Question

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

• A. $-0.0074V$
• B. $-1.00738V$
• C. $-0.0038V$
• D. None

Answer 1

The correct option is A $-0.0074V$

$Q=10\times 4825$
$=48250C$
no. of mole of $A{g}^{+}=\frac{48250}{96500}=0.5$
$Ag+\frac{1}{2}C{u}^{++}\to A{g}^{+}+\frac{1}{2}Cu$
2.00 2.00
2-0.25 2+0.50
$E_{cell}=E_{cell}^0-\frac{0.0591}{1}log\frac{\left[A{g}^{+}\right]}{[C{u}^{++}{]}^{1/2}}$
$E_1=E_{cell}^0-\frac{0.0591}{1}log\frac{2.00}{(2.00{)}^{1/2}}$
$E_2=E_{cell}^0-\frac{0.0591}{1}log\frac{2.50}{(1.75{)}^{1/2}}$
$\Delta E=E_2-E_1$

$=\frac{0.0591}{1}[log\sqrt{2}-log\frac{2.50}{\sqrt{1.75}}]$

$=\frac{0.0591}{1}[log1.41-log1.88]$
$=\frac{0.0591}{1}[0.1492-0.2742]$

$=-\frac{0.0591}{1}\times 0.125$

$=-0.0074V$