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

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(A) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)$.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,$E_1 > E_2$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Easy

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 $E_2$.

A
$E_1 > E_2$
B
$E_1 < E_2$
C
$E_1 = E_2$
D
$E_2 = 0 \neq E_1$

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Nernst equation and ECS

Consider the following cell reaction:
$2 Fe_{(s)} + O_{2_{(g)}} + 4 H_{(aq)}^{+} \rightarrow 2 Fe_{(aq)}^{2+} + 2 H_2O_{(l)} \quad E^{\circ} = 1.67 \ V$. At $[Fe^{2+}] = 10^{-3} \ M, P(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25^{\circ} C$ is (in $V$)

DifficultIIT 2011

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For a $Mg|Mg^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag$ cell,the correct Nernst Equation is $:$

MediumJEE MAIN 2025

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The hydrogen electrode is dipped in a solution of $pH = 3$ at $25\,^oC$. The potential of the cell would be ............. $V$ (the value of $2.303\,RT/F$ is $0.059\,V$).

Medium

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The cell potential for the following reaction is $0.03305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

Medium

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The standard $e.m.f.$ of a cell,involving one electron change is found to be $0.591 \ V$ at $25^{\circ} C$. The equilibrium constant of the reaction is :
$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$

Medium

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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 $E_2$.

Similar Questions

Consider the following cell reaction:$2 Fe_{(s)} + O_{2_{(g)}} + 4 H_{(aq)}^{+} \rightarrow 2 Fe_{(aq)}^{2+} + 2 H_2O_{(l)} \quad E^{\circ} = 1.67 \ V$. At $[Fe^{2+}] = 10^{-3} \ M, P(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25^{\circ} C$ is (in $V$)

For a $Mg|Mg^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag$ cell,the correct Nernst Equation is $:$

The hydrogen electrode is dipped in a solution of $pH = 3$ at $25\,^oC$. The potential of the cell would be ............. $V$ (the value of $2.303\,RT/F$ is $0.059\,V$).

The cell potential for the following reaction is $0.03305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

The standard $e.m.f.$ of a cell,involving one electron change is found to be $0.591 \ V$ at $25^{\circ} C$. The equilibrium constant of the reaction is :$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$

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Consider the following cell reaction:
$2 Fe_{(s)} + O_{2_{(g)}} + 4 H_{(aq)}^{+} \rightarrow 2 Fe_{(aq)}^{2+} + 2 H_2O_{(l)} \quad E^{\circ} = 1.67 \ V$. At $[Fe^{2+}] = 10^{-3} \ M, P(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25^{\circ} C$ is (in $V$)

The standard $e.m.f.$ of a cell,involving one electron change is found to be $0.591 \ V$ at $25^{\circ} C$. The equilibrium constant of the reaction is :
$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$