Question

Consider the following electrolytic cells
(i) $M_{(s)}\left|M_{(aq)}^{2+},0\cdot 1M\Vert X_{(aq)}^{2+},0\cdot 01M\right|X_{(s)}$
(ii) $M_{(s)}\left|M_{(aq)}^{2+},0\cdot 1M\Vert X_{(aq)}^{2+},0\cdot 1M\right|X_{(s)}$
(iii) $M_{(s)}\left|M_{(aq)}^{2+},0\cdot 01M\Vert X_{(aq)}^{2+},0\cdot 1M\right|X_{(s)}$
The cell $EMF$ 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

$M+X^{2+}⟶M^{2+}+X$

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 $E_3>E_2>E_1$