SJChO Notes · Chapter 6
6

Redox Chemistry

Singapore Junior Chemistry Olympiad

Iron rusts, a match burns, a battery runs down and a breathalyser changes colour. In each of these, electrons move from one substance to another. This chapter shows how to track those electrons with oxidation states, how to write and balance the equations, and how a titration measures an oxidising or reducing agent.

Learning objectives

By the end of this chapter, you can:

Foundation O Level
  • define oxidation and reduction in terms of oxygen, hydrogen, electrons and oxidation state
  • work out the oxidation state of an element in a compound or ion
  • decide whether a reaction is a redox reaction
  • identify the oxidising agent and the reducing agent
  • describe the tests with aqueous potassium iodide and acidified potassium manganate(VII)
Core A Level
  • write and balance half-equations in acidic and alkaline solution
  • combine half-equations into a balanced redox equation
  • recognise disproportionation
  • carry out manganate(VII), dichromate(VI) and iodine–thiosulfate titration calculations
  • deduce a change in oxidation state from the amounts that react
Prerequisites Chapter 2 (electrons in shells), Chapter 3 (ions, formulae and electronegativity), Chapter 4 (equations, ionic equations, moles and gas volumes) and Chapter 5 (reactions of acids, precipitation and titration). Fill in the blanks as the lesson goes.

6.1Oxidation and reduction

When iron rusts, it reacts with oxygen and water: 4Fe + 3O2 + 2xH2O → 2Fe2O3·xH2O. The iron gains oxygen, so it is oxidised. The reverse happens in a blast furnace, where carbon monoxide takes the oxygen away from iron(III) oxide:

Fe2O3 + 3CO → 2Fe + 3CO2

Iron(III) oxide loses oxygen, so it is reduced. At the same moment carbon monoxide gains that oxygen, so it is oxidised. Oxidation and reduction always happen together, and a reaction with both is a redox reaction (reduction–oxidation).

The first definitions used oxygen. Later they grew to cover hydrogen, then electrons, then oxidation states (Section 6.2). The last two work for every redox reaction, even one with no oxygen or hydrogen in it.

oxidationreduction
oxygengain of oxygenloss of oxygen
hydrogenloss of hydrogengain of hydrogen
electronsloss of electronsgain of electrons
oxidation stateincreasedecrease

Methane burning, CH4 + 2O2 → CO2 + 2H2O, shows the hydrogen definition: the carbon gains oxygen and loses all four of its hydrogen atoms, so methane is oxidised.

6.1.1Electron transfer

Put a strip of zinc into blue copper(II) sulfate solution. The zinc becomes coated with brown copper and the blue colour fades. Each zinc atom gives two electrons to a copper(II) ion (Figure 6.1):

Zn atom+2+Cu²⁺ ion2 electrons move2+Zn²⁺ ion+Cu atomZn loses electrons:oxidisedCu²⁺ gains electrons:reducedelectron
Figure 6.1 Zinc gives two electrons to a copper(II) ion. The sulfate ions are spectators and are not shown.

Each change can be written as a half-equation, which shows the electrons:

oxidation: Zn(s) → Zn2+(aq) + 2e−
reduction: Cu2+(aq) + 2e− → Cu(s)
overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

The electrons lost by zinc are exactly the electrons gained by copper(II) ions, so they cancel when the half-equations are added.

DefinitionOxidation is the loss of electrons. Reduction is the gain of electrons.
NoteA memory aid: OIL RIG, Oxidation Is Loss, Reduction Is Gain (of electrons).
Check your understanding
  1. Sodium burns in chlorine: 2Na + Cl2 → 2NaCl. Write the two half-equations and state which reactant is oxidised.
  2. In CuO + H2 → Cu + H2O, which substance is reduced? Explain in terms of oxygen.

6.2Oxidation states

In covalent compounds no electrons are fully transferred, so the electron definition is hard to use. Oxidation states solve this. Pretend every bond is ionic: in each bond the more electronegative atom (Chapter 3) takes both shared electrons. The charge each atom would then carry is its oxidation state.

The oxidation state (or oxidation number) of an atom is the charge it would have if the compound were made of ions.

An oxidation state is written with its sign first, as +2 or −1. A charge is written with the sign last, as 2+ or 1−. Work it out with these rules. An earlier rule wins over a later one.

  1. An uncombined element is 0: Mg, O2, P4.
  2. A simple ion has its charge: Na+ is +1, O2− is −2.
  3. The oxidation states in a compound add up to 0. In an ion they add up to the ion's charge.
  4. In compounds, Group 1 metals are +1, Group 2 metals are +2 and aluminium is +3.
  5. Fluorine is always −1.
  6. Hydrogen is +1, except in metal hydrides such as NaH, where it is −1.
  7. Oxygen is −2, except in peroxides such as H2O2 (−1) and in OF2 (+2).
  8. Chlorine is −1, except when it is bonded to oxygen or fluorine.

Elements that can have several oxidation states show it in the name with Roman numerals: iron(III) chloride contains iron in the +3 state, and potassium manganate(VII), KMnO4, contains manganese in the +7 state.

Worked exampleFind the oxidation state of (a) carbon in Na2CO3, (b) each nitrogen in NH4NO3, (c) manganese in MnO4−.

(a) 2(+1) + x + 3(−2) = 0, so x = +4
Na is +1, O is −2, the compound totals 0.
(b) NH4+: p + 4(+1) = +1, so p = −3. NO3−: q + 3(−2) = −1, so q = +5
Treat each ion on its own. The two nitrogen atoms differ.
(c) x + 4(−2) = −1, so x = +7
The ion totals its charge, −1.

One element can take many oxidation states. Nitrogen runs from −3 to +5 (Figure 6.2). Moving up the scale is oxidation and moving down is reduction.

+5HNO₃, NO₃⁻+4NO₂+3NO₂⁻+2NO+1N₂O0N₂−1−2−3NH₃, NH₄⁺oxidation stateoxidationreduction
Figure 6.2 The oxidation states of nitrogen.
Going furtherA calculated oxidation state can be a fraction. In the tetrathionate ion, S4O62−, 4x + 6(−2) = −2 gives x = +2.5. This is an average: the four sulfur atoms are bonded differently and do not all have the same oxidation state.
Check your understanding
  1. Find the oxidation state of (a) chromium in Cr2O72−, (b) sulfur in Na2S2O3, (c) hydrogen in NaH.
  2. Vanadium forms the ions VO2+ and VO2+. Find the oxidation state of vanadium in each.

6.3Is it a redox reaction?

Write the oxidation state of each element on both sides of the equation (Figure 6.3). If an element's oxidation state increases, it is oxidised. If it decreases, it is reduced. If no oxidation state changes, the reaction is not a redox reaction.

Worked exampleChlorine is bubbled into potassium iodide solution: Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq). What is oxidised and what is reduced?

1. Cl2: 0. KI: K +1, I −1. KCl: K +1, Cl −1. I2: 0
Elements are 0. K is in Group 1.
2. Iodine: −1 to 0, an increase
Iodide ions are oxidised.
3. Chlorine: 0 to −1, a decrease. Potassium does not change.
Chlorine is reduced. K+ is a spectator.
Cl2+2KI→2KCl+I20+1−1+1−10I: −1 → 0, oxidisedCl: 0 → −1, reducedK: +1 → +1, a spectator
Figure 6.3 The oxidation state of each atom, before and after. Potassium does not change.

Respiration, C6H12O6 + 6O2 → 6CO2 + 6H2O, is redox: carbon rises from 0 to +4 and oxygen falls from 0 to −2. In the Haber process, N2 + 3H2 ⇌ 2NH3, nitrogen falls from 0 to −3 and hydrogen rises from 0 to +1.

Two common kinds of reaction are not redox, because every ion keeps its charge:

  • Acid–base reactions (Chapter 5): H+(aq) + OH−(aq) → H2O(l). Hydrogen stays +1 and oxygen stays −2.
  • Precipitation (Chapter 5): Ag+(aq) + Cl−(aq) → AgCl(s). Silver stays +1 and chlorine stays −1.
Exam tipIf an element appears uncombined on one side of an equation and in a compound on the other, its oxidation state has changed from 0, so the reaction is redox.
Check your understanding
  1. Which of these are redox reactions? Explain using oxidation states. (a) NH3 + HNO3 → NH4NO3 (b) CuCO3 → CuO + CO2 (c) 2Cu(NO3)2 → 2CuO + 4NO2 + O2

6.4Oxidising and reducing agents

In a redox reaction one substance gives electrons and the other takes them (Figure 6.4).

DefinitionAn oxidising agent oxidises another substance by taking its electrons, so it is itself reduced. A reducing agent reduces another substance by giving it electrons, so it is itself oxidised.
Zn²⁺Cue⁻e⁻Zn: reducing agentCu²⁺: oxidising agentoxidised: loses e⁻reduced: gains e⁻
Figure 6.4 Electrons move from the reducing agent to the oxidising agent.

In Zn + Cu2+ → Zn2+ + Cu, zinc is the reducing agent and Cu2+ is the oxidising agent. Common agents, with the colours that show they have reacted:

oxidising agentbecomesreducing agentbecomes
O2, Cl2O2−, Cl−H2, C, CO, reactive metalsH+, CO2, metal ions
MnO4− in acid (purple)Mn2+ (very pale pink, looks colourless)I− (colourless)I2 (yellow-brown in water)
Cr2O72− in acid (orange)Cr3+ (green)Fe2+ (pale green)Fe3+ (yellow)
H2O2 in acidH2OSO2, SO32−SO42−

6.4.1Testing for oxidising and reducing agents

An unknown that reacts with a known reducing agent must be an oxidising agent, and the other way round. Two tests use this (Figure 6.5).

1Test for an oxidising agent: aqueous potassium iodide

Add a few drops of colourless KI(aq) to the unknown.

2I−(aq) → I2(aq) + 2e−

  • Positive: colourless to yellow-brown, as iodide ions are oxidised to iodine.
  • A lot of iodine may form a black solid. If the colour is too pale to judge, add starch: it turns dark blue with iodine.
  • For a gas, hold a strip of filter paper spotted with KI(aq) in it.
2Test for a reducing agent: acidified potassium manganate(VII)

Add a few drops of purple KMnO4(aq), acidified with dilute sulfuric acid, to the unknown.

MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l)

  • Positive: purple to colourless, as manganate(VII) ions are reduced to manganese(II) ions.
  • Acidified potassium dichromate(VI) works in the same way: orange to green.
  • For a gas, use filter paper spotted with the acidified solution.
KI(aq)+ oxidising agentcolourless → yellow-brownacidified KMnO₄+ reducing agentpurple → colourlessacidified K₂Cr₂O₇+ reducing agentorange → greentest for an oxidising agenttests for a reducing agent
Figure 6.5 Positive tests for oxidising and reducing agents.
Exam tipSome substances can act either way. Hydrogen peroxide oxidises iodide ions, but it reduces acidified manganate(VII) and is oxidised to oxygen. Iron(II) ions reduce manganate(VII), yet in FeSO4 + Mg → Fe + MgSO4 they are the oxidising agent. Decide the role from the reaction in front of you, not from a label.
Check your understanding
  1. Sulfur dioxide decolourises acidified potassium manganate(VII) and turns acidified potassium dichromate(VI) green. State whether each of the three substances is acting as an oxidising agent or a reducing agent.
  2. Iron(III) chloride solution oxidises iodide ions, and is reduced to iron(II). Write the ionic equation.

6.5Half-equations Core · A Level

A balanced redox equation must balance two things: the atoms and the charge. The surest way is to write the two half-equations, balance each, then add them.

Worked exampleWrite the equation for acidified manganate(VII) ions oxidising iodide ions: MnO4− + I− → Mn2+ + I2.

1. Reduction: MnO4− → Mn2+ (Mn +7 to +2). Oxidation: I− → I2 (I −1 to 0).
Split by the oxidation states that change.
2. MnO4− → Mn2+ and 2I− → I2
Balance atoms other than O and H.
3. MnO4− → Mn2+ + 4H2O
Add H2O to balance O.
4. MnO4− + 8H+ → Mn2+ + 4H2O
Add H+ to balance H.
5. MnO4− + 8H+ + 5e− → Mn2+ + 4H2O and 2I− → I2 + 2e−
Add electrons to balance charge: 7+ on the left becomes 2+.
6. Multiply the first by 2 and the second by 5: 10 electrons each.
Electrons lost must equal electrons gained.
7. 2MnO4− + 10I− + 16H+ → 2Mn2+ + 5I2 + 8H2O
Add, and cancel the electrons.

Check: the electrons in each half-equation equal the change in oxidation state. Mn changes by 5 and gains 5 electrons. Each I changes by 1, and two I atoms lose 2 electrons.

In alkaline solution, balance as if in acid, then add as many OH− ions as there are H+ ions to both sides. Each H+ + OH− becomes H2O, and spare water molecules cancel. In alkali, manganate(VII) is reduced only to MnO2:

2MnO4− + 6I− + 8H+ → 2MnO2 + 3I2 + 4H2O
add 8OH− to both sides: 2MnO4− + 6I− + 4H2O → 2MnO2 + 3I2 + 8OH−

These half-equations come up again and again. Learn them:

changehalf-equation
manganate(VII) in acidMnO4− + 8H+ + 5e− → Mn2+ + 4H2O
dichromate(VI) in acidCr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
iron(II) to iron(III)Fe2+ → Fe3+ + e−
iodide to iodine2I− → I2 + 2e−
thiosulfate to tetrathionate2S2O32− → S4O62− + 2e−
hydrogen peroxide as oxidising agentH2O2 + 2H+ + 2e− → 2H2O
hydrogen peroxide as reducing agentH2O2 → O2 + 2H+ + 2e−
ethanedioate to carbon dioxideC2O42− → 2CO2 + 2e−

6.5.1Finding an oxidation state from the amounts that react

Electrons lost always equal electrons gained. So for the atoms that change:

amount oxidised × rise in oxidation state = amount reduced × fall in oxidation state

Worked example30.0 cm3 of 0.0500 mol dm−3 iron(II) sulfate reacts exactly with 15.0 cm3 of 0.0250 mol dm−3 potassium manganate(VII). Iron ends as Fe3+. What is the final oxidation state of manganese?

1. nFe²⁺ = 0.0500 × 0.0300 = 1.500 × 10−3 mol. nMnO₄⁻ = 0.0250 × 0.0150 = 3.750 × 10−4 mol
n = c × V, volumes in dm3.
2. Each Fe2+ loses 1 electron: 1.500 × 10−3 mol of electrons
Fe +2 to +3.
3. Each Mn gains 1.500 × 10−3 ÷ 3.750 × 10−4 = 4 electrons
Electrons gained = electrons lost.
4. +7 − 4 = +3
Gaining electrons lowers the oxidation state.
Check your understanding
  1. Write the balanced equation for acidified dichromate(VI) ions oxidising iron(II) ions.
  2. 25.0 cm3 of 0.100 mol dm−3 NH2OH reacts exactly with 25.0 cm3 of 0.200 mol dm−3 Mn3+, which is reduced to Mn2+. Nitrogen in NH2OH is −1. What is its final oxidation state?

6.6Disproportionation Core · A Level

DefinitionDisproportionation is a redox reaction in which the same species is oxidised and reduced at the same time.

Chlorine reacts with aqueous sodium hydroxide: Cl2 + 2OH− → ClO− + Cl− + H2O. Half of the chlorine atoms rise from 0 to +1 in ClO−, and the other half fall from 0 to −1 in Cl− (Figure 6.6).

ClO⁻Cl is +1oxidised: 0 to +1Cl⁻Cl is −1reduced: 0 to −1Cl₂Cl is 0
Figure 6.6 Chlorine disproportionates in aqueous sodium hydroxide.

Two more examples:

  • 2H2O2 → 2H2O + O2: oxygen goes from −1 to −2 and to 0.
  • Cu2O + H2SO4 → CuSO4 + Cu + H2O: copper goes from +1 to +2 and to 0.
Exam tipTwo different elements changing, as in 2Pb(NO3)2 → 2PbO + 4NO2 + O2 (nitrogen reduced, oxygen oxidised), is ordinary redox, not disproportionation.
Check your understanding
  1. Is 2NO2 + H2O → HNO3 + HNO2 a disproportionation? Give the oxidation states of nitrogen.

6.7Redox titrations Core · A Level

A redox titration uses the same apparatus and method as an acid–base titration (Chapter 5). The difference is the reaction: electrons pass from the reducing agent to the oxidising agent, and the mole ratio comes from the balanced redox equation. Three kinds are common.

1Manganate(VII) titrations

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O

  • KMnO4 goes in the burette. It is a strong oxidising agent, used with Fe2+, H2O2 or ethanedioate.
  • No indicator is needed. Purple MnO4− is decolourised until all the reducing agent has reacted. The end-point is the first permanent pale pink from one excess drop.
  • The acid is dilute sulfuric acid. Hydrochloric acid is not used, because MnO4− would oxidise Cl− to Cl2. Nitric acid is not used, because it is itself an oxidising agent.
2Dichromate(VI) titrations

Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O

  • Orange to green, but the green appears long before the end-point, so an indicator is needed.
  • Dichromate(VI) is a weaker oxidising agent than manganate(VII). It does not oxidise Cl−, so hydrochloric or sulfuric acid can be used.
3Iodine–thiosulfate titrations

I2 + 2S2O32− → 2I− + S4O62−

  • An oxidising agent is first added to excess KI. It forms an equal amount of iodine every time, which is then titrated with standard sodium thiosulfate (Figure 6.7).
  • Excess KI makes sure all the oxidising agent reacts, and keeps the iodine dissolved.
  • The titre gives the amount of iodine present, and so the original amount of oxidising agent (e.g. 1 mol IO3− ≡ 3 mol I2 ≡ 6 mol S2O32−).
  • The brown iodine fades to pale yellow. Starch is then added, giving dark blue, and the end-point is when the blue just disappears. Starch is added only near the end-point, because at high iodine concentration the starch–iodine complex can come out as a solid.
oxidisingagent+ excess KIand acidbrowniodine formedtitrate withthiosulfatepale yellowmost I₂ usedadd starchdark bluestarch addedmorethiosulfatecolourlessend-point
Figure 6.7 The colours in an iodine–thiosulfate titration.

Worked example10.0 cm3 of hydrogen peroxide solution is diluted to 250 cm3. 25.0 cm3 of the diluted solution needs 14.60 cm3 of 0.0200 mol dm−3 acidified KMnO4. Find the concentration of the original hydrogen peroxide in g dm−3.

1. 2MnO4− + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
5 electrons × 2 and 2 electrons × 5.
2. nMnO₄⁻ = 0.0200 × 0.01460 = 2.920 × 10−4 mol, so nH₂O₂ = 5/2 × 2.920 × 10−4 = 7.300 × 10−4 mol
Ratio MnO4− : H2O2 = 2 : 5.
3. In 250 cm3: 7.300 × 10−3 mol, all from the 10.0 cm3 sample, so c = 7.300 × 10−3 ÷ 0.0100 = 0.7300 mol dm−3
Scale up, then divide by the original volume.
4. 0.7300 × 34.0 = 24.8 g dm−3
Mr(H2O2) = 34.0

Worked example0.875 g of potassium iodate(V), KIO3, is made up to 250 cm3. 25.0 cm3 is added to excess KI and dilute sulfuric acid: IO3− + 5I− + 6H+ → 3I2 + 3H2O. The iodine needs 22.70 cm3 of sodium thiosulfate. Find the concentration of the thiosulfate.

1. nKIO₃ = 0.875 ÷ 214.0 = 4.089 × 10−3 mol in 250 cm3, so 4.089 × 10−4 mol in 25.0 cm3
Mr(KIO3) = 214.0
2. nI₂ = 3 × 4.089 × 10−4 = 1.227 × 10−3 mol
1 IO3− gives 3I2.
3. nS₂O₃²⁻ = 2 × 1.227 × 10−3 = 2.453 × 10−3 mol
1 I2 reacts with 2S2O32−.
4. c = 2.453 × 10−3 ÷ 0.02270 = 0.108 mol dm−3
Each IO3− needs 6S2O32− in all.
Check your understanding
  1. 25.0 cm3 of an acidified iron(II) solution needs 20.00 cm3 of 0.0200 mol dm−3 KMnO4. Find the concentration of the iron(II) ions.
  2. 12.0 g of FeSO4·(NH4)2SO4·xH2O is made up to 250 cm3. 25.0 cm3 needs 25.50 cm3 of 0.0200 mol dm−3 K2Cr2O7. Find x.
  3. Why is no indicator needed in a manganate(VII) titration?

Answers to Check your understanding

6.1

  1. Na → Na+ + e− and Cl2 + 2e− → 2Cl−. Sodium is oxidised, because it loses electrons.
  2. Copper(II) oxide, because it loses oxygen.

6.2

  1. (a) 2x + 7(−2) = −2, so +6 (b) 2(+1) + 2x + 3(−2) = 0, so +2 (c) −1, because NaH is a metal hydride.
  2. VO2+: x − 2 = +2, so +4. VO2+: x − 4 = +1, so +5.

6.3

  1. (a) Not redox: nitrogen stays −3 and +5, hydrogen +1 and oxygen −2. (b) Not redox: copper stays +2, carbon +4 and oxygen −2. (c) Redox: nitrogen falls from +5 to +4, and some oxygen rises from −2 to 0.

6.4

  1. Sulfur dioxide is a reducing agent. Acidified potassium manganate(VII) and acidified potassium dichromate(VI) are oxidising agents, and both are reduced.
  2. 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq)

6.5

  1. Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
  2. nNH₂OH = 2.50 × 10−3 mol and nMn³⁺ = 5.00 × 10−3 mol, each gaining 1 electron. So each N loses 2 electrons: −1 + 2 = +1.

6.6

  1. Yes. Nitrogen is +4 in NO2, and goes up to +5 in HNO3 and down to +3 in HNO2.

6.7

  1. nMnO₄⁻ = 4.000 × 10−4 mol. MnO4− + 5Fe2+, so nFe²⁺ = 2.000 × 10−3 mol and c = 2.000 × 10−3 ÷ 0.0250 = 0.0800 mol dm−3.
  2. nCr₂O₇²⁻ = 5.100 × 10−4 mol, so nFe²⁺ = 3.060 × 10−3 mol in 25.0 cm3 and 0.03060 mol in 250 cm3. Mr = 12.0 ÷ 0.03060 = 392.2. 284.0 + 18.0x = 392.2, so x = 6.
  3. Purple manganate(VII) is decolourised as it reacts. The first excess drop leaves the solution pale pink, which marks the end-point.