Oxidation States in the Zinc-Copper Redox Reaction

Given

• Reaction: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)

• Rules: free elements have oxidation state 0; monatomic ions have oxidation state equal to their charge.

Find

The oxidation states of Zn and Cu on each side, which species is oxidized and reduced, and the oxidizing and reducing agents.

Principle

Oxidation is loss of electrons (oxidation state increases) and reduction is gain of electrons (oxidation state decreases). The species that is oxidized acts as the reducing agent because it donates electrons; the species that is reduced acts as the oxidizing agent because it accepts them.

Formula

Solution

1. Assign oxidation states on the left: Zn(s) is a free element, so 0; Cu2+ is a monatomic ion, so +2.

2. Assign oxidation states on the right: Zn2+ is +2; Cu(s) is a free element, so 0.

3. Zinc goes from 0 to +2: its oxidation state increased, meaning it lost 2 electrons — zinc is oxidized.

4. Copper goes from +2 to 0: its oxidation state decreased, meaning it gained 2 electrons — copper(II) is reduced.

5. The reducing agent is the species oxidized: Zn(s). The oxidizing agent is the species reduced: Cu2+(aq).

Unit check

Oxidation states are dimensionless numbers, and the electron balance checks out: zinc loses 2 electrons while copper gains 2, so charge is conserved (left: +2, right: +2) and atoms balance (1 Zn, 1 Cu each side).

What this means physically

Zinc metal spontaneously reduces copper(II) ions: zinc gives up electrons and dissolves as Zn2+ while Cu2+ plates out as copper metal. This electron transfer is the same chemistry that powers the Daniell cell — connect the two half-reactions through a wire and the electron flow becomes usable electric current.