Question

A zinc rod is placed in $0.1M$ solution of zinc sulphate$\left(ZnS{O}_4\right)$ at 25°C. Assuming that the salt is dissociated to the extent of 95 percent at this dilution, calculate the potential of the electrode at this temperature:
$(E_{Z{n}^{2+},Zn}^{\circ }=-0.76V)$

• A. +0.79 V
• B. -0.79 V
• C. +0.22 V
• D. -0.22 V

Answer 1

The correct option is B -0.79 V

Concentration of zinc sulphate solution = 0.1 M; dissociation = 95%

$\therefore \left[Z{n}^{2+}\right]=0.1\times 95/100=0.095M$

The electrode reaction in this case is

$Z{n}^{2+}\left(aq\right)+2e^{-}\rightleftharpoons Zn\left(s\right)$

According to the Nernst equation, the potential of the electrode is given by

$E_{el}=E_{el}^{\circ }-\frac{RT}{nF}\ln \frac{1}{\left[Z{n}^{2+}\right]}=E_{el}^{\circ }+\frac{RT}{nF}\ln \left[Z{n}^{2+}\right]$

$\Rightarrow E_{el}^{\circ }+\frac{0.0591}{n}\log \left[Z{n}^{2+}\right]$

Substituting the various values in the above equation, we have

$E_{el}=-0.76+\frac{0.0591}{2}\log 0.095=-0.79V$