(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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MediumJEE MAIN 2021

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Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

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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

$A$ hydrogen gas electrode is made by dipping platinum wire in a solution of $HCl$ of $pH = 10$ and by passing hydrogen gas around the platinum wire at $1 \ atm$ pressure. The oxidation potential of the electrode would be $.......... \ V$.

Calculate the cell potential for $Ag_{(s)}|Ag^{+} \, (0.01 \ M)||Ag^{+} \, (0.1 \ M)|Ag_{(s)}$ at $298 \ K$. (in $V$)

Which of the following is the correct representation of the Nernst equation for the reduction reaction $M^{n+} + ne^- \rightarrow M$?

$Emf$ of the following cell at $298 \,K$ in $V$ is $x \times 10^{-2}$. $Zn | Zn^{2+}(0.1 \,M) || Ag^{+}(0.01 \,M) | Ag$. The value of $x$ is .... . (Rounded off to the nearest integer) [Given: $E^{0}_{Zn^{2+}/Zn} = -0.76 \,V$; $E^{0}_{Ag^{+}/Ag} = +0.80 \,V$; $\frac{2.303 RT}{F} = 0.059$]

Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

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