Consider the following half-cell reactions. Al(s) $\to$ Al3+(aq) + 3e- Cu2+(aq) + 2e- $\to$ Cu(s) The overall equation for the reaction is

Analyze the question: You are given two half-reactions and asked to combine them into a balanced overall equation. The electrons lost by aluminium must equal the electrons gained by copper.

Concept: In a redox reaction, the electrons cancel out. To make this happen, find the LCM of the electrons in the two half-equations.

Solution:

• Oxidation: Al(s) → Al³⁺(aq) + 3e⁻ (3 electrons)
• Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s) (2 electrons)

The LCM of 3 and 2 is 6.

• Multiply the Al half-equation by 2: 2Al(s) → 2Al³⁺(aq) + 6e⁻
• Multiply the Cu half-equation by 3: 3Cu²⁺(aq) + 6e⁻ → 3Cu(s)

Add them together and cancel the 6e⁻ on both sides:

2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)

Check the balance:

• Atoms: 2 Al and 3 Cu on each side.
• Charge: left = 3(+2) = +6; right = 2(+3) = +6. ✓

Checking the options:

• A: charges are not balanced (+2 on the left, +3 on the right).
• B: Al is written without a charge, and the equation is unbalanced.
• C: matches our result. ✓
• D: charges are not balanced (+4 on the left, +9 on the right).

Tip: Cross-multiply the electron counts. The number of electrons in one half-equation becomes the multiplier for the other (Al × 2, Cu × 3). Then always check that the charges balance on both sides.