Question

if the concentration of Zn+2 ions is halved, then calculate change in potential of the cell at 25°C. Zn→ Zn+2 + 2e-

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Answer 1

8.9 mV would be the change in the potential of the cell.

Explanation:

Given that,

The congregation of Zn+2 ions is halved

To find,

Change in cell's potential at the temperature of 25°C

Procedure:

Assuming the primary concentration Zn+2 be 'a' and the new concentration as a/2 with 'n' to be the no. of moved electrons.

Using,

E = $\frac{0.0059}{n} log (\frac{Zn_{new} ^{+2} }{Zn}) - \frac{0.0059}{n} log (\frac{Zn_{old} ^{+2} }{Zn})$

= $\frac{0.0059}{n} log (\frac{x/2}{Zn}) - \frac{0.0059}{n} log (\frac{x }{Zn})$

= $\frac{0.0059}{n} (log \frac{x/2}{Zn} - log (\frac{x }{Zn})$

= $\frac{0.0059}{n} log (\frac{1}{2})$

By putting n = 2

= $\frac{0.0059}{2} (-0.301)$

= 0.0088 Volt

= 0.0088 Volt * 10^-3

= 8.88 mV

∵ E = 8.88 mV 0r 8.9 mV

Thus, 8.9 mV is the change in cell's potential.

Learn more: Potential of the cell

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