The emf of a Daniell cell at 298 , K is E_1:Z | vedclass

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(C) The cell reaction for the $Daniell$ cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.Using the $Nernst$ equation:$E_{cell} = E_{cell

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$E_1 > E_2$

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(C) The cell reaction for the $Daniell$ cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.Using the $Nernst$ equation:$E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$For $E_1$:$E_1 = E^o - \frac{0.0591}{2} \log \frac{0.01}{1.0} = E^o - \frac{0.0591}{2} \log(10^{-2}) = E^o + 0.0591 \, V$.For $E_2$:$E_2 = E^o - \frac{0.0591}{2} \log \frac{1.0}{0.01} = E^o - \frac{0.0591}{2} \log(10^2) = E^o - 0.0591 \, V$.Comparing the two values,$E^o + 0.0591 > E^o - 0.0591$,therefore $E_1 > E_2$.

The $emf$ 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 $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?

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The $emf$ 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 $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?