(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

Difficult

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

For a reaction,$A + B^{2+} \to B + A^{2+}; E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25 \ ^oC$ is:

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When a copper plate is kept in a $0.1 \ M$ solution of $CuSO_4$ at $298 \ K$ temperature and if $70 \ \%$ dissociation has occurred,then calculate the potential of the copper electrode.

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For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

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$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.

MediumAP EAMCET 2021

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Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

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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 $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

Similar Questions

For a reaction,$A + B^{2+} \to B + A^{2+}; E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25 \ ^oC$ is:

When a copper plate is kept in a $0.1 \ M$ solution of $CuSO_4$ at $298 \ K$ temperature and if $70 \ \%$ dissociation has occurred,then calculate the potential of the copper electrode.

For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.

Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

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