For a reaction,A + B^{2+} rightarrow B + A^{2 | vedclass

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Question

(B) The relationship between standard cell potential $(E^{\circ})$ and equilibrium constant $(K_{eq})$ is given by the Nernst equation at $25^{\circ}C

Vedclass · Accepted Answer

$10^{10}$

Vedclass · Answer

(B) The relationship between standard cell potential $(E^{\circ})$ and equilibrium constant $(K_{eq})$ is given by the Nernst equation at $25^{\circ}C$ $(298 \ K)$: $E^{\circ} = \frac{0.0591}{n} \log_{10} K_{eq}$ Here,the number of electrons transferred $(n)$ is $2$. Substituting the values: $0.2955 = \frac{0.0591}{2} \log_{10} K_{eq}$ $\log_{10} K_{eq} = \frac{0.2955 	imes 2}{0.0591} = 10$ $K_{eq} = 10^{10}$

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

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