The EMF of a Daniel cell at $298 ext{ K}$ is $E_1$: $Zn|ZnSO_4(0.01 ext{ M}) || CuSO_4(1.0 ext{ M})|Cu$. Wh

Question

The EMF of a Daniel cell at $298	ext{ K}$ is $E_1$: $Zn|ZnSO_4(0.01	ext{ M}) || CuSO_4(1.0	ext{ M})|Cu$. When the concentration of $ZnSO_4$ is $1.0	ext{ M}$ and that of $CuSO_4$ is $0.01	ext{ M}$, the EMF is changed to $E_2$. The correct relationship between $E_1$ and $E_2$ is:

Accepted Answer

The cell reaction for a Daniell cell is:
$Zn(s) + Cu^{2+}(aq) ightarrow Zn^{2+}(aq) + Cu(s)$

According to the Nernst equation at $298	ext{ K}$:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$

For the first case ($E_1$):
$[Zn^{2+}] = 0.01	ext{ M} = 10^{-2}	ext{ M}$ and $[Cu^{2+}] = 1.0	ext{ M}$
$E_1 = E^{\circ}_{cell} - \frac{0.0591}{2} \log \left(\frac{10^{-2}}{1}ight) = E^{\circ}_{cell} - \frac{0.0591}{2} (-2) = E^{\circ}_{cell} + 0.0591	ext{ V}$

For the second case ($E_2$):
$[Zn^{2+}] = 1.0	ext{ M}$ and $[Cu^{2+}] = 0.01	ext{ M} = 10^{-2}	ext{ M}$
$E_2 = E^{\circ}_{cell} - \frac{0.0591}{2} \log \left(\frac{1}{10^{-2}}ight) = E^{\circ}_{cell} - \frac{0.0591}{2} (2) = E^{\circ}_{cell} - 0.0591	ext{ V}$

Comparing $E_1$ and $E_2$, we get $E_1 > E_2$.

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