Question 1

How can a Cu+ species act as both an oxidising and a reducing agent?

Accepted Answer

In 2Cu⁺ → Cu²⁺ + Cu, one Cu⁺ loses an electron and changes from +1 to +2, so it is oxidised. The other Cu⁺ gains that electron and changes from +1 to 0, so it is reduced. Because the same species participates in both halves, Cu⁺ acts as both agents.

Question 2

Why is MgO represented as formula units rather than molecules?

Accepted Answer

MgO forms an extended ionic solid, not separate discrete molecules. In 2Mg + O₂ → 2MgO, the stoichiometry therefore produces two MgO formula units, and each formula unit is only part of the repeating lattice.

Question 3

What happens to the solution in the iron and copper sulfate reaction?

Accepted Answer

Iron dissolves as Fe²⁺ while Cu²⁺ ions are reduced to metallic copper on the iron surface. The blue colour of Cu²⁺ decreases, while the solution develops a pale-green Fe²⁺ colour. The copper is deposited on the nail, not floating as a separate product.

Question 4

What is the difference between disproportionation and comproportionation?

Accepted Answer

Disproportionation splits one element in a single oxidation number into higher and lower oxidation numbers, such as Cu(+1) forming Cu(+2) and Cu(0). Comproportionation is the reverse pattern: two different oxidation numbers merge into an intermediate one, such as Cu(+2) and Cu(0) forming Cu(+1).

Question 5

Why is copper(I) difficult to maintain in ordinary aqueous solution?

Accepted Answer

Copper(I) is unstable in many ordinary aqueous environments and tends to disproportionate into Cu²⁺ and copper metal. Cu(I) complexes can be stabilised with suitable ligands or under appropriate reaction conditions, so the model shows a ligand-stabilised Cu(I) pattern.

Redox Families in 3D — Combination, Decomposition, Displacement, and the 'Both' Species

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