2.303RT}{nF} \log \frac{[	ext{Zn}^{2+}]}{[	ext{Cu}^{2+}]}$
For the first case ($E_1$), $[	ext{Zn}^{2+}] = 0.01 	ext{ M}$ and $[	ext{Cu}^{2+}] = 1.0 	ext{ M}$, $n=2$.
$E_1 = E^\circ_{	ext{cell}} - \frac{0.059}{2} \log \left(\frac{0.01}{1.0}ight) = E^\circ_{	ext{cell}} - \frac{0.059}{2} (-2) = E^\circ_{	ext{cell}} + 0.059 	ext{ V}$
For the second case ($E_2$), $[	ext{Zn}^{2+}] = 1.0 	ext{ M}$ and $[	ext{Cu}^{2+}] = 0.01 	ext{ M}$.
$E_2 = E^\circ_{	ext{cell}} - \frac{0.059}{2} \log \left(\frac{1.0}{0.01}ight) = E^\circ_{	ext{cell}} - \frac{0.059}{2} (2) = E^\circ_{	ext{cell}} - 0.059 	ext{ V}$
Comparing the two equations, it is clear that $E_1 > E_2$.

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