Oxidation and reduction as electron transfer
Link to Std 9–10 : You learnt that oxidation is gain of oxygen and reduction is loss of oxygen; now we see the deeper idea behind it, the movement of electrons.
An iron nail left in rain turns into brown rust, and a cut apple turns brown. Both are slow oxidation in air, and in both the substance loses electrons.
Oxidation is loss of electrons (also gain of oxygen or loss of hydrogen). Reduction is gain of electrons (also loss of oxygen or gain of hydrogen). A redox reaction has both happening together.
In the figure : Zinc loses two electrons (oxidation) and copper ions gain them (reduction). Adding the two half-reactions gives the full reaction.
Rule 1: Classical meaning — Adding oxygen or removing hydrogen is oxidation; removing oxygen or adding hydrogen is reduction.
- Easy : 2Mg + O₂ → 2MgO: magnesium gains oxygen, so it is oxidised.
- Medium : CuO + H₂ → Cu + H₂O: CuO loses oxygen (reduced) and H₂ gains oxygen (oxidised).
- Exam-type : H₂S + Cl₂ → 2HCl + S: H₂S loses hydrogen, so it is oxidised, and Cl₂ gains hydrogen, so it is reduced.
Rule 2: Electron-transfer meaning — The substance that loses electrons is oxidised and is the reducing agent. The substance that gains electrons is reduced and is the oxidising agent.
- Easy : Zn → Zn²⁺ + 2e⁻ is oxidation; zinc is the reducing agent.
- Medium : Cu²⁺ + 2e⁻ → Cu is reduction; Cu²⁺ is the oxidising agent.
- Exam-type : In Zn + Cu²⁺ → Zn²⁺ + Cu, 2 electrons lost by one zinc atom are gained by one Cu²⁺ ion, so the electron count balances.
Rule 3: Half-reactions — Write the oxidation and the reduction separately, then add them after making the electrons equal.
- Easy : Oxidation Na → Na⁺ + e⁻; reduction Cl₂ + 2e⁻ → 2Cl⁻.
- Medium : To combine them multiply the oxidation by 2: 2Na + Cl₂ → 2Na⁺ + 2Cl⁻.
- Exam-type : The electrons in half-reactions never appear in the final equation.
Rule 4: Competition for electrons — A more reactive metal pushes electrons to the ion of a less reactive metal, so it displaces it from solution.
- Easy : A zinc rod in blue CuSO₄ solution turns reddish-brown with copper and the blue colour fades.
- Medium : A copper rod in ZnSO₄ solution shows no change, because copper is less reactive than zinc.
- Exam-type : Copper in AgNO₃ solution gets a grey coating of silver, so Cu gives electrons more readily than Ag.
Oxidation and reduction always go together; one cannot happen without the other.
Calling the oxidised substance the oxidising agent. The oxidised substance is the reducing agent.
Memory clue: "OIL RIG" — Oxidation Is Loss, Reduction Is Gain (of electrons).
Spot the difference
| Point | Oxidation | Reduction |
|---|---|---|
| Electrons | Lost | Gained |
| Oxygen | Gained | Lost |
| Agent | Substance is the reducing agent | Substance is the oxidising agent |
Practice questions
- 1Define oxidation in terms of electrons.
- 2In 2Mg + O₂ → 2MgO, which substance is oxidised?
- 3What is a reducing agent?
- 4In CuO + H₂ → Cu + H₂O name the oxidising agent and the reducing agent.
- 5Write the two half-reactions for Zn + Cu²⁺ → Zn²⁺ + Cu.
- 6Why does a copper rod not react with ZnSO₄ solution?
- 7Write the half-reactions and the balanced equation for sodium reacting with chlorine.
- 8Why do the electrons not appear in the final equation of a redox reaction?
Show answers (try them yourself first)
- 1Oxidation is the loss of electrons.
- 2Magnesium.
- 3A substance that gives electrons to another substance and is itself oxidised.
- 4CuO is the oxidising agent and H₂ is the reducing agent.
- 5Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu.
- 6Copper is less reactive than zinc and cannot give electrons to Zn²⁺.
- 7Na → Na⁺ + e⁻ and Cl₂ + 2e⁻ → 2Cl⁻; adding with the oxidation multiplied by 2 gives 2Na + Cl₂ → 2NaCl.
- 8The electrons lost in oxidation are exactly gained in reduction, so they cancel when the half-reactions are added.
Oxidation number and how to find it
Link to Std 9–10 : You know valency, the combining power of an atom; oxidation number is a similar idea that also tells us the sign and helps us follow electrons even in covalent compounds.
In a cricket match each player's score is tracked ball by ball. Oxidation numbers are the bookkeeping that tracks electrons atom by atom.
The oxidation number is the charge an atom would have if all shared electrons were given to the more electronegative atom. Oxidation means its number increases; reduction means its number decreases.
In the figure : Nitrogen can have oxidation numbers from −3 (NH₃) to +5 (HNO₃). Moving right means oxidation; moving left means reduction.
Rule 1: Basic rules — An element in its free state is 0. The sum for a neutral compound is 0; for an ion it equals the charge. Group 1 is +1, group 2 is +2, fluorine is always −1.
- Easy : S in S₈ is 0 and Na in NaCl is +1.
- Medium : Oxygen is usually −2 and hydrogen is usually +1.
- Exam-type : Exceptions: O is −1 in H₂O₂, +2 in OF₂, and H is −1 in metal hydrides such as NaH.
Rule 2: Calculate an unknown — Write the known numbers, let the unknown be x, and make the sum equal the charge.
- Easy : Mn in KMnO₄: +1 + x + 4(−2) = 0, so x = +7.
- Medium : Cr in K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0, so x = +6.
- Exam-type : S in Na₂S₄O₆: 2(+1) + 4x + 6(−2) = 0, so x = +2.5 (an average value).
Rule 3: Ions and special cases — For a polyatomic ion the sum equals its charge. The oxidation number can be a fraction when it is an average.
- Easy : N in NH₄⁺: x + 4(+1) = +1, so x = −3.
- Medium : N in NO₃⁻: x + 3(−2) = −1, so x = +5.
- Exam-type : In Fe₃O₄ the average oxidation number of Fe is 8/3 (it holds Fe²⁺ and Fe³⁺ in the ratio 1 : 2).
Rule 4: Reading a redox from oxidation numbers — An atom whose number rises is oxidised; one whose number falls is reduced.
- Easy : In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn goes 0 → +2 (oxidised) and Cu goes +2 → 0 (reduced).
- Medium : In 2H₂ + O₂ → 2H₂O, H goes 0 → +1 and O goes 0 → −2.
- Exam-type : AgNO₃ + NaCl → AgCl + NaNO₃ shows no change in any number, so it is not a redox reaction.
The oxidation number of an element in the free state is always 0.
Writing the sign after the number (3+) for oxidation numbers. Write +3, with the sign first.
Memory clue: "Sum to charge" — all numbers in a species add up to its charge.
Spot the difference
| Point | Valency | Oxidation number |
|---|---|---|
| Sign | None | Plus or minus |
| Value | Whole number | Can be a fraction |
| Free element | Not zero | Zero |
Practice questions
- 1Define oxidation number.
- 2What is the oxidation number of sulphur in S₈?
- 3What is the oxidation number of oxygen in H₂O₂?
- 4Find the oxidation number of Mn in KMnO₄.
- 5Find the oxidation number of N in NO₃⁻.
- 6Is AgNO₃ + NaCl → AgCl + NaNO₃ a redox reaction? Give a reason.
- 7Find the oxidation number of S in Na₂S₄O₆.
- 8Find the average oxidation number of Fe in Fe₃O₄ and explain the fraction.
Show answers (try them yourself first)
- 1The charge an atom would have if all shared electrons went to the more electronegative atom.
- 2Zero.
- 3−1.
- 4+7.
- 5+5.
- 6No, because no atom changes its oxidation number.
- 7+2.5.
- 88/3; the compound contains Fe²⁺ and Fe³⁺ in the ratio 1 : 2, and (2 + 3 + 3) ÷ 3 = 8/3.
Types of redox reactions
Link to Std 9–10 : You know the types of chemical reactions: combination, decomposition and displacement; now we see which of them are redox and one new type, disproportionation.
Hydrogen peroxide in a first-aid bottle slowly fizzes and loses strength. One part of it oxidises while another part is reduced, in the same bottle.
Disproportionation is a redox reaction in which the same element is both oxidised and reduced. Other types are combination, decomposition and displacement.
In the figure : The four types with an example of each.
Rule 1: Combination and decomposition — Combination reactions involving an element are redox. Decomposition reactions that produce an element are redox.
- Easy : C + O₂ → CO₂ is a combination redox (C goes 0 → +4, O goes 0 → −2).
- Medium : 2H₂O → 2H₂ + O₂ is a decomposition redox.
- Exam-type : CaCO₃ → CaO + CO₂ is decomposition but not redox, because no oxidation number changes.
Rule 2: Displacement — A more reactive element displaces a less reactive one: metal displacement, hydrogen displacement and non-metal displacement.
- Easy : Zn + 2HCl → ZnCl₂ + H₂: zinc displaces hydrogen from the acid.
- Medium : Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂: chlorine displaces bromine and the solution turns brown.
- Exam-type : Br₂ cannot displace Cl⁻, since chlorine is a stronger oxidising agent than bromine.
Rule 3: Disproportionation — The element must be in an intermediate oxidation state, so that it can go both up and down.
- Easy : 2H₂O₂ → 2H₂O + O₂: oxygen goes −1 → −2 and −1 → 0.
- Medium : Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O: chlorine goes 0 → −1 and 0 → +1.
- Exam-type : An element in its highest oxidation state cannot disproportionate. For example, N in HNO₃ (+5) can only be reduced.
Rule 4: Not redox — Precipitation, neutralisation and ion exchange have no change in oxidation numbers.
- Easy : HCl + NaOH → NaCl + H₂O is not a redox reaction.
- Medium : BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl is not a redox reaction.
- Exam-type : The test is simple: write the oxidation number of every atom on both sides; if nothing changes, it is not redox.
A reaction with a free element on one side only is always a redox reaction.
Treating every decomposition as redox. CaCO₃ → CaO + CO₂ is not.
Memory clue: "Free element in or out means redox".
Spot the difference
| Point | Displacement | Disproportionation |
|---|---|---|
| Species oxidised and reduced | Two different species | The same element |
| Example | Zn + Cu²⁺ | Cl₂ + OH⁻ |
| Element state | Free element and ion | Intermediate state |
Practice questions
- 1Define disproportionation.
- 2Is C + O₂ → CO₂ a redox reaction?
- 3Give one example of displacement.
- 4Is CaCO₃ → CaO + CO₂ a redox reaction? Give a reason.
- 5Show that 2H₂O₂ → 2H₂O + O₂ is a disproportionation.
- 6Will Br₂ displace Cl⁻ from solution?
- 7Show that Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O is a disproportionation.
- 8Why can N in HNO₃ not undergo disproportionation?
Show answers (try them yourself first)
- 1A redox reaction in which the same element is both oxidised and reduced.
- 2Yes; C goes 0 → +4 and O goes 0 → −2.
- 3Zn + CuSO₄ → ZnSO₄ + Cu.
- 4No; no oxidation number changes.
- 5O in H₂O₂ is −1; it becomes −2 in H₂O (reduced) and 0 in O₂ (oxidised).
- 6No; chlorine is a stronger oxidising agent than bromine.
- 7Chlorine goes 0 → −1 in Cl⁻ and 0 → +1 in ClO⁻.
- 8N is already at its highest oxidation state, +5, so it can only be reduced, not oxidised.
Balancing redox equations
Link to Std 9–10 : You balanced simple equations by counting atoms; redox equations also need the electron count to be equal.
A trade fair keeps accounts so that what is paid equals what is received. A redox equation likewise keeps electrons lost equal to electrons gained.
In the oxidation number method, the total increase in oxidation number equals the total decrease. In the ion-electron (half-reaction) method, two half-reactions are balanced separately and added after equalising electrons.
In the figure : Five steps for balancing. The orange step (matching electron loss and gain) is the key one.
Rule 1: Oxidation number method — Find the changes, multiply to equalise, place coefficients, then balance O with H₂O and H with H⁺ in acid.
- Easy : MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺: Mn falls by 5 and Fe rises by 1, so multiply Fe by 5.
- Medium : In acid the balanced result is MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
- Exam-type : Check the charge: left −1 + 10 + 8 = +17; right +2 + 15 = +17. The O and H counts also match.
Rule 2: Ion-electron method in acidic medium — Balance atoms other than O and H, then O with H₂O, then H with H⁺, then charge with electrons.
- Easy : Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O is the balanced reduction half.
- Medium : Oxidation half: 2I⁻ → I₂ + 2e⁻; multiply by 3 to match 6 electrons.
- Exam-type : Adding gives Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O (charge +6 on both sides).
Rule 3: Basic medium — Balance as in acid, then add OH⁻ equal to the H⁺ on both sides to turn H⁺ into water.
- Easy : The H⁺ are changed into H₂O by adding the same number of OH⁻ on both sides.
- Medium : MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic medium: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
- Exam-type : Check: charge −8 on each side; oxygen 12 on each side; hydrogen 8 on each side.
Rule 4: Another worked example — Dilute nitric acid with copper: 3Cu + 8H⁺ + 2NO₃⁻ → 3Cu²⁺ + 2NO + 4H₂O.
- Easy : Cu goes 0 → +2 (loses 2) and N goes +5 → +2 (gains 3).
- Medium : Make electrons equal: 3 Cu atoms lose 6 electrons and 2 N atoms gain 6.
- Exam-type : Check the charge: left +8 − 2 = +6; right +6.
Always check both atoms and charge at the end.
Balancing atoms but not charge, so the electrons do not cancel.
Memory clue: "O with water, H with H⁺, charge with electrons".
Spot the difference
| Point | Acidic medium | Basic medium |
|---|---|---|
| Balance O with | H₂O | H₂O |
| Balance H with | H⁺ | H⁺, then add OH⁻ both sides |
| Final species | H⁺ present | OH⁻ present |
Practice questions
- 1What must be equal in a balanced redox equation besides atoms?
- 2What is added to balance oxygen atoms?
- 3What is added to balance hydrogen atoms in acidic medium?
- 4Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic medium.
- 5Balance the half-reaction Cr₂O₇²⁻ → 2Cr³⁺ in acid.
- 6By how many units does N change in HNO₃ → NO?
- 7Balance Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ in acid.
- 8Balance MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic medium.
Show answers (try them yourself first)
- 1Total charge, and the electrons lost and gained.
- 2H₂O.
- 3H⁺.
- 4MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
- 5Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
- 6N goes from +5 to +2, a decrease of 3.
- 7Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O.
- 82MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
Redox in a cell: electrode potential and spontaneity
Link to Std 9–10 : You know that a metal reacts with salt solution of a less reactive metal; here we measure the push of electrons as a voltage.
A torch cell gives electricity because a redox reaction pushes electrons through the wire. Place the two half-reactions in separate containers and electrons must flow through a wire.
A galvanic cell turns the energy of a spontaneous redox reaction into electricity. The tendency of a half-cell to gain electrons is its standard electrode potential (E°). E°cell = E°cathode − E°anode.
In the figure : The Daniell cell: zinc is the anode (oxidation), copper the cathode (reduction), electrons flow through the wire and ions move through the salt bridge.
Rule 1: Anode, cathode and flow — Oxidation takes place at the anode, reduction at the cathode. Electrons flow from anode to cathode in the wire. The salt bridge completes the circuit.
- Easy : In the Daniell cell Zn is the anode and Cu is the cathode.
- Medium : Electrons travel from the zinc rod to the copper rod.
- Exam-type : Without the salt bridge, charge would build up around the electrodes and the current would stop quickly.
Rule 2: Cell potential and spontaneity — Use E° (reduction values): Zn²⁺/Zn −0.76 V, Fe²⁺/Fe −0.44 V, Cu²⁺/Cu +0.34 V, Ag⁺/Ag +0.80 V. If E°cell is positive the reaction is spontaneous.
- Easy : Zn–Cu cell: E°cell = 0.34 − (−0.76) = +1.10 V.
- Medium : Fe + Cu²⁺ → Fe²⁺ + Cu: E°cell = 0.34 − (−0.44) = +0.78 V, so it is spontaneous.
- Exam-type : Cu + Zn²⁺ → Cu²⁺ + Zn: E°cell = −0.76 − 0.34 = −1.10 V, so it does not occur.
Rule 3: Strength of oxidising and reducing agents — A species with a larger positive E° is a stronger oxidising agent; one with a more negative E° is a stronger reducing agent.
- Easy : F₂ (+2.87 V) is a stronger oxidising agent than Cl₂ (+1.36 V).
- Medium : Cl₂ can oxidise Br⁻ because E°cell = 1.36 − 1.09 = +0.27 V.
- Exam-type : Mg (−2.36 V) is a stronger reducing agent than Zn (−0.76 V), so Mg displaces Zn²⁺.
Rule 4: Cell potential, energy and K — ΔG° = −nFE°cell with F = 96 500 C mol⁻¹, and log K = nE°cell ÷ 0.059 at 298 K.
- Easy : For the Zn–Cu cell n = 2, so ΔG° = −2 × 96 500 × 1.10 = −212 300 J = −212.3 kJ.
- Medium : log K = 2 × 1.10 ÷ 0.059 = 37.3, so K is about 2 × 10³⁷, which means the reaction goes almost to completion.
- Exam-type : A positive E°cell gives a negative ΔG° and K > 1, which all describe a spontaneous reaction.
The more positive the reduction potential, the greater the tendency to be reduced.
Changing the sign of E° when the half-reaction is multiplied. E° does not depend on the amount.
Memory clue: "Cathode minus anode" gives E°cell, and a positive value means go.
Spot the difference
| Point | Anode | Cathode |
|---|---|---|
| Process | Oxidation | Reduction |
| Sign in galvanic cell | Negative | Positive |
| Electrons | Leave | Arrive |
Practice questions
- 1At which electrode does oxidation occur?
- 2What is the use of a salt bridge?
- 3Write the formula for E°cell.
- 4Calculate E°cell for Zn + Cu²⁺ → Zn²⁺ + Cu using Zn²⁺/Zn = −0.76 V and Cu²⁺/Cu = +0.34 V.
- 5Will Fe displace Cu²⁺ from solution? (Fe²⁺/Fe = −0.44 V)
- 6Will Cu displace Zn²⁺? Give E°cell.
- 7Find ΔG° for the Zn–Cu cell (n = 2, F = 96 500 C mol⁻¹).
- 8Can Cl₂ oxidise Br⁻? (Cl₂/Cl⁻ = +1.36 V, Br₂/Br⁻ = +1.09 V)
Show answers (try them yourself first)
- 1At the anode.
- 2It completes the circuit and keeps the solutions electrically neutral.
- 3E°cell = E°cathode − E°anode.
- 40.34 − (−0.76) = +1.10 V.
- 5Yes; E°cell = 0.34 − (−0.44) = +0.78 V, which is positive.
- 6No; E°cell = −0.76 − 0.34 = −1.10 V.
- 7ΔG° = −2 × 96 500 × 1.10 = −212 300 J, which is −212.3 kJ.
- 8Yes; E°cell = 1.36 − 1.09 = +0.27 V.