For a Daniel cell Zn | ZnSO_{4(0.01 M)} || C | vedclass

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(A) The cell reaction is: $Zn(s) + Cu^{2+}(aq) 	o Zn^{2+}(aq) + Cu(s)$.Using the Nernst equation: $E = E^0 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[

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$E_1 > E_2$

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(A) The cell reaction is: $Zn(s) + Cu^{2+}(aq) 	o Zn^{2+}(aq) + Cu(s)$.Using the Nernst equation: $E = E^0 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.For $E_1$: $[Zn^{2+}] = 0.01 \ M$ and $[Cu^{2+}] = 1 \ M$,so $E_1 = E^0 - \frac{0.0592}{2} \log(0.01) = E^0 - \frac{0.0592}{2} (-2) = E^0 + 0.0592$.For $E_2$: $[Zn^{2+}] = 1 \ M$ and $[Cu^{2+}] = 0.01 \ M$,so $E_2 = E^0 - \frac{0.0592}{2} \log(100) = E^0 - \frac{0.0592}{2} (2) = E^0 - 0.0592$.Comparing the two,$E_1 > E_2$.

For a $Daniel$ cell $Zn | ZnSO_{4(0.01 \ M)} || CuSO_{4(1 \ M)} | Cu$ at $298 \ K$,the cell potential is $E_1$. When the concentrations of $ZnSO_4$ and $CuSO_4$ are changed to $1 \ M$ and $0.01 \ M$ respectively,the cell potential becomes $E_2$. Determine the relationship between $E_1$ and $E_2$.

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