Question

If the concentration of Zn2+ ions is halved, then
calculate change in potential of the cell at 25°C
Zn gives Zn^2+ + 2e-​

Answer 1

Given:

Concentration of Zn+2 ions is halved.

To find:

Change in potential of the cell.

Calculation:

Let initial concentration of Zn+2 be x and the new concentration be x/2. Let change in potential be E.

n refers to number of electrons transferred.

$\therefore \: E = \dfrac{0.059}{n} log( \dfrac{{Zn_{new}}^{ + 2} }{Zn} ) - \dfrac{0.059}{n} log( \dfrac{{Zn_{old}}^{ + 2} }{Zn} )$

$= > E = \dfrac{0.059}{n} log( \dfrac{ \frac{x}{2} }{Zn} ) - \dfrac{0.059}{n} log( \dfrac{x }{Zn} )$

$= > E = \dfrac{0.059}{n} \bigg \{log( \dfrac{ \frac{x}{2} }{Zn} ) - log( \dfrac{x }{Zn} ) \bigg \}$

$= > E = \dfrac{0.059}{n} \bigg \{log (\frac{1}{2} )\bigg \}$

$= > E = \dfrac{0.059}{2} \bigg \{log (\frac{1}{2} )\bigg \} \: \: \: .....(n = 2)$

$= > E = \dfrac{0.059}{2} \bigg \{ - 0.301\bigg \}$

$= > E = 0.0088 \: volt$

$= > E = 8.8 \times {10}^{ - 3} \: volt$

$= > E = 8.8 \: mV$

So , final answer is :

$\boxed{ \bold{ \red{ \large{ E = 8.8 \: mV}}}}$