At 298 K the equilibrium constant for the re | vedclass

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(A) The relationship between the standard cell potential $E_{cell}^{\ominus}$ and the equilibrium constant $K_c$ is given by the Nernst equation at eq

Vedclass · Accepted Answer

$0.45$

Vedclass · Answer

(A) The relationship between the standard cell potential $E_{cell}^{\ominus}$ and the equilibrium constant $K_c$ is given by the Nernst equation at equilibrium: $E_{cell}^{\ominus} = \frac{2.303 RT}{nF} \log K_c$ Here,$n = 2$ (number of electrons transferred in the balanced equation). Given: $\frac{2.303 RT}{F} = 0.06 \ V$,$K_c = 10^{15}$,and $n = 2$. Substituting these values: $E_{cell}^{\ominus} = \frac{0.06}{2} \log(10^{15})$ $E_{cell}^{\ominus} = 0.03 	imes 15$ $E_{cell}^{\ominus} = 0.45 \ V$

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

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