Question

A cell $\text{Cu }|\text{ C}\text{u}^{\text{++}}\ ||\text{ A}\text{g}^{+}\ |$Ag initially contains 2M Ag+ and 2M Cu++ ions in 1L electrolyte. The change in cell potential after the passage of 10 amp current for 4825 sec is :

• A. – 0.00738 V
• B. – 1.00738 V
• C. – 0.0038 V
• D. None

Answer 1

Answer: (A)

– 0.00738 V

Answer 2

Q = 10 × 4825 = 48250 C

no. of faraday = $\frac{48250}{96500} = 0.5$

Ag+$\frac{1}{2}$Cu++ $\longrightarrow$Ag+

+ $\frac{1}{2}$Cu

2.0

2.00

20.250

$E_{cel}l = {E^{0}}_{cell}\frac{0.0591}{1}\log\frac{\left\lbrack Ag^{+} \right\rbrack}{\left\lbrack Cu^{+ +} \right\rbrack^{1/2}}$

$E_{1}\text{ = }{\text{E}^{0}}_{\text{Cell}}\text{ - }\frac{\text{0.0591}}{1}\text{log log}\frac{\text{2.00}}{\left( \text{2.00} \right)^{1/2}}$

$E_{2} = {E^{0}}_{cell} - \frac{0.0591}{1}\log\frac{2.50}{(1.75)^{1/2}}$

$\Delta\text{ E = }\text{E}_{2}\text{ - }\text{E}_{1}\text{ = }\frac{\text{0.0591}}{1}\left\lbrack \text{log}\sqrt{2} - \log\frac{2.50}{\sqrt{1.75}} \right\rbrack\text{ = }\frac{\text{0.0591}}{1}\ \lbrack\text{log 1.41 - log 1.88}\rbrack\ = \frac{\text{0.0591}}{1}\text{ }\lbrack\text{0.1492 - 0.2742}\rbrack\text{ = -}\frac{\text{0.0591}}{1}\text{ }\text{×}\text{ 0.125 = - 0.00738 V.}$