Which one of the following species is stable in aqueous solution?

We have to find which ion or oxo-anion can remain as it is when it is put into ordinary water. In aqueous medium an ion will be called “stable” only when it does not undergo self-oxidation, self-reduction or any other spontaneous redox change such as disproportionation.

The simplest way to judge this behaviour is to compare the standard electrode potentials $$E^\circ$$ of the redox couples that involve the given species. We recall the general principle:

$$\text{If }E^\circ(\text{oxidation step})\lt E^\circ(\text{reduction step}),$$ then the intermediate oxidation state will undergo disproportionation because the reduction to the lower state is easier than its own re-oxidation back to the higher state.

Now we examine each option one by one.

Option A : $$Cr^{2+}$$

The standard potentials for chromium are

$$Cr^{3+}+e^- \rightarrow Cr^{2+},\;E^\circ=-0.41\;{\rm V}$$ $$Cr^{2+}+2e^- \rightarrow Cr,\;E^\circ=-0.91\;{\rm V}$$

Because both potentials are highly negative, $$Cr^{2+}$$ is a very strong reducing agent. In water it is readily oxidised by dissolved oxygen to the more stable $$Cr^{3+}$$ state. Hence $$Cr^{2+}$$ is not stable in aqueous solution.

Option B : $$MnO_4^{2-}$$

For the manganate series the relevant redox couples are

$$MnO_4^- + e^- \rightarrow MnO_4^{2-},\;E^\circ=+0.56\;{\rm V}$$ $$MnO_4^{2-} + e^- \rightarrow MnO_4^{3-},\;E^\circ=+0.37\;{\rm V}$$

Here we notice two points:

(i) The first reduction potential $$+0.56{\rm V}$$ is not so large that water will automatically reduce $$MnO_4^-$$ to $$MnO_4^{2-}$$, but it is large enough to keep $$MnO_4^{2-}$$ from getting oxidised back by dissolved oxygen.

(ii) The second potential $$+0.37{\rm V}$$ is smaller; therefore $$MnO_4^{2-}$$ does not have a strong tendency to get further reduced to $$MnO_4^{3-}$$ in neutral water.

Consequently $$MnO_4^{2-}$$ persists in alkaline or even mildly neutral medium as a green manganate solution. Thus it is reasonably stable in water.

Option C : $$MnO_4^{3-}$$

The step that could destroy this ion is

$$2\,MnO_4^{3-} \rightarrow MnO_4^{2-} + MnO_4^{4-}$$

Because the potential for $$MnO_4^{3-} + e^- \rightarrow MnO_4^{4-}$$ is extremely small (in fact negative when calculated), the ion is an exceedingly powerful reducing agent. It instantly reacts with even the solvent water and disproportionates. Hence $$MnO_4^{3-}$$ is not stable in aqueous solution.

Option D : $$Cu^+$$

The copper redox data are

$$Cu^{2+}+e^- \rightarrow Cu^+,\;E^\circ=+0.153\;{\rm V}$$ $$Cu^++e^- \rightarrow Cu,\;E^\circ=+0.521\;{\rm V}$$

To test disproportionation of $$Cu^+$$ we write the composite reaction

$$2\,Cu^+ \rightarrow Cu^{2+}+Cu$$

For this overall process the net potential is

$$E^\circ = \dfrac{E^\circ_{\text{red}}(Cu^{2+}/Cu^+) + E^\circ_{\text{red}}(Cu^+/Cu)}{2} = \dfrac{0.153\;{\rm V} + 0.521\;{\rm V}}{2}=0.337\;{\rm V}>0$$

Since $$E^\circ$$ is positive, the reaction is spontaneous, so $$Cu^+$$ disproportionates in water to $$Cu$$ metal and $$Cu^{2+}$$ ion. Therefore $$Cu^+$$ is definitely not stable.

Summarising all four cases we find that only $$MnO_4^{2-}$$ resists any spontaneous redox change in aqueous solution.

Hence, the correct answer is Option B.