Question

For the disproportionation of copper 2Cu+------ Cu2+ + Cu, E(standard)cell is (given E for Cu2+/Cu is 0.34V and E for Cu2+/Cu+ is 0.15V)
1- 0.49V 2- -0.19V
3- 0.38V 4- -0.38V

Answer 1

Answer : (3) E(standard)cell is 0.38 V.

Solution : Given,

$E_{Cu^{2+}/Cu}=0.34V$

$E_{Cu^{2+}/Cu^+}=0.15V$

The disproportionation reaction of copper is,

$2Cu^+\rightarrow Cu^{2+}+Cu$           $E^o_{cell}$ = ?

(A) $Cu^{2+}+2e^-\rightarrow Cu$        $E_{Cu^{2+}/Cu}=0.34V$

(B) $Cu^{2+}+e^-\rightarrow Cu^+$       $E_{Cu^{2+}/Cu^+}=0.15V$

Step 1 : Reverse the equation(B) and then adding equation(A) & (B).

$Cu^{2+}+2e^-\rightarrow Cu$        $E_{Cu^{2+}/Cu}=0.34V$

$Cu^{+}\rightarrow Cu^{2+}+e^-$    $E_{Cu^{+}/Cu^{2+}}=-0.15V$

Net reaction : $Cu^++e^-\rightarrow Cu$

Step 2 : Now we have to calculate the value of $E_{Cu^{+}/Cu}$.

Formula used : $\Delta G=-nFE^o$

where,

$\Delta G$ = gibbs free energy

n = number of electrons

F = Faraday constant

$E^o$ = electrode potential

The expression for $E_{Cu^{+}/Cu}$ by using above formula, we get

$-nFE_{Cu^{+}/Cu}=[(-nFE_{Cu^{2+}/Cu})+(-nFE_{Cu^{+}/Cu^{2+}})]$

$-1\times E_{Cu^{+}/Cu}=[(-2\times 0.34)+(-1\times (-0.15))]$

$E_{Cu^{+}/Cu}=0.53V$

Step 3 : Now we have to calculate $E^o_{cell}$ by using equations,

$Cu^{+}\rightarrow Cu^{2+}+e^-$            $E_{Cu^{+}/Cu^{2+}}=-0.15V$

$Cu^++e^-\rightarrow Cu$              $E_{Cu^{+}/Cu}=0.53V$

Net reaction : $2Cu^+\rightarrow Cu^{2+}+Cu$

Formula used :

$E^o_{cell}=E_{oxidation}+E_{reduction}$

$E^o_{cell}=E_{{Cu^{+}/Cu^{2+}}}+E_{Cu^{+}/Cu}$ = -0.15V +0.53V = 0.38 V

Therefore, the value of E(standard)cell is 0.38 V.

Answer 2

Answer:

__________ 0.38 V __________