Quick answer: Redox reactions transfer electrons: oxidation is loss of electrons and an increase in oxidation number, reduction is gain of electrons and a decrease. This Chemistry 11-12 worksheet on redox reactions gives 20 problems from easy to hard: oxidation numbers of common species, sparkler chemistry and zinc coating on a nail. The worked answers are on a separate page, and the worksheet is a free printable PDF.
These problems train you to assign oxidation numbers, identify oxidising and reducing agents, balance half-equations in acid and alkali, and use redox titrations. They begin with recall and finish with multi-step analysis of real samples.
The level suits A Level, IB Chemistry, CBSE Class 11-12 and senior WAEC and NECO students. Show the change in oxidation number, then check atoms and charge in every equation.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.2
Problem 1 Oxidation numbers of common species ★★★
Assign the oxidation number of the named element in each species.1. Sulfur in H₂SO₄
2. Manganese in KMnO₄
3. Nitrogen in NH₄⁺
4. Oxygen in H₂O₂
5. Chlorine in ClO₃⁻
Problem 2 Sparkler chemistry ★★★
Kofi in Accra watches a magnesium sparkler burn at a festival: 2Mg + O₂ → 2MgO.1. Write the oxidation number of each element before and after.
2. State which substance is oxidised and which is reduced.
3. Name the oxidising agent and the reducing agent.
Problem 3 Zinc coating on a nail ★★★
Zinc on a galvanised nail loses atoms to the solution as shown in the diagram.1. Write the half-equation for zinc forming Zn²⁺.
2. State whether this is oxidation or reduction and why.
3. State how many electrons are released by 3 mol of zinc.
Problem 4 Redox or not? ★★★
Decide whether each reaction is a redox reaction. Give a reason using oxidation numbers.1. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
2. 2KI + Cl₂ → 2KCl + I₂
3. AgNO₃ + NaCl → AgCl + NaNO₃
Problem 5 Roman numerals in names ★★★
Use oxidation numbers to answer.1. Name Fe₂O₃.
2. Name CuSO₄.
3. Write the formula of chromium(III) oxide.
4. Name SnCl₂.
Problem 6 Harder oxidation numbers ★★★
Find the oxidation number of the named element.1. Chromium in Cr₂O₇²⁻
2. Sulfur in S₂O₃²⁻
3. Carbon in C₂H₅OH (take both carbons together, then give the average)
4. Iron in Fe₃O₄ (give the average)
Problem 7 Balancing two half-equations in acid ★★★
Balance each half-equation in acidic solution by adding H⁺, H₂O and e⁻.1. NO₃⁻ → NO
2. H₂O₂ → O₂
Problem 8 Rhubarb leaves and permanganate ★★★
Oxalate ions in rhubarb leaves are oxidised by acidified permanganate. Half-equations: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and C₂O₄²⁻ → 2CO₂ + 2e⁻.1. State how the half-equations must be multiplied.
2. Write the overall ionic equation.
3. Check the charge balance.
Problem 9 Thermite reaction ★★★
Railway workers in Lahore weld rails with thermite: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe.1. Give the oxidation numbers of Al and Fe before and after.
2. Identify the oxidising agent and the reducing agent.
3. State the number of electrons transferred per formula unit of Fe₂O₃.
Problem 10 Silver recovery from waste ★★★
A photo lab in Singapore recovers silver by placing copper in silver nitrate solution: Cu + 2Ag⁺ → Cu²⁺ + 2Ag.1. Write the two half-equations.
2. Calculate the moles of Cu in 0.635 g (Ar Cu = 63.5).
3. Calculate the mass of silver formed (Ar Ag = 107.9).
Problem 11 Making bleach ★★★
Emma in Leeds studies the manufacture of household bleach: Cl₂ + 2NaOH → NaCl + NaClO + H₂O.1. Give the oxidation number of chlorine in Cl₂, NaCl and NaClO.
2. Explain why this is called disproportionation.
3. State the number of electrons transferred per Cl₂ molecule.
Problem 12 Balancing by oxidation numbers ★★★
Wanjiru in Nairobi dissolves copper in dilute nitric acid: Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O.1. Give the change in oxidation number of Cu and of N (the N in NO).
2. Choose coefficients so that the electron changes are equal.
3. Balance the full equation.
Problem 13 Permanganate in alkali ★★★
In alkaline solution MnO₄⁻ is reduced to solid MnO₂.1. First balance the half-equation as if it were in acid.
2. Convert it to alkaline conditions by adding OH⁻.
3. Check atoms and charge.
Problem 14 Reading oxidation numbers from colours ★★★
Mei in Singapore reduces permanganate with different reagents and records the colours in the diagram (left to right).1. Assign the oxidation number of manganese in each species.
2. State the number of electrons gained per Mn in going from MnO₄⁻ to MnO₂ and to Mn²⁺.
3. Predict the colour change when MnO₄⁻ is reduced to Mn²⁺ in acid.
Problem 15 Iron supplement tablet ★★★
Chiamaka dissolves one iron(II) supplement tablet and makes the solution up to 250 cm³. Samples of 25.0 cm³ need 22.40 cm³ of 0.00200 mol/dm³ MnO₄⁻ in acid (Ar Fe = 55.8).1. Write the overall ionic equation.
2. Calculate the moles of MnO₄⁻ used.
3. Calculate the moles of Fe²⁺ in 25.0 cm³ and in the whole 250 cm³.
4. Calculate the mass of iron in the tablet in mg.
Problem 16 Breathalyser chemistry ★★★
Diego in Texas studies a dichromate breath test: orange Cr₂O₇²⁻ turns green Cr³⁺ when it oxidises ethanol to ethanoic acid. Half-equations: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and C₂H₅OH + H₂O → CH₃COOH + 4H⁺ + 4e⁻.1. Find the oxidation number change of chromium per Cr₂O₇²⁻ and of the carbon of the CH₂OH group.
2. Combine the half-equations into an overall equation.
3. Check that H, O and charge balance.
Problem 17 Analysing bleach with thiosulfate ★★★
Bilal in Lahore dilutes 10.0 cm³ of bleach to 250 cm³. A 25.0 cm³ portion is mixed with excess KI and acid: ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O. The iodine needs 21.50 cm³ of 0.100 mol/dm³ thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The colour sequence is shown.1. Calculate the moles of thiosulfate and of I₂.
2. Calculate the moles of ClO⁻ in the 25.0 cm³ portion and in the original 10.0 cm³.
3. Calculate the concentration of ClO⁻ in the bleach in mol/dm³ and in g/dm³ (Mr NaClO = 74.4).
Problem 18 Spot the error in a redox equation ★★★
A student writes: MnO₄⁻ + 8H⁺ + Fe²⁺ → Mn²⁺ + 4H₂O + Fe³⁺.1. Show that the equation is not balanced for charge.
2. Find the correct coefficient for Fe²⁺ and Fe³⁺.
3. Write the corrected equation and check the charge.
Problem 19 Iodine in table salt ★★★
Kofi tests iodised salt. He dissolves 50.0 g of salt, adds excess iodide and acid: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. The iodine formed needs 12.60 cm³ of 0.00500 mol/dm³ thiosulfate (I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻). Ar I = 126.9.1. Show that the equation is balanced for atoms and charge.
2. Calculate the moles of I₂ and of IO₃⁻.
3. Calculate the mass of iodine in the form of iodate in mg per kg of salt.
Problem 20 Oxidation numbers in electron count ★★★
Aarav in Pune oxidises iodide with dichromate: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂.1. Show that the equation balances for charge.
2. Calculate the total change in oxidation number for chromium and for iodine.
3. Calculate the mass of iodine formed from 25.0 cm³ of 0.0200 mol/dm³ dichromate in excess iodide (Ar I = 126.9).
