Question

Given the equilibrium constant :
$KC$ of the reaction :
$\mathrm{Cu}(\mathrm{s})+\mathrm{Ag}{}^{+}(\mathrm{aq})\to \mathrm{Cu}{}^{2+}(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})$ is $10\times 10^{15}$, calculate the $E_{cell}^0$ of this reaction at $298K$ $\left[2.303\frac{RT}{F}\text{at}298K=0.059V\right]$

• A. 0.04736 V
• B. 0.4736 V
• C. 0.4736 mV
• D. 0.04736 mV

Answer 1

Answer: (B)

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.059}{n}\log Q$
At equilibrium
$E_{\text{cell}}^{\circ }=\frac{0.059}{2}\log 10^{16}$
$=0.059\times 8$
$=0.472V$