Redox Chemistry

Singapore Junior Chemistry Olympiad

By the end of this chapter, you can answer

1

Rusting and burning both use oxygen. What do they have in common with a reaction that has no oxygen at all?

2

Water has no ions. How can its hydrogen be +1 and its oxygen −2?

3

Which reactions are not redox, and how can you tell in seconds?

4

How do potassium iodide and acidified potassium manganate(VII) show whether something is an oxidising or a reducing agent?

5

How do you balance an equation whose charges do not match?

6

Why does a manganate(VII) titration need no indicator, while an iodine titration needs starch?

Recall: Chapters 3 to 5

1What is the charge on a magnesium ion, and why?

2+: magnesium loses its two valence electrons to reach an octet.

2In an O–H bond, which atom attracts the shared electrons more strongly?

Oxygen: it is more electronegative than hydrogen.

3How many moles are in 20.0 cm³ of 0.100 mol dm⁻³ solution?

0.100 × 0.0200 = 2.00 × 10⁻³ mol

1

Oxidation and Reduction

Blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂

Fe₂O₃ loses oxygen: reduced
CO gains oxygen: oxidised

Both happen at the same time: a redox reaction.

	oxidation	reduction
oxygen	gain	loss
hydrogen	loss	gain
electrons	loss	gain
oxidation state	increase	decrease

1

Oxidation and Reduction

Zinc in copper(II) sulfate

The zinc gets coated with brown copper and the blue colour fades.

Zn → Zn²⁺ + 2e⁻

Cu²⁺ + 2e⁻ → Cu

OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.

1

Oxidation and Reduction

Your turnSodium burns in chlorine: 2Na + Cl₂ → 2NaCl. Write the two half-equations and state which reactant is oxidised.

Na → Na⁺ + e⁻ and Cl₂ + 2e⁻ → 2Cl⁻[1]

Sodium is oxidised: it loses electrons.[1]

Your turnIn CuO + H₂ → Cu + H₂O, which substance is reduced? Explain in terms of oxygen.

Copper(II) oxide: it loses oxygen.[1]

2

Oxidation States

In a covalent bond no electrons are fully transferred. So pretend the bond is ionic.

AnalogyA tug of war over each shared pair.

The more electronegative atom wins the tug of war and takes both shared electrons.

The charge each atom would then have is its oxidation state: O is −2, each H is +1.

Written sign first (+2, −1). A charge is written sign last (2+, 1−).

2

Oxidation States

The rules (an earlier rule wins)

Uncombined element: 0
Simple ion: its charge
Total: 0 in a compound, the charge in an ion
Group 1 +1, Group 2 +2, Al +3
F always −1
H +1, but −1 in metal hydrides (NaH)
O −2, but −1 in peroxides (H₂O₂), +2 in OF₂
Cl −1, unless bonded to O or F

Worked exampleFind the oxidation state of Mn in MnO₄⁻ and of each N in NH₄NO₃.

x + 4(−2) = −1

the ion totals −1

Mn = +7

so KMnO₄ is manganate(VII)

NH₄⁺: N = −3

p + 4 = +1

NO₃⁻: N = +5

q − 6 = −1

2

Oxidation States

One element can take many oxidation states. Nitrogen runs from −3 to +5.

Moving up the scale is oxidation. Moving down is reduction.

Roman numerals in a name give the state: iron(III) is +3, manganate(VII) contains Mn at +7.

Going furtherIn S₄O₆²⁻, 4x − 12 = −2 gives +2.5: an average over sulfur atoms that are bonded differently.

2

Oxidation States

Your turnFind the oxidation state of (a) Cr in Cr₂O₇²⁻, (b) S in Na₂S₂O₃, (c) H in NaH, (d) V in VO₂⁺.

(a)

+6[1]

(b)

+2[1]

(c)

−1, a metal hydride[1]

(d)

+5[1]

Quick checkIn which species does nitrogen have an oxidation state of +3?

ANO₂

BNO₂⁻

CNH₄⁺

DN₂O

NoteB: x + 2(−2) = −1, so x = +3. NO₂ is +4, NH₄⁺ is −3 and N₂O is +1.

3

Is It a Redox Reaction?

Worked exampleCl₂(g) + 2KI(aq) → 2KCl(aq) + I₂(aq). What is oxidised and what is reduced?

Not redox: every ion keeps its charge

Acid–base: H⁺ + OH⁻ → H₂O
Precipitation: Ag⁺ + Cl⁻ → AgCl

Exam tipAn element uncombined on one side and combined on the other has changed from 0: redox.

3

Is It a Redox Reaction?

Your turnWhich are redox? Explain using oxidation states.
(a) NH₃ + HNO₃ → NH₄NO₃
(b) CuCO₃ → CuO + CO₂
(c) 2Cu(NO₃)₂ → 2CuO + 4NO₂ + O₂

(a)

Not redox: N stays −3 and +5[1]

(b)

Not redox: Cu +2, C +4, O −2 throughout[1]

(c)

Redox: N +5 to +4, some O −2 to 0[1]

Quick checkWhich reaction is not a redox reaction?

AMg + 2HCl → MgCl₂ + H₂

BCaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

CCl₂ + 2KBr → 2KCl + Br₂

DCuO + H₂ → Cu + H₂O

NoteB: an acid with a carbonate. Ca stays +2, C +4, O −2, H +1, Cl −1. Each of the others has an uncombined element.

4

Oxidising and Reducing Agents

DefinitionAn oxidising agent takes electrons, so it is itself reduced. A reducing agent gives electrons, so it is itself oxidised.

4

Oxidising and Reducing Agents

oxidising agent	becomes	reducing agent	becomes
O₂, Cl₂	O²⁻, Cl⁻	H₂, C, CO, reactive metals	H⁺, CO₂, metal ions
MnO₄⁻ in acid (purple)	Mn²⁺ (looks colourless)	I⁻ (colourless)	I₂ (yellow-brown)
Cr₂O₇²⁻ in acid (orange)	Cr³⁺ (green)	Fe²⁺ (pale green)	Fe³⁺ (yellow)
H₂O₂ in acid	H₂O	SO₂, SO₃²⁻	SO₄²⁻

Exam tipSome substances act either way. H₂O₂ oxidises I⁻ but reduces MnO₄⁻. Fe²⁺ reduces MnO₄⁻, but in FeSO₄ + Mg → Fe + MgSO₄ it is the oxidising agent. Judge the reaction, not the label.

4

Oxidising and Reducing Agents

1Oxidising agent? Add KI(aq)

2I⁻ → I₂ + 2e⁻

Positive: colourless to yellow-brown. Starch turns dark blue with iodine.

2Reducing agent? Add acidified KMnO₄

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Positive: purple to colourless. Acidified K₂Cr₂O₇: orange to green.

4

Oxidising and Reducing Agents

Your turnSulfur dioxide decolourises acidified KMnO₄ and turns acidified K₂Cr₂O₇ green. State the role of each of the three substances.

SO₂: reducing agent[1]

KMnO₄ and K₂Cr₂O₇: oxidising agents, both reduced[1]

Your turnIron(III) ions oxidise iodide ions and become iron(II). Write the ionic equation.

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂[1]

Quick checkA colourless gas turns a filter paper spotted with KI(aq) yellow-brown. What can be deduced?

AThe gas is an oxidising agent.

BThe gas is a reducing agent.

CThe iodide ions are reduced.

DThe gas is iodine.

NoteA: iodide ions lose electrons to form iodine, so they are oxidised, and the gas that took the electrons is an oxidising agent.

5

Half-EquationsCore · A Level

Worked exampleBalance MnO₄⁻ + I⁻ → Mn²⁺ + I₂ in acid.

MnO₄⁻ → Mn²⁺ 2I⁻ → I₂

1. split, balance atoms other than O and H

MnO₄⁻ → Mn²⁺ + 4H₂O

2. add H₂O to balance O

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

3. add H⁺ to balance H

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 2I⁻ → I₂ + 2e⁻

4. add electrons to balance charge

× 2 and × 5: 10 electrons each

5. electrons lost = electrons gained

2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O

5

Half-EquationsCore · A Level

Check, then alkaline

Electrons in a half-equation = change in oxidation state. Mn: +7 to +2, so 5 electrons.

In alkali: balance as in acid, then add as many OH⁻ as there are H⁺ to both sides. H⁺ + OH⁻ = H₂O.

2MnO₄⁻ + 6I⁻ + 8H⁺ → 2MnO₂ + 3I₂ + 4H₂O

2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻

Learn these

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Fe²⁺ → Fe³⁺ + e⁻ 2I⁻ → I₂ + 2e⁻

2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻

H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O H₂O₂ → O₂ + 2H⁺ + 2e⁻

C₂O₄²⁻ → 2CO₂ + 2e⁻

5

Half-EquationsCore · A Level

Worked example30.0 cm³ of 0.0500 mol dm⁻³ Fe²⁺ reacts with 15.0 cm³ of 0.0250 mol dm⁻³ MnO₄⁻. Fe ends as Fe³⁺. Find the final oxidation state of Mn.

n_Fe²⁺ = 1.500 × 10⁻³ mol, each losing 1 electron
n_MnO₄⁻ = 3.750 × 10⁻⁴ mol
electrons per Mn = 1.500 × 10⁻³ ÷ 3.750 × 10⁻⁴ = 4
+7 − 4 = +3

amount oxidised × rise = amount reduced × fall

Your turnWrite the equation for acidified Cr₂O₇²⁻ oxidising Fe²⁺.

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O[2]

Your turn25.0 cm³ of 0.100 mol dm⁻³ NH₂OH (N is −1) reacts with 25.0 cm³ of 0.200 mol dm⁻³ Mn³⁺, which becomes Mn²⁺. Find the final oxidation state of N.

2 electrons per N: +1[2]

6

DisproportionationCore · A Level

DefinitionIn disproportionation the same species is oxidised and reduced at the same time.

Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O

2H₂O₂ → 2H₂O + O₂: O −1 to −2 and to 0

Cu₂O + H₂SO₄ → CuSO₄ + Cu + H₂O: Cu +1 to +2 and to 0

6

DisproportionationCore · A Level

Your turnIs 2NO₂ + H₂O → HNO₃ + HNO₂ a disproportionation? Give the oxidation states of nitrogen.

Yes: N is +4 in NO₂, +5 in HNO₃ and +3 in HNO₂.[2]

Quick checkWhich reaction is a disproportionation?

A2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

BCl₂ + 2NaOH → NaCl + NaClO + H₂O

CZn + CuSO₄ → ZnSO₄ + Cu

DMgO + 2HCl → MgCl₂ + H₂O

NoteB: chlorine goes from 0 to −1 and to +1. In A two different elements change (N down, O up), C is ordinary redox and D is not redox.

7

Redox TitrationsCore · A Level

1Manganate(VII): no indicator

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

KMnO₄ in the burette is decolourised until the reducing agent is used up. End-point: the first permanent pale pink.

Acidify with dilute H₂SO₄. Not HCl (Cl⁻ is oxidised to Cl₂), not HNO₃ (an oxidising agent).

2Dichromate(VI): needs an indicator

Orange to green, but the green shows long before the end-point. Weaker than MnO₄⁻, so HCl can be used.

3Iodine–thiosulfate: starch near the end

I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

Oxidising agent + excess KI gives I₂, then titrate with thiosulfate.

7

Redox TitrationsCore · A Level

Excess KI makes sure all the oxidising agent reacts and keeps the iodine dissolved.

Starch goes in only near the end: at high iodine concentration the starch–iodine complex can come out as a solid.

7

Redox TitrationsCore · A Level

Worked example10.0 cm³ of H₂O₂ solution is diluted to 250 cm³. 25.0 cm³ needs 14.60 cm³ of 0.0200 mol dm⁻³ acidified KMnO₄. Find the original concentration in g dm⁻³.

2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O
n_MnO₄⁻ = 2.920 × 10⁻⁴ mol
n_H₂O₂ = 5/2 × 2.920 × 10⁻⁴ = 7.300 × 10⁻⁴ mol

in 250 cm³: 7.300 × 10⁻³ mol, all from 10.0 cm³
c = 0.7300 mol dm⁻³ × 34.0 = 24.8 g dm⁻³

7

Redox TitrationsCore · A Level

Worked example0.875 g of KIO₃ is made up to 250 cm³. 25.0 cm³ + excess KI + acid: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. The iodine needs 22.70 cm³ of thiosulfate. Find its concentration.

n_KIO₃ in 25.0 cm³ = 0.875 ÷ 214.0 ÷ 10 = 4.089 × 10⁻⁴ mol
n_I₂ = 3 × that = 1.227 × 10⁻³ mol
n_{S₂O₃²⁻} = 2 × that = 2.453 × 10⁻³ mol
c = 2.453 × 10⁻³ ÷ 0.02270 = 0.108 mol dm⁻³

Your turn25.0 cm³ of acidified Fe²⁺(aq) needs 20.00 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find [Fe²⁺].

n_MnO₄⁻ = 4.000 × 10⁻⁴ mol, so n_Fe²⁺ = 2.000 × 10⁻³ mol[1]

[Fe²⁺] = 0.0800 mol dm⁻³[1]

7

Redox TitrationsCore · A Level

Challenge12.0 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.

n_{Cr₂O₇²⁻} = 5.100 × 10⁻⁴ mol, so n_Fe²⁺ = 3.060 × 10⁻³ mol in 25.0 cm³[1]

0.03060 mol in 250 cm³, so M_r = 392.2[1]

284.0 + 18.0x = 392.2, so x = 6[1]

Quick checkHow many moles of MnO₄⁻ react with 1.0 × 10⁻³ mol of Fe²⁺ in acid?

A2.0 × 10⁻⁴

B5.0 × 10⁻⁴

C1.0 × 10⁻³

D5.0 × 10⁻³

NoteA: each MnO₄⁻ takes 5 electrons and each Fe²⁺ gives 1, so MnO₄⁻ : Fe²⁺ = 1 : 5. Option D multiplies instead of dividing.

Exit ticket

1Define oxidation in terms of electrons.

Oxidation is the loss of electrons.[1]

2What is the oxidation state of sulfur in H₂SO₄?

+6[1]

3What colour change shows a reducing agent with acidified KMnO₄?

Purple to colourless[1]

4An oxidising agent is itself oxidised or reduced?

Reduced: it gains electrons.[1]