Question

Calculate the cell potential ($E_{cell}$) of the following cell at $298K$.
$Ag\left(s\right)|AgN{O}_3\left(0.01M\right)||AgN{O}_3\left(1.0M\right)|Ag\left(s\right)$

• A. $0.1182V$
• B. $0.591V$
• C. $0.2364V$
• D. $0V$

Answer 1

The correct option is A $0.1182V$

As we know that,
$E_{cell}={E^{\circ }}_{cell}-\frac{0.0591}{n}\log \frac{C_1}{C_2}$
Given:- ${Ag}_{\left(s\right)}|{AgN{O}_3}_{\left(0.001M\right)}||{AgN{O}_3}_{\left(1.0M\right)}|{Ag}_{\left(s\right)}$
Here,
$C_1={10}^{-2}M$
$C_2=1M$
$n=1$
When both electrode are of same metal,
${E^{\circ }}_{cell}=0$
$\therefore E_{cell}=0-\frac{0.0591}{1}\log \frac{{10}^{-2}}{1}$
$\Rightarrow E_{cell}=-0.0591\times (-2)\log 10$
$\Rightarrow E_{cell}=0.1182V$
Hence the cell potential of given cell is $0.1182V$.