Question

Consider the following electrolytic cells
(i) $M _{( s )}\left| M _{( aq )}^{2+}, 0 \cdot 1\, M \| X _{( aq )}^{2+}, 0 \cdot 01 \,M \right| X _{( s )}$
(ii) $M _{( s )}\left| M _{( aq )}^{2+}, 0 \cdot 1 \,M \| X _{( aq )}^{2+}, 0 \cdot 1 \,M \right| X _{( s )}$
(iii) $M _{( s )}\left| M _{( aq )}^{2+}, 0 \cdot 01\, M \| X _{( aq )}^{2+}, 0 \cdot 1\, M \right| X _{( s )}$
The cell $E M F$ of the above cells are $E_{1}, E_{2}$ and $E_{3}$ respectively. Which one of the following is true?

• A. $E_1 > E_2 > E_3$
• B. $E_2 > E_3 > E_1$
• C. $ E_3 >E_1 > E_2$
• D. $E_1 > E_3 > E_2 $
• E. $E_3 > E_2 > E_1 $

Answer 1

Answer: (E)

For the given cell, cell reaction is

$M+X^{2+} \longrightarrow M^{2+}+X$

Nernst equation is

$E =E^{\circ}-\frac{0.059}{2} \log \frac{\left[M^{2+}\right]}{\left[X^{2+}\right]} $

$=E^{\circ}+\frac{0.059}{2} \log \frac{\left[X^{2+}\right]}{\left[M^{2+}\right]}$

For cell (i),

$ E_{1} =E^{\circ}+\frac{0.059}{2} \log \frac{(0.01)}{(0.1)} $

$=E^{\circ}+\frac{0.059}{2}(-1) $

$=E^{\circ}-\frac{0.059}{2} $

$ E_{2} =E^{\circ}+\frac{0.059}{2} \log \frac{(0.1)}{(0.1)} $

$=E^{\circ} (\because \log 1=0)$

$ E_{3} =E^{\circ}+\frac{0.059}{2} \log \frac{(0.1)}{(0.01)}$

$=E^{\circ}+\frac{0.059}{2} \log 10$

$=E^{\circ}+\frac{0.059}{2} $

Thus, the order of $E_{1}, E_{2}$ and $E_{3}$ is $E_{3} > E_{2} > E_{1}$