Oxidising/reducing pairs and electron half-reactions

The oxidising agent/reducing agent pair

Metallic copper (Cu) and the Cu²⁺ ion form a pair denoted as Cu²⁺/Cu: these are the two forms – oxidised and reduced – of the same element. The oxidising agent is always written first.

Pair	Oxidised form	Reduced form
Cu²⁺/Cu	Cu²⁺	Cu
Fe³⁺/Fe²⁺	Fe³⁺	Fe²⁺
MnO₄⁻/Mn²⁺	MnO₄⁻	Mn²⁺

The electron half-reaction

Each pair has a half-reaction that links its two forms via an electron transfer:

Cu2+ + 2 e- = Cu

This notation must always be checked in two respects:

1. The elements: there must be the same number of atoms of each type on the left as on the right.
2. The electric charge: the sum of the charges on the left must equal that on the right.

Let’s check: on the left, (+2) + 2×(-1) = 0; on the right, Cu is neutral, so 0 = 0. The half-reaction is balanced.

A more complex example: the MnO₄⁻/Mn²⁺ pair

In an acidic medium, the permanganate ion MnO₄⁻ (purple) is converted into the manganese ion Mn²⁺ (almost colourless):

MnO4- + 8 H+ + 5 e- = Mn2+ + 4 H2O

Checking the elements: 1 Mn = 1 Mn; 4 O (in MnO₄⁻) = 4 O (in 4 H₂O); 8 H (in 8 H⁺) = 8 H (in 4 H₂O). Checking the charges: on the left, (-1) + 8×(+1) + 5×(-1) = -1 + 8 - 5 = +2; on the right, +2. Both sides equal +2: the half-reaction is correctly balanced.

General method for balancing a half-reaction

1. First, balance the element whose oxidation number changes.
2. Balance the oxygen with H₂O molecules.
3. Balance the hydrogen with H+ ions (in an acidic medium).
4. Finally, balance the charges with e- electrons.

Always check both balances at the end; never check just one in isolation.