Question

A zinc electrode is dipped in a 0.1 M solution at${25}^{0}C$. Assuming that salt is dissociated to 20% at this dilution. Calculate the electrode potential. $E{{}^0}_{Z{n}^{2+}/Zn}=-0.76V$.

• A. +0.81 V
• B. -0.81 V
• C. +0.4 V
• D. -0.4 V

Answer 1

Answer: (B)

-0.81 V

The expression for the electrode potential is $E_{el}=E_{el}^0+\frac{0.0592}{2}log\left[{Zn}^{2+}\right].$
Substitute values in the above expression.
$E_{el}=-0.76V+\frac{0.0592}{2}log[0.1\times 0.2]=-0.81V.$