Given the equilibrium constant: $\mathrm{K}_{\mathrm{C}}$ of the reaction: $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) ightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ is $10 imes 10^{15}$ calculate the $E_{ ext {cell }}^{0}$ of this reaction at $298 \mathrm{~K}$…

Question

Given the equilibrium constant: $\mathrm{K}_{\mathrm{C}}$ of the reaction: $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) ightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ is $10 	imes 10^{15}$ calculate the $E_{	ext {cell }}^{0}$ of this reaction at $298 \mathrm{~K}$ $\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}}ight.$ at $\left.298 \mathrm{~K}=0.059 \mathrm{~V}ight]$

Aicharya · Accepted Answer

$E_{	ext {cell }}^{0}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{C}}$ or $\mathrm{E}_{	ext {cell }}^{0}=\frac{0.059 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}_{\mathrm{C}}$ $ =\frac{0.059 \mathrm{~V}}{2} \log 10^{16}=0.4736 \mathrm{~V} $

Given the equilibrium constant: $\mathrm{K}_{\mathrm{C}}$ of the reaction: $\mathrm{Cu}(\mathrm{s})+2…

Solution