Question

Given,
$C{u}^{2+}+2e^{-}\to Cu;E^o=0.337$V
$C{u}^{2+}+e^{-}\to C{u}^{+};E^o=-0.153$V
$E^o$ value for the reaction, $C{u}^{+}+e\to Cu$ will be:

• A. $0.38$V
• B. $0.52$V
• C. $0.83$V
• D. $0.30$V

Answer 1

Answer: (C)

$0.83$V

As we know that Gibb's free energy is an additive property and can be calculated as-

${\Delta G}^{\circ }=-nF{E}^{\circ }$

For reaction-

${Cu}^{2+}+2e^{-}⟶Cu;E^{\circ }=0.337V$

${\Delta G}^{\circ }=-2\times F\times 0.337=-0.674F$

Therefore,

${Cu}^{2+}+2e^{-}⟶Cu;{\Delta G}^{\circ }=-0.674F.....\left(1\right)$

Given:-

${Cu}^{2+}+e^{-}⟶{Cu}^{+};E^{\circ }=-0.153V$

For reaction-

${Cu}^{+}⟶e^{-}+{Cu}^{+2};E^{\circ }=+0.153V$

${\Delta G}^{\circ }=-1\times F\times \left(0.153\right)=-0.153F$

Therefore,

${Cu}^{+}⟶e^{-}+{Cu}^{2+};{\Delta G}^{\circ }=-0.153F.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

${Cu}^{+}+e^{-}⟶Cu;{\Delta G}^{\circ }=-0.827F$

Now again,

${\Delta G}^{\circ }=-nF{E}^{\circ }$

$\Rightarrow E^{\circ }=\frac{-0.827F}{-1\times F}=0.827$

Hence the $E^{\circ }$ value for the given reaction will be $0.827V$.