The following reaction takes place at 298 , K | vedclass

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Question

(A) For the reaction,$A^{2+}_{(aq)} + B_{(s)} \longrightarrow B^{2+}_{(aq)} + A_{(s)}$According to the Nernst equation at $298 \, K$:$E_{cell} = E^{\c

Vedclass · Accepted Answer

$4 	imes 10^{36}$

Vedclass · Answer

(A) For the reaction,$A^{2+}_{(aq)} + B_{(s)} \longrightarrow B^{2+}_{(aq)} + A_{(s)}$According to the Nernst equation at $298 \, K$:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[B^{2+}]}{[A^{2+}]}$Given $n = 2$,$E_{cell} = 1.091 \, V$,$[A^{2+}] = 4 	imes 10^{-3} \, M$,and $[B^{2+}] = 2 	imes 10^{-3} \, M$.$1.091 = E^{\circ}_{cell} - \frac{0.0591}{2} \log \left( \frac{2 	imes 10^{-3}}{4 	imes 10^{-3}} ight)$$1.091 = E^{\circ}_{cell} - 0.02955 \log(0.5)$$1.091 = E^{\circ}_{cell} - 0.02955 	imes (-0.301)$$1.091 = E^{\circ}_{cell} + 0.00889$$E^{\circ}_{cell} = 1.091 - 0.00889 = 1.08211 \, V$Now,using the relation $E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$:$\log K_{eq} = \frac{E^{\circ}_{cell} 	imes n}{0.0591} = \frac{1.08211 	imes 2}{0.0591} \approx 36.61$$K_{eq} = 10^{36.61} \approx 4.07 	imes 10^{36} \approx 4 	imes 10^{36}$

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