Question

A zinc electrode is placed in a 0.1M solution at ${25}^{0}C$. assuming that the salt is 20% dissociated at this dilutions calculate the electrode reduction potential.$E^0\left({Zn}^{2+}|Zn\right)=-0.76V$

Answer 1

${Zn}^{2+}+{2e}^{-}\to Zn$

$\alpha =\frac{20}{100}=0.2,C=0.1M$

$\therefore$ concentration of ${Zn}^{2+}=C\alpha$

$=0.1\times 0.2$

$=0.02M$

$\therefore$ From nernst equation,

$E_{cell}=E^0-\frac{0.059}{n}log\left[\frac{1}{{Zn}^{2+}}\right]$

$E_{cell}=-0.76-\frac{0.059}{2}log\left[\frac{1}{0.02}\right]$

$E_{cell}=-0.76-\frac{0.059\times 1.698}{2}$

$E_{cell}=-0.76-0.05$

$E_{cell}=-0.81V$