How to Use This Worksheet
Work in order. Section A checks the definitions, the oxidation number rules and the agent names. Section B practises assigning oxidation numbers, writing half-reactions and balancing in acid, including a disproportionation. Section C combines balancing with redox titration calculations in exam style.
Reference facts used throughout. OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). Oxidation numbers: free element 0; monatomic ion equal to its charge; fluorine −1; oxygen −2 except in peroxides (−1) and with fluorine (positive); hydrogen +1 except in metal hydrides (−1); the numbers in a species sum to 0 for a neutral compound and to the charge for an ion. Molar masses: Zn 65.38 g/mol, Cu 63.55 g/mol. Carry unrounded values through each calculation and round only the final answer, to three significant figures.
Three tools recur in nearly every question. Assign oxidation numbers before and after to find what changed. Name agents by the partner’s fate: the oxidising agent is itself reduced, the reducing agent is itself oxidised. Balance redox equations in halves, and let the charge check tell you how many electrons each half needs.
One habit protects every answer in Section C: never assume a 1 to 1 ratio. Permanganate takes five electrons where dichromate takes six and iodine takes two, so the mole ratios differ each time. Read the ratio from the balanced equation in front of you before touching the numbers.
Do not look at the answer key until you have attempted every question.
Section A: Recall and Concepts
A1. Define each term in one sentence. (a) Oxidation. (b) Reduction. (c) Oxidising agent. (d) Reducing agent. (e) Oxidation number. (f) Disproportionation.
A2. Give the oxidation number of the element named in each species. (a) S in H₂SO₄ (b) N in NO₃⁻ (c) Mn in MnO₄⁻ (d) O in H₂O₂ (e) H in NaH (f) Fe in Fe₂O₃ (g) C in CO₂ (h) S in H₂S
A3. For each change, state whether it is oxidation or reduction, and give the number of electrons transferred per particle. (a) Zn → Zn²⁺ (b) Cu²⁺ → Cu (c) Fe²⁺ → Fe³⁺ (d) Cl₂ → 2Cl⁻ (e) 2I⁻ → I₂ (f) MnO₄⁻ → Mn²⁺
A4. Consider 2Na + Cl₂ → 2NaCl. (a) Which species is oxidised? (b) Which species is reduced? (c) Name the oxidising agent. (d) Name the reducing agent. (e) State the oxidation number of chlorine before and after the reaction.
A5. Decide whether each reaction is redox, justifying your answer with oxidation numbers. (a) CaCO₃ → CaO + CO₂ (b) CH₄ + 2O₂ → CO₂ + 2H₂O (c) AgNO₃ + NaCl → AgCl + NaNO₃ (d) Mg + 2HCl → MgCl₂ + H₂ (e) HCl + NaOH → NaCl + H₂O
Section B: Core Practice
B1. Show the sum equation and work out each oxidation number. (a) Cr in Cr₂O₇²⁻ (b) Cl in ClO₃⁻ (c) S in SO₃²⁻ (d) P in H₃PO₄ (e) Mn in MnO₂ (f) N in NH₃
B2. Zinc metal is placed in copper(II) sulfate solution: Zn + Cu²⁺ → Zn²⁺ + Cu. (a) Write the oxidation half-reaction. (b) Write the reduction half-reaction. (c) Name the oxidising agent and the reducing agent. (d) A 5.00 g piece of zinc reacts completely with excess Cu²⁺. Calculate the mass of copper formed. (e) State the oxidation number of zinc and of copper before and after the reaction.
B3. Iron(II) ions are oxidised by acidified permanganate. (a) State the oxidation number of manganese in MnO₄⁻ and in Mn²⁺, and hence the number of electrons gained per MnO₄⁻ ion. (b) State the number of electrons lost per Fe²⁺ ion as it becomes Fe³⁺. (c) Deduce the ratio Fe²⁺ : MnO₄⁻ in the balanced equation. (d) Write the two half-reactions. (e) Combine them into the full balanced equation, and check the total charge on each side.
B4. Build the reduction half-reaction for dichromate in acidic solution, Cr₂O₇²⁻ → Cr³⁺, step by step. (a) Balance the chromium atoms. (b) Balance the oxygen atoms using H₂O. (c) Balance the hydrogen atoms using H⁺. (d) Work out how many electrons must be added, and on which side. (e) Verify the final half-reaction by checking atoms and charge. (f) Combine your half-reaction with Fe²⁺ → Fe³⁺ + e⁻ to write the full balanced equation for dichromate oxidising iron(II).
B5. Hydrogen peroxide decomposes: 2H₂O₂ → 2H₂O + O₂. (a) Give the oxidation number of oxygen in H₂O₂, in H₂O and in O₂. (b) Explain why this reaction is a disproportionation. (c) Chlorine reacts with cold dilute alkali: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Give the oxidation number of chlorine in Cl₂, in Cl⁻ and in ClO⁻, and state whether this is also a disproportionation. (d) In OF₂, oxygen has oxidation number +2. Explain why the usual −2 rule fails in this compound.
B6. Two more transfers. (a) In Cu + 2Ag⁺ → Cu²⁺ + 2Ag, identify the species oxidised, the species reduced, and both agents. (b) How many electrons does each copper atom transfer? (c) In 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu, how many electrons are lost per aluminium atom, how many gained per copper ion, and how many electrons are transferred in the equation as written? (d) In Mg + 2H⁺ → Mg²⁺ + H₂, name the oxidising agent and justify your choice. (e) In Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, identify the species oxidised and reduced, and name both agents.
Section C: Exam Style
C1. A 25.0 mL sample of an iron(II) solution is acidified and titrated with 0.0200 mol/L potassium permanganate. The endpoint, the first permanent purple tinge, comes at a titre of 24.0 mL. The equation is MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. (a) Calculate the moles of MnO₄⁻ used. (b) Calculate the moles of Fe²⁺ in the sample. (c) Calculate the concentration of the iron(II) solution. (d) Explain why no indicator is needed. (e) Explain why the sample must be acidified for this equation, and its 5 to 1 ratio, to apply. (f) Calculate the mass of iron in the sample. (Molar mass of Fe: 55.85 g/mol.)
C2. Acidified potassium dichromate also oxidises iron(II): Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺. A 25.0 mL sample of an iron(II) solution needs 15.0 mL of 0.0200 mol/L dichromate solution to reach the endpoint. (a) Calculate the moles of Cr₂O₇²⁻ used. (b) Calculate the moles of Fe²⁺ in the sample. (c) Calculate the concentration of the iron(II) solution. (d) State the colour change at the endpoint region, and the oxidation number change for chromium that it signals. (e) A student uses a 1 to 1 ratio for part (b). By what factor is the student’s answer wrong?
C3. Iodine solution is titrated with sodium thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. A 25.0 mL sample of iodine solution reacts exactly with 20.0 mL of 0.100 mol/L thiosulfate solution. (a) Write the oxidation half-reaction and the reduction half-reaction. (b) Name the oxidising agent and the reducing agent. (c) Calculate the moles of thiosulfate used. (d) Calculate the moles of I₂ in the sample. (e) Calculate the concentration of the iodine solution. (f) Starch solution is added near the endpoint. Describe the colour change seen at the endpoint and name the species responsible for the colour.
Show the answer key and full working (only after you have attempted every question)
A1. (a) The loss of electrons by a species; its oxidation number increases. (b) The gain of electrons by a species; its oxidation number decreases. (c) The species that takes electrons from another species, causing oxidation; it is itself reduced. (d) The species that gives electrons to another species, causing reduction; it is itself oxidised. (e) A bookkeeping number assigned by fixed rules that tracks how electrons are distributed in a species; the values in a species sum to its overall charge. (f) A redox reaction in which the same species is both oxidised and reduced.
A2. (a) +6: 2(+1) + S + 4(−2) = 0. (b) +5: N + 3(−2) = −1. (c) +7: Mn + 4(−2) = −1. (d) −1: peroxide exception, 2(+1) + 2(O) = 0. (e) −1: metal hydride exception, (+1) + H = 0. (f) +3: 2(Fe) + 3(−2) = 0. (g) +4: C + 2(−2) = 0. (h) −2: 2(+1) + S = 0.
A3. (a) Oxidation: the atom loses 2 electrons, number rising 0 to +2. (b) Reduction: the ion gains 2 electrons, number falling +2 to 0. (c) Oxidation: 1 electron lost, +2 to +3. (d) Reduction: each chlorine atom gains 1 electron, 0 to −1, so 2 electrons per Cl₂ molecule. (e) Oxidation: each iodide loses 1 electron, −1 to 0, so 2 electrons per I₂ formed. (f) Reduction: 5 electrons gained, manganese falling +7 to +2.
A4. (a) Sodium, Na: it loses electrons, 0 to +1. (b) Chlorine, Cl₂: it gains electrons, 0 to −1. (c) Cl₂, the species that is reduced. (d) Na, the species that is oxidised. (e) 0 in Cl₂, −1 in NaCl.
A5. (a) Not redox: Ca is +2, C is +4 and O is −2 on both sides; no number changes. (b) Redox: carbon rises from −4 in CH₄ to +4 in CO₂, and oxygen falls from 0 in O₂ to −2 in the products. (c) Not redox: Ag stays +1, Na +1, Cl −1 and N +5 throughout; the ions only swap partners. (d) Redox: magnesium rises 0 to +2 and hydrogen falls +1 to 0. (e) Not redox: H is +1, Cl −1, Na +1 and O −2 on both sides; a proton moves, but no oxidation number changes.
B1. (a) 2Cr + 7(−2) = −2, so 2Cr = +12 and Cr = +6. (b) Cl + 3(−2) = −1, so Cl = +5. (c) S + 3(−2) = −2, so S = +4. (d) 3(+1) + P + 4(−2) = 0, so P = +5. (e) Mn + 2(−2) = 0, so Mn = +4. (f) N + 3(+1) = 0, so N = −3.
B2. (a) Zn → Zn²⁺ + 2e⁻. (b) Cu²⁺ + 2e⁻ → Cu. (c) Oxidising agent: Cu²⁺, because it gains the electrons and is reduced. Reducing agent: Zn, because it loses the electrons and is oxidised. (d) Moles of Zn = 5.00 ÷ 65.38 = 0.0765 mol. The equation runs 1 to 1, so moles of Cu = 0.0765 mol. Mass of Cu = 0.0765 × 63.55 = 4.86 g. (e) Zinc: 0 in Zn metal, +2 in Zn²⁺. Copper: +2 in Cu²⁺, 0 in Cu metal.
B3. (a) Manganese is +7 in MnO₄⁻ (Mn + 4(−2) = −1) and +2 in Mn²⁺, a fall of 5, so each MnO₄⁻ gains 5 electrons. (b) Each Fe²⁺ loses 1 electron. (c) Electrons must balance: 5 Fe²⁺ supply the 5 electrons one MnO₄⁻ takes, so the ratio Fe²⁺ : MnO₄⁻ is 5 : 1. (d) Oxidation: Fe²⁺ → Fe³⁺ + e⁻. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. (e) MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Charge check: left, −1 + 8 + 5(+2) = +17; right, +2 + 5(+3) = +17. Atoms also match: Mn 1, O 4, H 8, Fe 5 on each side.
B4. (a) Cr₂O₇²⁻ → 2Cr³⁺. (b) Add 7H₂O to the right: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O. (c) Add 14H⁺ to the left: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O. (d) Charge before electrons: left −2 + 14 = +12; right 2(+3) = +6. The left must fall by 6, so add 6e⁻ to the left. (e) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Atoms: Cr 2, O 7, H 14 on each side. Charge: −2 + 14 − 6 = +6 on the left, +6 on the right. Balanced. (f) The iron(II) half gives 1 electron, so multiply it by 6 and add: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺. The 6 electrons cancel, and the charge is +24 on each side.
B5. (a) H₂O₂: −1 (peroxide). H₂O: −2. O₂: 0. (b) The same element, oxygen, in the same reactant moves in both directions at once: some atoms fall from −1 to −2 and are reduced, others rise from −1 to 0 and are oxidised. One species doing both jobs is disproportionation. (c) Cl₂: 0. Cl⁻: −1. ClO⁻: +1, since Cl + (−2) = −1. Yes: chlorine starts at 0 and splits to −1 and +1, so it is both reduced and oxidised in the same reaction. (d) Fluorine is more electronegative than oxygen and always takes −1. With two fluorines at −1, oxygen must be +2 for the sum to reach 0: O + 2(−1) = 0. The −2 rule assumes oxygen is the most electronegative partner present, which is false here.
B6. (a) Oxidised: Cu, 0 to +2. Reduced: Ag⁺, +1 to 0. Oxidising agent: Ag⁺. Reducing agent: Cu. (b) 2 electrons per copper atom, matching the two Ag⁺ ions each taking one. (c) Each aluminium atom loses 3 electrons; each copper ion gains 2; in the equation as written, 2 × 3 = 6 electrons are transferred, equal to 3 × 2 on the gaining side. (d) H⁺: it gains electrons to form H₂ (falling from +1 to 0) and is therefore reduced, which makes it the oxidising agent; Mg is the reducing agent. (e) Oxidised: Br⁻, rising from −1 to 0 as Br₂. Reduced: Cl₂, falling from 0 to −1. Oxidising agent: Cl₂. Reducing agent: Br⁻.
C1. (a) Moles of MnO₄⁻ = 0.0200 × (24.0 ÷ 1000) = 0.0200 × 0.0240 = 4.80 × 10⁻⁴ mol. (b) The ratio Fe²⁺ : MnO₄⁻ is 5 : 1, so moles of Fe²⁺ = 5 × 4.80 × 10⁻⁴ = 2.40 × 10⁻³ mol. (c) Concentration = 2.40 × 10⁻³ ÷ 0.0250 = 0.0960 mol/L. (d) MnO₄⁻ is purple and its product Mn²⁺ is effectively colourless, so the titrant is its own indicator: the endpoint is the first permanent purple tinge from one excess drop. (e) H⁺ is a reactant in the equation, consumed in an 8 to 1 ratio to MnO₄⁻. Without acid in good excess the reduction does not follow this equation to Mn²⁺, so the 5 to 1 ratio would not hold. (f) Mass of Fe = 2.40 × 10⁻³ × 55.85 = 0.134 g.
C2. (a) Moles of Cr₂O₇²⁻ = 0.0200 × 0.0150 = 3.00 × 10⁻⁴ mol. (b) The ratio Fe²⁺ : Cr₂O₇²⁻ is 6 : 1, so moles of Fe²⁺ = 6 × 3.00 × 10⁻⁴ = 1.80 × 10⁻³ mol. (c) Concentration = 1.80 × 10⁻³ ÷ 0.0250 = 0.0720 mol/L. (d) The orange colour of Cr₂O₇²⁻ fades as green Cr³⁺ forms; chromium falls from +6 to +3. (e) Six times too small: the student reports 3.00 × 10⁻⁴ mol instead of 1.80 × 10⁻³ mol, an answer wrong by a factor of 6.
C3. (a) Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻. Reduction: I₂ + 2e⁻ → 2I⁻. (b) Oxidising agent: I₂, which gains the electrons. Reducing agent: S₂O₃²⁻, which loses them. (c) Moles of thiosulfate = 0.100 × 0.0200 = 2.00 × 10⁻³ mol. (d) The ratio I₂ : S₂O₃²⁻ is 1 : 2, so moles of I₂ = 2.00 × 10⁻³ ÷ 2 = 1.00 × 10⁻³ mol. (e) Concentration = 1.00 × 10⁻³ ÷ 0.0250 = 0.0400 mol/L. (f) The blue-black colour of the starch and iodine complex disappears, leaving the solution colourless, at the drop where the last of the I₂ is reduced to I⁻; the colour is caused by I₂ held in the starch.