At 25,^{circ}C,calculate the equilibrium cons | vedclass

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(C) The cell reaction is $X_{(s)} + Y_{(aq)}^{2+} \rightleftharpoons Y_{(s)} + X_{(aq)}^{2+}$.The standard cell potential $E_{cell}^{o}$ is given by $

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$10^{20}$

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(C) The cell reaction is $X_{(s)} + Y_{(aq)}^{2+} \rightleftharpoons Y_{(s)} + X_{(aq)}^{2+}$.The standard cell potential $E_{cell}^{o}$ is given by $E_{cathode}^{o} - E_{anode}^{o}$.Here,$Y^{2+}/Y$ is the cathode and $X^{2+}/X$ is the anode.$E_{cell}^{o} = E_{Y^{2+}/Y}^{o} - E_{X^{2+}/X}^{o} = -0.76\,V - (-1.36\,V) = 0.60\,V$.The relationship between equilibrium constant $K_{eq}$ and $E_{cell}^{o}$ is $E_{cell}^{o} = \frac{0.059}{n} \log K_{eq}$.Using the given value $\frac{2.303\,RT}{F} = 0.06$,the formula becomes $E_{cell}^{o} = \frac{0.06}{n} \log K_{eq}$.Here,$n = 2$ (number of electrons transferred).$0.60 = \frac{0.06}{2} \log K_{eq} \implies 0.60 = 0.03 \log K_{eq}$.$\log K_{eq} = \frac{0.60}{0.03} = 20$.$K_{eq} = 10^{20}$.

At $25\,^{\circ}C$,calculate the equilibrium constant for the cell reaction,
$X_{(s)} + Y_{(aq)}^{2+} \rightleftharpoons Y_{(s)} + X_{(aq)}^{2+}$
Given:
$E_{X^{2+}/X}^{o} = -1.36\,V$;
$E_{Y^{2+}/Y}^{o} = -0.76\,V$; $\frac{2.303\,RT}{F} = 0.06$

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At $25\,^{\circ}C$,calculate the equilibrium constant for the cell reaction,$X_{(s)} + Y_{(aq)}^{2+} \rightleftharpoons Y_{(s)} + X_{(aq)}^{2+}$Given:$E_{X^{2+}/X}^{o} = -1.36\,V$;$E_{Y^{2+}/Y}^{o} = -0.76\,V$; $\frac{2.303\,RT}{F} = 0.06$