Chemistry - Electrochemistry Question with Solution | TestHub

Concept: Disproportionation occurs if $E^0$ for the oxidation half-reaction is less than $E^0$ for the reduction half-reaction.

Solution:

Given $E_{C{u}^{2+}/Cu}^0=0.3V$ and $E_{C{u}^{+}/Cu}^0=0.5V$.

We need $E_{C{u}^{2+}/C{u}^{+}}^0$ for disproportionation of Cu⁺.

Using Latimer diagram principles, we can calculate $E_{C{u}^{2+}/C{u}^{+}}^0$.

For the reaction $C{u}^{2+}+2e^{-}\to Cu$, $\Delta G_1^0=-2F{E}_{C{u}^{2+}/Cu}^0=-2F(0.3)=-0.6F$.

For the reaction $C{u}^{+}+e^{-}\to Cu$, $\Delta G_2^0=-1F{E}_{C{u}^{+}/Cu}^0=-1F(0.5)=-0.5F$.

The reaction $C{u}^{2+}+e^{-}\to C{u}^{+}$ corresponds to $\Delta G_3^0=\Delta G_1^0-\Delta G_2^0=-0.6F-(-0.5F)=-0.1F$.

Since $\Delta G_3^0=-nF{E}_{C{u}^{2+}/C{u}^{+}}^0$, for $n=1$, we have $-0.1F=-1F{E}_{C{u}^{2+}/C{u}^{+}}^0$.

So, $E_{C{u}^{2+}/C{u}^{+}}^0=0.1V$.

For disproportionation of Cu⁺:

Oxidation: $C{u}^{+}\to C{u}^{2+}+e^{-}$, $E_{ox}^0=-E_{C{u}^{2+}/C{u}^{+}}^0=-0.1V$

Reduction: $C{u}^{+}+e^{-}\to Cu$, $E_{red}^0=E_{C{u}^{+}/Cu}^0=0.5V$

The overall cell potential for disproportionation is $E_{cell}^0=E_{red}^0+E_{ox}^0=0.5V+(-0.1V)=0.4V$.

Since $E_{cell}^0>0$, the disproportionation of Cu⁺ is spontaneous.

Therefore, Cu⁺ disproportionates.