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AS · Practice questions · Redox equations and half-equations

Two halves, one whole.

Six original Cambridge-style questions on redox equations: half-equations for common oxidising agents, combining halves, and a simple redox titration ratio.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Atoms, O, H, then charge.

01
Analysis
[2 marks]

Write the half-equation for the reduction of manganate(VII) ions to manganese(II) ions in acidic solution.

  • MnO4− + 8H+ + 5e− → Mn2+ + 4H2O. ✓
  • Check the charge: −1 + 8 − 5 = +2 on both sides. ✓
02
Analysis
[2 marks]

Write the half-equation for the oxidation of sulfite ions, SO32−, to sulfate ions in water.

  • SO32− + H2O → SO42− + 2H+ + 2e−. ✓
  • S goes +4 → +6, losing two electrons; charge −2 on both sides. ✓
03
Analysis
[3 marks]

Combine the manganate(VII) half-equation with H2O2 → O2 + 2H+ + 2e− to give the overall equation.

  • LCM of 5 and 2 is 10: double the manganate half, multiply the peroxide half by 5. ✓
  • 2MnO4− + 16H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2 + 10H+. ✓
  • Cancel H+: 2MnO4− + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2. ✓
04
Analysis
[3 marks]

Write the overall equation for the reaction of acidified dichromate(VI) with iron(II) ions, and give the mole ratio of Cr2O72− to Fe2+.

  • Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O; Fe2+ → Fe3+ + e− (× 6). ✓
  • Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O. ✓
  • Ratio 1 : 6. ✓
05
Calculation
[3 marks]

25.0 cm3 of an iron(II) solution needs 22.50 cm3 of 0.0200 mol dm−3 KMnO4 in acid. Calculate the concentration of Fe2+.

  • n(MnO4−) = 0.0200 × 0.02250 = 4.50 × 10−4 mol. ✓
  • n(Fe2+) = 5 × 4.50 × 10−4 = 2.25 × 10−3 mol. ✓
  • c = 2.25 × 10−3 ÷ 0.0250 = 0.0900 mol dm−3. ✓
06
Analysis
[3 marks]

Use oxidation number changes to balance: Cu + HNO3 → Cu(NO3)2 + NO2 + H2O.

  • Cu goes 0 → +2 (loses 2); N goes +5 → +4 in NO2 (gains 1), so 2NO2 per Cu. ✓
  • Two more HNO3 supply the nitrate ions in Cu(NO3)2: 4HNO3 in total. ✓
  • Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.