(D) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.For this reaction,$n = 2$.The Nernst equation is: $E_{cell} = E^{\circ

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(D) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.For this reaction,$n = 2$.The Nernst equation is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$,where $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}$.At complete discharge,$E_{cell} = 0$.Therefore,$0 = 1.10 - \frac{0.0591}{2} \log Q$.$\log Q = \frac{1.10 \times 2}{0.0591} \approx 37.225$.$Q = 10^{37.225} \approx 10^{37.3}$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

A
$9.65 \times 10^{4}$
B
$Antilog(24.08)$
C
$37.3$
D
$10^{37.3}$

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Nernst equation and ECS

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For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

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Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)

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$E_{cell}^{0}$ of the reaction $Mg_{(s)} + 2 Ag_{(0.0001 \ M)}^{+} \rightleftharpoons Mg_{(0.01 \ M)}^{2+} + 2 Ag_{(s)}$ is $3.17 \ V$. The $E_{cell}$ of the reaction and its cell notation respectively are :

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