Question

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

• A. $0.159V$
• B. $1.09V$
• C. $0.52V$
• D. $0.03V$

Answer 1

The correct option is A $0.159V$

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

Cell reaction will be,
$A{g}^{+}(aq,0.5M)\to A{g}^{+}(aq,0.2M)$

By nernst equation,

$E_{cell}=E_{cell}^0-\frac{0.0591}{1}log\frac{\left[0.001\right]}{\left[0.5\right]}$

$E_{cell}=E_{cell}^0+\frac{0.0591}{1}log\frac{\left[0.5\right]}{\left[0.001\right]}$

For concentration cells, cathode and anode consist of same metal and its solution​.
Thus,
$E^0=0$

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

$E_{cell}=\frac{0.0591}{1}log5\times {10}^2$

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

$E_{cell}\approx 0.159V$