Question

$C{u}^{+}$ ion is not stable in aqueous solution because of disproportionation reaction whose $E^o$ is: (Given; $E_{C{u}^{2+}/C{u}^{+}}^o=0.15,E_{C{u}^{2+}/C{u}_s}^o=0.34V$ )

• A. - 0.49 V
• B. 0.49 V
• C. - 0.38 V
• D. 0.38 V

Answer 1

Answer: (D)

0.38 V

$\text{The required reaction}(C{u}^{2+}+Cu\to 2C{u}^{+})\text{can be obtained by using the following reactions.}$

$C{u}^{2+}+e^{-}\to C{u}^{+},E_{C{u}^{2+}/C{u}^{+}}^o=0.15V...\left(i\right)$

$C{u}^{2+}+2^e-\to Cu,E_{C{u}^{2+}/Cu}^o=0.34V..\left(ii\right)$

$\text{Multiplying eq. (i) by 2 we get}$

$2C{u}^{2+}+2e^{-}\to 2C{u}^{+}...\left(iii\right)$

$\Delta G_1=-nFE=-2\times F\times 0.15$

$C{u}^{2+}+2e^{-}\to Cu...\left(iv\right)$

$\text{Subtract the eq. (iv) from (iii)}$

$C{u}^{2+}+Cu\to 2C{u}^{+}$

$\Delta G_3=-nFE=-1\times F\times E_o$

Also

$\Delta G_3=\Delta G_1-\Delta G_2$

$-1F{E}_o=(-2F\times 0.15)-(-2F\times 0.34)$

$E_o=-0.38$

$\text{this is the value of the reaction}$

$C{u}^{2+}+Cu\to 2C{u}^{+}$

$\text{but the given reaction is just reverse of it}$

$\therefore E_{cell}forgivenreaction=+0.38V.$