Ionic Equations - Question 4

Question: Zinc powder is added to blue copper(II) sulfate solution. The blue colour fades and a brown solid forms. Write the ionic equation and the two half-equations.Step 1: full equation.Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)Step 2: split and cancel. The sulfate ion appears unchanged on both sides, so it is the spectator ion.Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)Step 3: half-equations. Zinc atoms lose electrons, so zinc is oxidised. Copper(II) ions gain electrons, so they are reduced.Zn(s) → Zn²⁺(aq) + 2e^- (oxidation)
Cu²⁺(aq) + 2e^- → Cu(s) (reduction)Adding the half-equations gives the ionic equation, because the electrons cancel.Extension: aluminium with copper(II) sulfate. The half-equations are:Al(s) → Al³⁺(aq) + 3e^-
Cu²⁺(aq) + 2e^- → Cu(s)The number of electrons lost must equal the number gained. Multiply the first by 2 and the second by 3, so that 6 electrons are transferred, then add:2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)Check: charge on the left is 3 × (2+) = 6+ and on the right 2 × (3+) = 6+.