Question

The e.m.f. of cell for spontaneous cell reaction having two electrodes as $Ag|A{g}^{+}||C{u}^{2+}|Cu$ containing $1MC{u}^{2+}$ and $1MA{g}^{+}$ ions after the passage of 9.65 ampere current for 1 hour will :

• A. increase by $0.010V$
• B. decrease by $0.010V$
• C. increase by $0.020V$
• D. decrease by $0.020V$

Answer 1

The correct option is A increase by $0.010V$

Given,
$C{u}^{2+}+e\to C{u}^{+};-\Delta G_1^o=E_1^{o}F\times 1$
$C{u}^{+}+e\to Cu;-\Delta G_2^o=E_2^{o}F\times 1$
----------------------------
by adding these equations
$C{u}^{2+}+2e\to Cu;-\Delta G_3^o=-[\Delta G_1^o+\Delta G_2^o]=-[0.15F+0.5F]$
or $2\times E_3^{o}F=0.65F$
(as $\Delta G=-nF{E}^o$)
$\therefore E_3^o=0.325V$
So for reverse reaction.
$\therefore Cu\to C{u}^{2+}+2e{E}^o=-0.325V$
$Cu\to C{u}^{2+}+2e{E}^o=-0.325V$
$2A{g}^{+}+2e\to 2Ag{E}^o=0.799V$
$\therefore E_{cell}^o=E_{O{P}_{Cu/C{u}^{2+}}}^o+E_{R{P}_{A{g}^{+}/Ag}}^o$
$=-0.325+0.799=0.474V$
On passing the current through the given cell Ag will oxidise to Ag${}^{+}$ and Cu${}^{2+}$ will be reduced to Cu.
As same charge is passing throuth both solutions so
According to Faraday's law:
$\frac{w}{E}=\frac{i.t}{96500}=\frac{9.65\times 3600}{96500}$
$=0.36$ eq. of $A{g}^{+}=0.36$ mol of $A{g}^{+}$
$=0.36$ eq. of $C{u}^{2+}$
$=\frac{0.36}{2}$ mole of $C{u}^{2+}$ (as mole $=$ equivalent/valency)
$\therefore \left[A{g}^{+}\right]=1+0.36=1.36M$
$\left[C{u}^{2+}\right]=1.0-0.18=0.82M$
For $Ag|A{g}^{+}||C{u}^{2+}|Cu$ cell,
$E^o_{cell} = - 0.474 $
And according to Nernst equation,
$E=E_{O{P}_{Ag}}^o+E_{R{P}_{Cu}}^o+\frac{0.059}{2}log\frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}$
Initially $C{u}^{2+}=1M$ and $A{g}^{+}=1M$
$\therefore E_{cell}=E_{cell}^o=-0.474$
After electrolysis $C{u}^{2+}=0.82M,A{g}^{+}=1.36M$
$\therefore E_{cell}=E_{cell}^o+\frac{0.059}{2}log\frac{0.82}{(1.36{)}^2}$
$=E_{cell}^o-0.010=-0.474-0.010=-0.484V$
$E_{cell}^o$ is negative for $A{g}^{+}|Ag||C{u}^{2+}|Cu$ and is $-0.474V$
As for spontaneous reaction E should be positive and it was 0.474 V in first case and 0.484 V in II case so it increases by 0.010 V.