Question

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

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

Answer 1

Answer: (B)

$E_{\text{cell}} =E^{\circ}_{\text{cell}} - \frac{0.059}{n}\log Q$
At equilibrium
$ E^{\circ}_{\text{cell}} = \frac{0.059}{2} \log10^{16} $
$ = 0.059 \times8 $
$ = 0.472\, V $