Question (3 marks): In acidic solution, dichromate(VI) ions are reduced to chromium(III) ions. Construct the ionic half-equation for this change.Step 1: balance the element that changes. There are two chromium atoms on the left, so put 2 in front of Cr3+:Cr2O72− → 2 Cr3+
Step 2: balance oxygen with water. Seven oxygen atoms need 7 H2O on the right.Step 3: balance hydrogen with H+. 7 H2O contains 14 hydrogen atoms, so add 14 H+ on the left:Cr2O72− + 14 H+ → 2 Cr3+ + 7 H2O
Step 4: balance charge with electrons. Left side: (−2) + (+14) = +12. Right side: 2 × (+3) = +6. Add 6 electrons to the left:Cr2O72− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O
Check with oxidation numbers: each Cr goes from +6 to +3, a fall of 3. Two chromium atoms fall by 6 in total, which matches the 6 electrons gained. Electrons on the left means reduction, so dichromate(VI) is acting as an oxidising agent.Examiner tip: follow the same order every time: the element, then oxygen (H2O), then hydrogen (H+), then charge (e−). Marks are usually given for the correct species and for the fully balanced equation.
Step 2: balance oxygen with water. Seven oxygen atoms need 7 H2O on the right.Step 3: balance hydrogen with H+. 7 H2O contains 14 hydrogen atoms, so add 14 H+ on the left:
Step 4: balance charge with electrons. Left side: (−2) + (+14) = +12. Right side: 2 × (+3) = +6. Add 6 electrons to the left:
Check with oxidation numbers: each Cr goes from +6 to +3, a fall of 3. Two chromium atoms fall by 6 in total, which matches the 6 electrons gained. Electrons on the left means reduction, so dichromate(VI) is acting as an oxidising agent.Examiner tip: follow the same order every time: the element, then oxygen (H2O), then hydrogen (H+), then charge (e−). Marks are usually given for the correct species and for the fully balanced equation.