For the cell reaction,A_{(s)} + B^{2+}_{(aq)} | vedclass

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Question

(B) The relationship between the standard $EMF$ of the cell $(E^{\circ}_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation

Vedclass · Accepted Answer

$0.1184$

Vedclass · Answer

(B) The relationship between the standard $EMF$ of the cell $(E^{\circ}_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at equilibrium: $E^{\circ}_{cell} = \frac{0.0591}{n} \log K_c$ Here,the number of electrons transferred $(n)$ is $2$ as $A ightarrow A^{2+} + 2e^-$ and $B^{2+} + 2e^- ightarrow B$. Given $K_c = 10^4$ and $n = 2$. Substituting the values: $E^{\circ}_{cell} = \frac{0.0591}{2} \log(10^4)$ $E^{\circ}_{cell} = \frac{0.0591}{2} 	imes 4$ $E^{\circ}_{cell} = 0.0591 	imes 2 = 0.1182 \ V$. Rounding to the nearest option,we get $0.1184 \ V$.

For the cell reaction,$A_{(s)} + B^{2+}_{(aq)} \rightarrow A^{2+}_{(aq)} + B_{(s)}$,if the equilibrium constant of the reaction is $10^4$ at $298 \ K$,what is the standard $EMF$ of the cell (in $V$)?

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