Animated Solution for Chemistry - Electrochemistry: Given that the standard potentials ($E^{\circ }$) of $C{u}^{2+}/Cu$ and $C{u}^{+}/Cu$ are $0.34\text{ V}$ and $0.522\text{ V}$ respectively, the $E^{\circ }$ of $C{u}^{2+}/C{u}^{+}$ is

Let's visualize the reduction pathways of copper using a Latimer diagram. We have three oxidation states of copper: $C{u}^{2+}$, $C{u}^{+}$, and solid $Cu$. We are given the standard reduction potentials for the direct reduction of $C{u}^{2+}$ to $Cu$, and $C{u}^{+}$ to $Cu$. We need to find the potential for the intermediate step, which is the reduction of $C{u}^{2+}$ to $C{u}^{+}$.

Now, a very common trap is to simply add or subtract the potentials algebraically. Don't make a silly mistake here! Standard electrode potentials ($E^{\circ }$) are intensive properties, meaning they do not depend on the amount of substance. They cannot be added directly. Instead, we must use standard Gibbs free energy ($\Delta G^{\circ }$), which is an extensive property and is perfectly additive.

Let's write down the half-reactions to see how they relate to each other:

Reaction 1: $C{u}^{2+}+2e^{-}⟶Cu(E_1^{\circ }=0.34\text{ V})$

Reaction 2: $C{u}^{+}+e^{-}⟶Cu(E_2^{\circ }=0.522\text{ V})$

Reaction 3: $C{u}^{2+}+e^{-}⟶C{u}^{+}(E_3^{\circ }=x)$

Notice how subtracting Reaction 2 from Reaction 1 gives us exactly Reaction 3. Since the reactions add up, their Gibbs free energies do too. So, we can write:

Substituting the fundamental thermodynamic relation $\Delta G^{\circ }=-nF{E}^{\circ }$, we get:

We can cancel out Faraday's constant ($F$) from all terms. For Reaction 1, the number of electrons transferred ($n$) is $2$. For Reactions 2 and 3, $n$ is $1$.

Let's substitute the values and get the answer:

This is the standard potential for the $C{u}^{2+}$ to $C{u}^{+}$ half-cell.

Disproportionation

Now, let's think beyond the question. Is $C{u}^{+}$ stable in an aqueous solution? We can check this by calculating the standard cell potential for its disproportionation into $C{u}^{2+}$ and solid $Cu$:

A positive potential means the reaction is spontaneous. Therefore, $C{u}^{+}$ is actually unstable in aqueous solutions and readily disproportionates!