Question

A cell $Cu|C{u}^{2+}||A{g}^{\oplus }|Ag$ initially contains 2 M $A{g}^{\oplus }$ and 2 M $C{u}^{2+}$ ions in 1 L solution each. The change in cell potential after it has supplied 1 A current for 96500 s is :

• A. $-0.003V$
• B. $-0.02V$
• C. $-0.04V$
• D. None of these

Answer 1

Answer: (B)

$-0.02V$

$Cu|C{u}^{2+}\left(2M\right)||A{g}^{\oplus }\left(2M\right)|Ag$
Number of faradays supplied = $\frac{It}{96500}=\frac{1\times 96500}{96500}=1$
After of faradays supplied 1.0 F electricity:
At anode : $Cu\to C{u}^{2+}+2e^{-}$
$\Rightarrow$ 2 F electricity $\equiv$ 1 mol $C{u}^{2+}$ produced.
$\Rightarrow$ 1.0 F electricity $\equiv$ 0.50 mol $C{u}^{2+}$ produced.
$\Rightarrow [C{u}^{2+}{]}_{new}=\frac{2\times 1+0.50}{1}=2.5M$
At cathode: $A{g}^{\oplus }+e^{-}\to Ag$
$\Rightarrow$ 1 F electricity $\equiv$ mole $A{g}^{\oplus }$ used
$\Rightarrow [A{g}^{\oplus }{]}_{new}=\frac{2\times 1-1.0}{1}=1.0M$
$E_{cell}=E_{cell}^{\ominus }-\frac{0.059}{2}log\frac{\left[C{u}^{2+}\right]}{[A{g}^{\oplus }{]}^2}$
$\Rightarrow (E_{cell}{)}_1=E_{cell}^{\ominus }-\frac{0.059}{2}log\frac{2}{2^2}$
$and(E_{cell}{)}_1=E_{cell}^{\ominus }-\frac{0.059}{2}log\frac{2.5}{1^2}$
$\Rightarrow changein{E}_{cell}=(E_{cell}{)}_2-(E_{cell}{)}_1$
$=-\frac{0.059}{2}log2.5-\frac{0.059}{2}log2$
$=-\frac{0.059}{2}log5=-0.02V$