Question

Calculate the cell potential of the following cell at ${25}^{\circ }\text{C}$.
$Ag\left(s\right)|A{g}^{+}(aq,0.001\text{M})||A{g}^{+}(aq,0.5\text{M})|Ag\left(s\right)$

• A. $0.03\text{V}$
• B. $0.159\text{V}$
• C. $1.09\text{V}$
• D. $0.52\text{V}$

Answer 1

The correct option is B $0.159\text{V}$

From cell representation,
$\left[A{g}^{+}\right]=0.001\text{M}\Rightarrow$ Anode
$\left[A{g}^{+}\right]=0.5\text{M}\Rightarrow$ Cathode

Cell reaction will be,
$A{g}^{+}(aq,0.5\text{M})\to A{g}^{+}(aq,0.001\text{M})$

By nernst equation,

$E_{\text{cell}}=E_{cell}^{\circ }-\frac{0.0591}{1}\log \frac{\left[0.001\right]}{\left[0.5\right]}$

$E_{\text{cell}}=E_{cell}^{\circ }+\frac{0.0591}{1}\log \frac{\left[0.5\right]}{\left[0.001\right]}$

For concentration cells, cathode and anode consist of same metal and its solution​.
Thus,
$E_{cell}^{\circ }=0$

$\Rightarrow E_{\text{cell}}=\frac{0.0591}{1}\log \frac{\left[0.5\right]}{\left[0.001\right]}$

$\Rightarrow E_{\text{cell}}=\frac{0.0591}{1}\log 5\times {10}^2$

$\Rightarrow E_{\text{cell}}=\frac{0.0591}{1}\times 0.69+\frac{0.0591}{1}\times 2$

$\Rightarrow E_{\text{cell}}\approx 0.159\text{V}$