Question

A cell, $Ag|A{g}^{\oplus }||C{u}^{2+}|Cu,$ initially contains 1 M $A{g}^{\oplus }$ and $1MC{u}^{2+}$ ions. The change in the cell potential after the passage of 9.65 A of current for 1 h is:

• A. $-0.010V$
• B. $-0.020V$
• C. $-0.015V$
• D. $Noneofthese$

Answer 1

Answer: (A)

$-0.010V$

For a cell $Ag|A{g}^{\oplus }||C{u}^{2+}|Cu$

Cell reaction is:

$Ag\left(s\right)+\frac{1}{2}C{u}^{2+}\left(aq\right)\to A{g}^{\oplus }\left(aq\right)+\frac{1}{2}Cu\left(s\right)$ (n = 1)

As we know that:

$m=Z\times I\times t=\frac{Ew}{96500}\times I\times t$

$\frac{W}{Ew}$ or gram equivalent of $C{u}^{2+}=\frac{1}{96500}\times It$

$=\frac{1}{96500}\times 9.65\times 60\times 60=0.36$

Decrease in the concentration of copper ion is $\frac{0.36}{2}M.$ Remaining concentration of copper = 1 0.18 = 0.82 M So, increase in the concentration of silver ion is 0.36 M Now concentration of silver ion = 1 + 0.36 = 1.36 M

On applying Nernst equation

$E_{cell}=E_{cell}^{\ominus }-\frac{0.0591}{n}log\frac{\left[A{g}^{\oplus }\right]}{\sqrt{\left[C{u}^{2+}\right]}}$

$\therefore$ Change in cell potential,

$\Delta E=\frac{0.0591}{1}log\frac{1.36}{\sqrt{0.82}}=-0.010V$