Chemistry - Electrochemistry Question with Solution | TestHub

The reaction is ${\mathrm{Cu}}^{2+}(\mathrm{aq})+{\mathrm{Sn}}^{2+}(\mathrm{aq})\to {\mathrm{Sn}}^{4+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$.

The standard cell potential, $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$.

Here, Cu²⁺/Cu is reduced (cathode) and Sn²⁺/Sn⁴⁺ is oxidized (anode).

$E_{cell}^{\circ }=E_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\circ }-E_{{\mathrm{Sn}}^{4+}/{\mathrm{Sn}}^{2+}}^{\circ }=0.34\mathrm{V}-0.15\mathrm{V}=0.19\mathrm{V}$.

The number of electrons transferred, $n=2$.

Using the Nernst equation at equilibrium: $E_{cell}^{\circ }=\frac{0.0591}{n}{\log }_{10}{K}_C$.

$0.19=\frac{0.0591}{2}{\log }_{10}{K}_C$.

${\log }_{10}{K}_C=\frac{0.19\times 2}{0.0591}=\frac{0.38}{0.0591}\approx 6.43$.

Since ${\log }_{10}{K}_C=a$, then $a\approx 6.43$.

Rounding to the nearest integer, $a=6$.