Question

Calculate the concentration of electrolyte ($C_1$) at anode for the given cell:

$Ag\left(s\right)|AgN{O}_3(aq,C_{1}M)||AgN{O}_3(aq,C_2=0.2M)|Ag\left(s\right)$

The emf of the cell is $0.8V$

Take
$Antilog\left(13.5\right)=3.16\times {10}^{13}$

• A. $C_1=0.63\times {10}^{-14}$
• B. $C_1=3.16\times {10}^{13}$
• C. $C_1=3.16\times {10}^{-13}$
• D. $C_1=6.32\times {10}^{13}$

Answer 1

The correct option is A $C_1=0.63\times {10}^{-14}$

$Ag\left(s\right)|AgN{O}_3(aq,C_{1}M)||AgN{O}_3(aq,C_2=0.2M)|Ag\left(s\right)$

For the given cell representation, the cell reaction will be,
$A{g}^{+}(aq,C_2=0.2M)\to A{g}^{+}(aq,C_{1}M)$

For concentration cell, we know that,
$E_{cell}=E_{cell}^0-\frac{0.0591}{n}log\frac{\left[C_1\right]}{\left[C_2\right]}$

$E_{cell}=E_{cell}^0+\frac{0.0591}{n}log\frac{\left[C_2\right]}{\left[C_1\right]}$

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

$0.8=\frac{0.0591}{1}log\frac{\left[0.2\right]}{\left[C_1\right]}$

$\frac{0.8}{0.0591}=log\frac{\left[0.2\right]}{C_1}$

$13.5\approx log\frac{\left[0.2\right]}{C_1}$

$Antilog\left(13.5\right)=\frac{0.2}{C_1}$

$3.16\times {10}^{13}=\frac{0.2}{C_1}$

$C_1=0.063\times {10}^{-13}$

$C_1=0.63\times {10}^{-14}$