(A) The relationship between the standard cell potential $(E^{\circ}_{cell})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation

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(A) The relationship between the standard cell potential $(E^{\circ}_{cell})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation at equilibrium: $E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$ Given: $E^{\circ}_{cell} = 0.591 \ V$ $n = 1$ (one-electron change) Substituting the values: $0.591 = \frac{0.0591}{1} \log K_{eq}$ $\log K_{eq} = \frac{0.591}{0.0591} = 10$ $K_{eq} = 10^{10} = 1.0 \times 10^{10}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

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

A
$1.0 \times 10^{10}$
B
$1.0 \times 10^{5}$
C
$1.0 \times 10^{1}$
D
$1.0 \times 10^{30}$

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What is the reduction electrode potential $E$ of a $0.1 \, M$ solution of $M^{+}$ ions,given that the standard reduction potential $E^o_{RP} = -2.36 \, V$?

Medium

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For the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

Easy

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What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

MediumAP EAMCET 2025

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Calculate the cell potential for $Ni_{(s)} | Ni^{2+}(0.036 \ M) || Co^{2+}(0.018 \ M) | Co_{(s)}$ at $298 \ K$.

Easy

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An oxidation-reduction reaction in which $3$ electrons are transferred has a $\Delta G^{\circ}$ of $17.37 \ kJ \ mol^{-1}$ at $25^{\circ} C$. The value of $E_{\text{cell}}^{\circ}$ (in $V$) is........ $\times 10^{-2} \ V$.
$(1 \ F = 96,500 \ C \ mol^{-1})$

MediumJEE MAIN 2020

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

Similar Questions

What is the reduction electrode potential $E$ of a $0.1 \, M$ solution of $M^{+}$ ions,given that the standard reduction potential $E^o_{RP} = -2.36 \, V$?

For the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

Calculate the cell potential for $Ni_{(s)} | Ni^{2+}(0.036 \ M) || Co^{2+}(0.018 \ M) | Co_{(s)}$ at $298 \ K$.

An oxidation-reduction reaction in which $3$ electrons are transferred has a $\Delta G^{\circ}$ of $17.37 \ kJ \ mol^{-1}$ at $25^{\circ} C$. The value of $E_{\text{cell}}^{\circ}$ (in $V$) is........ $\times 10^{-2} \ V$.$(1 \ F = 96,500 \ C \ mol^{-1})$

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What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

An oxidation-reduction reaction in which $3$ electrons are transferred has a $\Delta G^{\circ}$ of $17.37 \ kJ \ mol^{-1}$ at $25^{\circ} C$. The value of $E_{\text{cell}}^{\circ}$ (in $V$) is........ $\times 10^{-2} \ V$.
$(1 \ F = 96,500 \ C \ mol^{-1})$