These nine original questions cover the six lessons in SPM Chemistry redox equilibrium. Attempt each question with full working before opening the answer.
Questions
Question 1. Find the oxidation number of chromium in Cr₂O₇²⁻.
Answer
Seven oxygens give −14. Two chromium atoms together are 2x, and the ion charge is −2.
2x − 14 = −2, so 2x = 12 and x = +6.
Revisit assigning oxidation numbers if you forgot to multiply by the number of chromium atoms.
Question 2. Find the oxidation number of nitrogen in NO₃⁻ and in NH₃.
Answer
NO₃⁻: x + 3(−2) = −1, so x = +5.
NH₃: x + 3(+1) = 0, so x = −3.
Question 3. In Zn + 2HCl → ZnCl₂ + H₂, state what is oxidised, what is reduced, and the reducing agent.
Answer
Zinc goes from 0 to +2, so zinc is oxidised.
Hydrogen goes from +1 to 0, so hydrogen ions are reduced.
The reducing agent is the substance oxidised, which is zinc.
Question 4. Is Fe₂O₃ + 3CO → 2Fe + 3CO₂ a redox reaction? Explain using oxidation numbers.
Answer
Yes. Iron goes from +3 to 0, so Fe₂O₃ is reduced and is the oxidising agent.
Carbon goes from +2 in CO to +4 in CO₂, so CO is oxidised and is the reducing agent.
Question 5. Zinc is added to lead(II) nitrate solution. Write the two half-equations and the ionic equation.
Answer
Oxidation: Zn → Zn²⁺ + 2e⁻.
Reduction: Pb²⁺ + 2e⁻ → Pb.
Ionic equation: Zn + Pb²⁺ → Zn²⁺ + Pb. Charge: +2 on both sides.
Question 6. Predict the products at each electrode when concentrated sodium chloride solution is electrolysed with carbon electrodes, and repeat for dilute sodium chloride solution.
Answer
Cathode, both solutions: hydrogen gas, 2H⁺ + 2e⁻ → H₂, because H⁺ is lower than Na⁺ in the series.
Anode, concentrated: chlorine, 2Cl⁻ → Cl₂ + 2e⁻.
Anode, dilute: oxygen, 4OH⁻ → 2H₂O + O₂ + 4e⁻.
The concentration decides which anion is discharged. See predicting electrolysis products.
Question 7. Silver nitrate solution is electrolysed using silver electrodes. State what happens at each electrode.
Answer
Anode: silver is not inert, so it dissolves, Ag → Ag⁺ + e⁻, and the anode becomes thinner.
Cathode: Ag⁺ + e⁻ → Ag, so a silver deposit forms. This is the idea behind silver electroplating.
Question 8. A cell is made from magnesium and copper. State the negative terminal, the direction of electron flow, and the observations at each electrode.
Answer
Magnesium is higher in the series, so it is the negative terminal: Mg → Mg²⁺ + 2e⁻.
Electrons flow from magnesium to copper through the external wire.
Observations: the magnesium plate becomes thinner, a brown deposit forms on copper, and the blue colour of the copper(II) solution fades.
Question 9. A galvanised roof sheet is scratched to show the iron. Explain why the iron does not rust quickly.
Answer
Zinc is higher than iron in the electrochemical series, so zinc loses electrons more readily.
Zn → Zn²⁺ + 2e⁻ takes place in preference to Fe → Fe²⁺ + 2e⁻, so zinc is oxidised and the iron is protected until the zinc is used up.
If you got these wrong
For Questions 1 and 2, see assigning oxidation numbers. Questions 3 and 4 link to identifying oxidation and reduction, and Question 5 to writing half-equations. For Question 8, see explaining simple chemical cells, and for Question 9, corrosion and prevention.
The chemical equation balance checker helps you verify ionic equations. For a teacher to work through the questions you missed, see online one-to-one Chemistry tuition.