Chemistry · Redox Equilibrium · NEET
The full d10 shell would suggest Cu+ should be stable, and in dry or solid form some Cu(I) compounds are. But in water, stability is decided by hydration enthalpy, not just electron configuration. Cu2+ is a smaller ion with a higher charge (+2), so water molecules are held far more strongly around it. NCERT states the extra hydration enthalpy released when Cu+ becomes Cu2+ more than pays for the energy needed to remove the second electron. So in water Cu2+(aq) wins, and Cu+ collapses into Cu2+ and Cu.
Disproportionation is when one element in one oxidation state is oxidised and reduced at the same time. In 2Cu+ -> Cu2+ + Cu, copper starts at +1. One Cu+ loses an electron and goes up to +2 (oxidised). The other Cu+ gains an electron and goes down to 0 as Cu metal (reduced). Same element, same starting state, two different fates - that is disproportionation.
Write the two half-reactions. Cu+ + e- -> Cu has E = +0.52 V, and Cu2+ + e- -> Cu+ has E = +0.16 V. For disproportionation, the ion sits in the middle. The reduction step to its lower form (+0.52 V) must have a higher E than the step that formed it from above (+0.16 V). Since 0.52 > 0.16, the overall cell EMF is positive (+0.36 V), so Cu+ disproportionating is spontaneous. NCERT simply says the E value for this is favourable.
For the aqueous stability of Cu+ versus Cu2+, hydration enthalpy is the deciding factor named by NCERT. The d10 argument explains why Cu+ can exist at all (and in solids or non-aqueous media). But once you put it in water, the much more negative hydration enthalpy of Cu2+ drives the disproportionation. In NEET questions, quote hydration enthalpy as the reason.
No, that is the opposite: comproportionation. There two different oxidation states of the same element (+2 and +7) come together to give one middle state (+4). Disproportionation is one state splitting into two; comproportionation is two states merging into one. NEET 2019 tested exactly this trap in the Mn options.
Which of the following reactions are disproportionation reactions? (a) 2Cu+ -> Cu2+ + Cu (b) 3MnO4^2- + 4H+ -> 2MnO4- + MnO2 + 2H2O (c) 2KMnO4 -> K2MnO4 + MnO2 + O2 (d) 2MnO4- + 3Mn2+ + 2H2O -> 5MnO2 + 4H+ Select the correct option.
From the Latimer diagram BrO4- (1.82 V) BrO3- (1.5 V) HBrO (1.595 V) Br2 (1.0652 V) Br-, the species undergoing disproportionation is:
E values: Al+/Al = +0.55 V, Tl+/Tl = -0.34 V, Al3+/Al = -1.66 V, Tl3+/Tl = +1.26 V. Identify the incorrect statement.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
2Cu+(aq) -> Cu2+(aq) + Cu(s). Copper in the +1 state goes to +2 and to 0.
Cu2+ is smaller and carries a higher +2 charge, so it attracts and binds water molecules more strongly, releasing more energy on hydration. This extra energy makes Cu2+(aq) the favoured form.
No. Many copper(I) compounds are stable as solids or in non-aqueous media (for example CuCl, Cu2O). It is mainly in aqueous solution that Cu+ disproportionates. Cu2I2 is stable because I- also reduces Cu2+.
Disproportionation: one oxidation state splits into a higher and a lower state (one becomes two). Comproportionation: a higher and a lower state combine into one middle state (two become one).
Yes. Using Cu+ + e- -> Cu (0.52 V) as reduction and Cu+ -> Cu2+ + e- (reverse of 0.16 V) as oxidation, the cell EMF is +0.36 V, which is positive, so the reaction is spontaneous.
