Redox Reactions

— Chemistry · AS Level (7405) · Crash Course —

★ Exam Tip: Redox balancing is the concept most students never get clear — it shows up as MCQs in Paper 1 (find the coefficients) and sometimes in Paper 2. The other big source of lost marks: forgetting formulae of compounds/ions, so learn the 7 polyatomic ions cold.

1. Oxidation & Reduction

	Oxidation	Reduction
Always works ✓	loss of electrons increase in oxidation number	gain of electrons decrease in oxidation number
Sometimes only	gain of oxygen / loss of hydrogen	gain of hydrogen / loss of oxygen

Warning: the O / H way "might not always work" — it can even fail you. The concrete proof is always the change in oxidation number; write all explanations and answers in oxidation numbers, not in O/H terms.

Moving right on the oxidation-number line = oxidised; moving left = reduced

2. What is an Oxidation Number?

Definition (write exactly this in the exam):Oxidation number = the number of electrons an atom has gained control of or lost control of.

Remember: + = lost control of e⁻, − = gained control of e⁻. e.g. Al³⁺ → +3 ⇒ Al has lost control of 3 electrons.

It is NOT the same as charge — oxidation numbers exist in covalent compounds too, where nothing is actually an ion.

NaCl (ionic): e⁻ is fully transferred Na → Cl. Na = +1 (lost 1 e⁻), Cl = −1 (gained 1 e⁻).
HCl (covalent): pair is shared, but Cl has much larger electronegativity so it "bullies" the pair towards itself → Cl = −1 (gained control), H = +1 (lost control).

Same oxidation numbers in both — ionic transfer vs shared pair dragged towards the more electronegative atom

3. Rules for Oxidation Numbers

Eleven pointers — the teacher put a star on each, ★★ on rule 11. Periodic table = your best friend.

#	Rule	Examples / notes
1	Oxidation number of an element is 0	Na, Cl₂, O₂, H₂, Ca, F₂ — as long as it is in its elemental form (nothing bonded to it)
2	Oxidation number of an ion = its charge	+1 for Na⁺, −1 for Cl⁻, −2 for O²⁻, +3 for Al³⁺
3	Group 1 elements are always +1	K in KMnO₄ = +1
4	Group 2 elements are always +2	Ba in Ba(OH)₂ = +2
5	Group 3 (13) elements are always +3	Al = +3
6	Fluorine is always −1 (in compounds)	Most electronegative → wins the tug-of-war even against O. F₂ itself = 0 (rule 1)
7	Oxygen is −2 unless combined with F, or in a peroxide / superoxide	H₂O₂ (peroxide, 2 O joined) → O is not −2. O bonded to F becomes positive. Superoxide not in our course
8	Hydrogen is nearly always +1	Unless the examiner tells you otherwise

⑨ Neutral compoundΣ oxidation numbers of all atoms = 0
⑩ Complex (poly-atomic) ionΣ oxidation numbers of all atoms = charge on the ion

Trick: groups 14–17 give only usual values — but if there is no other way to find it, assume the usual one.

Group patterns — use the periodic table if you forget:

Groups 1, 2, 13, 18 = certain. Groups 14–17 = usual value only (F always −1)

Common mistake: not all of Group 17 is −1 — only F is always −1. Cl, Br, I are usually −1 (e.g. Cl is +1 in HClO, +5 in ClO₃⁻). Work it out whenever there is another way. Same for oxygen (H₂O₂) and H.

4. Important Polyatomic Ions ★★

★★ Memorise — no way round it: 7 names, 7 formulae, 7 charges. The examiner can ask formulae without telling you the ion. Poly-atomic = more than one atom (e.g. N + O) together forming an ion.

Ion	Formula	Charge	Compound you already know (to work it out)
nitrate	NO₃⁻	−1	nitric acid HNO₃
carbonate	CO₃²⁻	−2	H₂CO₃, CaCO₃ (limewater milky, marble, chips)
sulfate (sulphate)	SO₄²⁻	−2	sulfuric acid H₂SO₄
hydroxide	OH⁻	−1	NaOH, Ba(OH)₂
ammonium	NH₄⁺	+1	ammonium chloride NH₄Cl
hydrogencarbonate	HCO₃⁻	−1	sodium hydrogencarbonate NaHCO₃ (a little obscure — learn NaHCO₃)
phosphate	PO₄³⁻	−3	phosphoric acid H₃PO₄

Trick (forgot a charge?): take any compound you know that has the ion and work it out with rules 8 & 9. Don't panic!

Working out a forgotten charge (H is +1, compound neutral, ion as a whole = x):

Sulfate in H₂SO₄2(+1) + x = 0x = −2

Phosphate in H₃PO₄3(+1) + x = 0x = −3

Barium in Ba(OH)₂ (back-working)x + 2(−1) = 0 (OH⁻ = −1)x(Ba) = +2

5. Writing Formulae of Compounds

Metal + Roman numeral ⇒ that is the charge: Ti(IV) = Ti⁴⁺. (Metals always give away e⁻ so it is +.)
Non-metal: no other info ⇒ take the usual charge (Cl⁻, O²⁻); polyatomic ions from your memorised list.
Simplify the two numbers if you can (÷ common factor), then swap them — charge of one becomes the subscript of the other.
Polyatomic ion needing a number > 1 goes in brackets: Fe₂(SO₄)₃, Mg(NO₃)₂.

Ti(IV) oxide: simplify 4 & 2 first, then swap

Fe(III) sulfate: nothing simplifies → swap → brackets

Q. Write the formulae:

#	Compound	Ions	Formula
1	Titanium(IV) chloride	Ti⁴⁺ Cl⁻	TiCl₄
2	Titanium(IV) oxide	Ti⁴⁺ O²⁻	TiO₂ (simplified)
3	Vanadium(III) bromide	V³⁺ Br⁻	VBr₃
4	Vanadium(V) oxide	V⁵⁺ O²⁻	V₂O₅
5	Chromium(III) oxide	Cr³⁺ O²⁻	Cr₂O₃
6	Chromium(VI) oxide	Cr⁶⁺ O²⁻	CrO₃ (6 & 2 simplify to 3 & 1)
7	Iron(II) sulfate	Fe²⁺ SO₄²⁻	FeSO₄ (2+ and 2− cancel)
8	Copper(I) oxide	Cu⁺ O²⁻	Cu₂O
9	Copper(II) oxide	Cu²⁺ O²⁻	CuO
10	Iron(III) sulfate	Fe³⁺ SO₄²⁻	Fe₂(SO₄)₃
11	Sodium hydrogencarbonate	Na⁺ HCO₃⁻	NaHCO₃
12	Magnesium nitrate	Mg²⁺ NO₃⁻	Mg(NO₃)₂

Where students lose marks: (1) polyatomic ions — be observant, use brackets; (2) Group 2 salts — Group 1 salts are usually right, but Group 2 salts lose the "2" (Mg(NO₃)₂, Ba(OH)₂). Not sure? Do the working — it takes under a minute.

Sir's habit: he wrote his own usual mistakes on a paper, read it before the exam, and used spare time to check every answer for those mistakes. Keep a personal “my mistakes” list and do the same.

6. Neutralisation & Forming Salts

Exam questions make you write products, so you must get the salt formula right.

acid + base → salt + water (neutralisation)

Never forget: neutralisation is NEVER a redox reaction — no oxidation number changes.

① HCl + Ba(OH)₂ → ?

Water is easy. Salt = Ba²⁺ + Cl⁻ (Ba is Group 2 = +2; or back-work: x + 2(−1) = 0 ⇒ +2; Cl usually −1) ⇒ BaCl₂

HCl + Ba(OH)₂ → BaCl₂ + H₂OCl: 2 ⇒ 2HCl ; H, O: 2 H₂O2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

② H₂SO₄ + NaOH → ? (forgot sulfate's charge?)

H₂SO₄: 2(+1) + x = 0 ⇒ SO₄ = −2Na⁺ + SO₄²⁻ ⇒ Na₂SO₄H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

③ H₃PO₄ + Al₂O₃ → ? (Al₂O₃ acts as the base)

H₃PO₄: 3(+1) + x = 0 ⇒ PO₄ = −3Al₂O₃ back-worked: Al³⁺, O²⁻ ✓Al³⁺ + PO₄³⁻ ⇒ 3 & 3 simplify ⇒ AlPO₄2H₃PO₄ + Al₂O₃ → 2AlPO₄ + 3H₂Ocheck: Al 2=2, PO₄ 2=2, H 6=6, O 8+3 = 8+3 ✓

Note: NaOH + (NH₄)₂CO₃ — both are bases so no neutralisation. Ammonium salt + alkali releases NH₃ — that is done in the Nitrogen & Sulfur chapter (wait for it).

7. Calculating Oxidation Numbers

Method: ① call the unknown atom x ② write the usual value for every other atom (rules 3–8) ③ set the total = 0 (neutral) or = charge (ion) ④ solve the algebra.

Careful: for an ion the total equals the charge, not zero. The −2 on SO₃²⁻ is the whole-ion charge, not O's number. And don't slip in the algebra!

Q. Oxidation number of the underlined atom:

1) H in H₂O2x − 2 = 0x = +1

2) Cu in Cu₂O2x − 2 = 0x = +1

3) N in HNO₃+1 + x + 3(−2) = 0x = +5

4) S in H₂SO₄+2 + x + 4(−2) = 0x = +6

5) Cl in Cl₂O2x − 2 = 0x = +1

6) Cl in Cl₂O₇2x + 7(−2) = 0x = +7

7) Cl in NaClO+1 + x − 2 = 0x = +1

8) Mn in KMnO₄+1 + x + 4(−2) = 0x = +7

9) Cr in K₂Cr₂O₇2(+1) + 2x + 7(−2) = 0x = +6

10) S in Na₂S₂O₃2(+1) + 2x + 3(−2) = 0x = +2

11) Cl in KClO₃+1 + x + 3(−2) = 0x = +5

12) Cl in Cl₂element ⇒ x = 0

13) Mn in MnO₄²⁻x + 4(−2) = −2x = +6

14) Cr in Cr₂O₇²⁻2x + 7(−2) = −2x = +6

15) S in SO₃²⁻x + 3(−2) = −2x = +4

Pattern: the same atom can take many values — Cl: −1 (Cl⁻), 0 (Cl₂), +1 (HClO, NaClO), +5 (ClO₃⁻), +7 (Cl₂O₇). Mn: +7 (MnO₄⁻), +6 (MnO₄²⁻), +4 (MnO₂), +2 (Mn²⁺).

8. Redox: Who is Oxidised / Reduced?

Reducing agentA species which reduces other species in the reaction and gets oxidised itself.
Oxidising agentA species which oxidises other species in the reaction and gets reduced itself.

Trick: look at what happened to the species itself — the agent is the opposite of it. Oxidised ⇒ reducing agent. Reduced ⇒ oxidising agent.

Steps: work out the oxidation number of every element on both sides (elements = 0, ions = charge) ⇒ the one whose number goes up is oxidised, the one that goes down is reduced ⇒ no change = neither. Change = final − initial.

★ How to write it in the exam: give the species (just the element) and the reason in words: “Mn is reduced as its oxidation number decreased from +4 to +2.” Arrows (+4 → +2) are only for quick notes on the board — not allowed in answers. Naming the whole compound (MnO₂, HCl) as the agent is also acceptable.

Board Q ①identify oxidised / reduced / agents

MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Ox. nos: Mn in MnO₂: x + 2(−2) = 0 → +4; H +1; Cl in HCl −1; Mn in MnCl₂: x + 2(−1) = 0 → +2; Cl₂ = 0. H and O unchanged (+1, −2).

Oxidised: Cl — oxidation number increased from −1 to 0 Reduced: Mn — oxidation number decreased from +4 to +2 Oxidising agent: Mn (MnO₂) | Reducing agent: Cl (HCl)

Board Q ②ions & elements

Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂

Elements = 0, ions = charge.

Oxidised: Br — oxidation number −1 → 0 (increased) Reduced: Cl — oxidation number 0 → −1 (decreased) Oxidising agent: Cl | Reducing agent: Br

Board Q ③neutralisation → NOT redox

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

H +1 → +1, O −2 → −2, Na +1 → +1, S +6 → +6 (sulfate is −2 as a whole; S +6, O −2; hydroxide −1 as a whole). Nothing changes.

Not a redox reaction — so oxidised / reduced / agents: none. (Neutralisation is never redox.)

Board Q ④organic acid

Br₂(aq) + HCO₂H → 2HBr + CO₂

C in HCO₂H: x + 1 + 1 + 2(−2) = 0x = +2

Oxidised: C — +2 → +4 | Reduced: Br — 0 → −1 Oxidising agent: Br | Reducing agent: C

Note: when asked for a species, give the single element (Mn, Cl), not the whole compound — sir asked for elements in this exercise.

9. Half-Equations & Combining Them

Useful for A2 too, and for the MCQs now. These are examples to understand, not to memorise — once the method is clear, any redox equation can be written.

Rules• Oxidation = loss of e⁻ ⇒ electrons on the RIGHT (products)
• Reduction = gain of e⁻ ⇒ electrons on the LEFT (reactants)
• Ions are (aq); elements/molecules in their standard state (Cl₂ gas, Na solid, Br₂ liquid)

Why? More positive ⇒ push negative e⁻ out (lose). More negative ⇒ take e⁻ in (gain).

Combining half-equations: electrons must be equal before you add

① Oxidation of bromide ions by chlorine molecules

2Br⁻(aq) → Br₂(l) + 2e⁻ (bromide oxidised: loses e⁻ → Br₂ liquid, so 2Br⁻ needed)Cl₂(g) + 2e⁻ → 2Cl⁻(aq) (chlorine reduced, gains 1 e⁻ per Cl)2Br⁻ + Cl₂ + 2e⁻ → Br₂ + 2e⁻ + 2Cl⁻ (add reactants with reactants, products with products)2Br⁻(aq) + Cl₂(aq) → Br₂(aq) + 2Cl⁻(aq)

e⁻ are the same (2 = 2) so they cancel directly — no multiplying needed.

② Oxidation of sodium atoms by oxygen molecules

Na(s) → Na⁺(aq) + e⁻ × 4O₂(g) + 4e⁻ → 2O²⁻(aq) (each O needs 2 e⁻; two O ⇒ 4 e⁻)4Na + O₂ + 4e⁻ → 4Na⁺ + 2O²⁻ + 4e⁻4Na(s) + O₂(g) → 4Na⁺ + 2O²⁻

Watch out: when multiplying by 4, multiply EVERY term (Na, Na⁺, e⁻). Students often multiply only the first and then can't see why it won't balance. (Board typo: Na²⁺ was written — it should be Na⁺.)

③ Oxidation of chloride ions to chlorine by acidified manganate(VII) ions

Remember: oxidising agents like KMnO₄ (MnO₄⁻), K₂Cr₂O₇ / Na₂Cr₂O₇ (Cr₂O₇²⁻) must be acidified (H⁺ added) — they oxidise far better (strong oxidising agents need acid).

Oxidation: Cl⁻ loses e⁻, neutral Cl atoms don't exist alone — chlorine is Cl₂ ⇒ 2Cl⁻ → Cl₂ + 2e⁻

Reduction — build the MnO₄⁻ half-equation yourself:

1. Find the e⁻ neededMn in MnO₄⁻: x + 4(−2) = −1x = +7Mn: +7 → +2 ⇒ gains 5e⁻

2. Balance the restMnO₄⁻ + 5e⁻ → Mn²⁺4 O on left ⇒ 4H₂O on right8 H on right ⇒ 8H⁺ on left

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Charge check: LHS −1 + 8 − 5 = +2 ; RHS = +2 ✓

Charge check: total charge left = total charge right — a second way to confirm a balance (use overall charge, not individual oxidation numbers).

Why acidify? (sir's explanation) MnO₄⁻ and Cl⁻ are both negative so they repel each other; H⁺ helps the reaction go, and it acts as a charge balancer — the better the charge balance, the further the forward reaction goes. (Note: H⁺ is actually used up in the equation — 8H⁺ appear on the left — so in an answer say “H⁺ is needed / is a reactant”.)

Making the electrons equal: 2 e⁻ (Cl half) and 5 e⁻ (Mn half) → LCM = 10.

(2Cl⁻ → Cl₂ + 2e⁻) × 5(MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O) × 210Cl⁻ + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Cl₂ + 10e⁻ + 2Mn²⁺ + 8H₂O10Cl⁻(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) → 5Cl₂(g) + 2Mn²⁺(aq) + 8H₂O(l)

(Board slip: the combined line shows 4H₂O and one Mn²⁺ — after ×2 it must be 2Mn²⁺ + 8H₂O.)

10. Disproportionation

DefinitionDisproportionation = simultaneous oxidation and reduction of the same element (one species goes up and down in the same reaction). A special kind of redox.

Exam: asked to define ⇒ write the above; asked “is X disproportionating?” ⇒ work out ox. nos. of X on both sides.

④ Disproportionation of Cu(I) ions in solution

Cu⁺(aq) → Cu²⁺(aq) + e⁻ (oxidation: +1 → +2)Cu⁺(aq) + e⁻ → Cu(s) (reduction: +1 → 0)e⁻ already equal, cancel2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)

⑤ Disproportionation of chlorate(I) to chloride and chlorate(V)

Given half-equations without electrons: find the e⁻ from the oxidation numbers.

Oxidation: ClO⁻ → ClO₃⁻Cl in ClO⁻: x − 2 = −1 ⇒ +1Cl in ClO₃⁻: x + 3(−2) = −1 ⇒ +5+1 → +5 ⇒ lose 4e⁻ (right)balance O: 2H₂O ; then H: 4H⁺ClO⁻ + 2H₂O → ClO₃⁻ + 4H⁺ + 4e⁻

Reduction: ClO⁻ → Cl⁻Cl in Cl⁻ = −1+1 → −1 ⇒ gain 2e⁻ (left)O out: H₂O ; H in: 2H⁺ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O× 2 to make 4e⁻

3ClO⁻ + 4H⁺ + 4e⁻ + 2H₂O → ClO₃⁻ + 4H⁺ + 4e⁻ + 2Cl⁻ + 2H₂O3ClO⁻ → ClO₃⁻ + 2Cl⁻H⁺, e⁻ and H₂O cancel · also a disproportionation: Cl +1 → +5 (oxidised) and +1 → −1 (reduced)

Cu⁺: +1 → +2 and +1 → 0 (same species goes both ways)

ClO⁻: +1 → +5 and +1 → −1

Trap: “describe the type of reaction” does not always mean disproportionation! If two different elements are oxidised and reduced it is just redox (see the KI + CuSO₄ question below). Prove it with ox. nos. first.

To prove a disproportionationFind ox. nos. of every element → one species (the same element) is found to go up (oxidised) and down (reduced). Answers must name it: “N is oxidised from +4 to +5 and reduced from +4 to +2”.

Mn +7 → +2: five steps down = 5 e⁻ gained (+3 → +2 is the 5th)

Going up the scale = removing an e⁻ each step (4 steps = 4 e⁻)

11. Balancing Redox Equations

The hard bit. First decide whether to use redox (oxidation-number) balancing at all — often simple balancing is faster and safer.

Which method? — “Don't waste time”: these balancing questions are long, so pick fast

Two rules: ① use redox balancing only if (a) the examiner says so, or (b) a known oxidising / reducing agent is involved — oxidising: MnO₄⁻ / KMnO₄, Cr₂O₇²⁻ / K₂Cr₂O₇ (or Na₂Cr₂O₇); reducing: Fe²⁺ → Fe³⁺ (KI too, but obscure at AS/A2). ② It works only on redox reactions — never on neutralisation.

Method (using changes in oxidation number)

Find which species are oxidised and reduced (H and O usually stay −2/+1, so they are not the players).
Write incomplete half-equations for just those species: balance the atom (Fe & Fe, Mn & Mn) and the e⁻ count and side. Ignore H⁺, H₂O for now.
Multiply to make e⁻ equal, add, cancel e⁻.
Copy the coefficients you got into the full equation, then balance the rest (O with H₂O, H with H⁺, or just by counting) — now it is easy.

Mohr's salt (d)(i)structured · use ox. no. changes

MnO₄⁻(aq) + _Fe²⁺(aq) + _H⁺(aq) → _Mn²⁺(aq) + _Fe³⁺(aq) + _H₂O(l) — balance using changes in oxidation numbers. [1]

Fe (oxidised)Fe²⁺ → Fe³⁺ : more +, lose e⁻Fe²⁺ → Fe³⁺ + e⁻

Mn (reduced)Mn: x + 4(−2) = −1 ⇒ +7+7 → +2 ⇒ gain 5e⁻MnO₄⁻ + 5e⁻ → Mn²⁺

(Fe²⁺ → Fe³⁺ + e⁻) × 5MnO₄⁻ + 5e⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 5e⁻Now O: 4 O ⇒ 4H₂O ; H: 8 ⇒ 8H⁺

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

(ii) Role of Fe²⁺: reducing agent, as they reduce Mn from +7 to +2 [2]. (The titration calculation for x (0.784 g salt, 20.0 cm³ of 0.0200 mol dm⁻³ MnO₄⁻) was not worked in this class.)

No known agent / no instruction ⇒ balance simply

Golden rule: start with the element that appears in only ONE place on each side (e.g. H in the urea question) — it is the easiest to balance. Leave elements (O, N) that appear in 2–3 places till last. Then use the answer options to eliminate: an option that fails one relation is wrong immediately.

Tip: if a species is in several places, write an atom count for each (“stash”) — sir did it once, then showed it can be done in 2 minutes without.

If nothing is unique, do the algebra and use trial and error on the options. The examiner often builds options so that several of them pass the first check.

12. Past-Paper Questions

2(a)structured · KI + CuSO₄

Aqueous KI added to aqueous copper(II) sulfate ⇒ white precipitate of CuI in a brown solution of iodine + potassium sulfate.

(i) Balance + state symbols [2] — “precipitate” ⇒ (s); “brown solution” ⇒ I₂ dissolved ⇒ (aq); all solutions (aq).

1 CuSO₄(aq) + 2 KI(aq) → 1 CuI(s) + ½ I₂(aq) + 1 K₂SO₄(aq)

Why ½? 2 I on left, 1 already in CuI, so 1 I left ⇒ ½ I₂. Doubling everything (2, 4, 2, 1, 2) is equally right. Don't multiply by 3/2 etc. — that changes SO₄ and Cu too and you get stuck.

(ii) Cu in CuSO₄: sulfate is 2− ⇒ Cu = +2 (also “copper(II)” tells you). Cu in CuI: I is −1 ⇒ Cu = +1.

(iii) Type of reaction, in terms of electron transfer [2]:

Redox reaction. Cu gained an electron and went from +2 to +1. Iodine lost an electron and went from −1 to 0.

Not disproportionation — different elements (Cu, I) are reduced/oxidised, none does both.

Wording: “in terms of oxidation numbers” ⇒ +2 → +1 (Cu), −1 → 0 (I) is enough. “in terms of electron transfer” ⇒ say which species gained / lost electrons and how many, plus the numbers. “The examiner doesn't do kanjoosi (stinginess) with marks” — write it out in full.

(c)(i)(ii)structured · stage 3, nitric acid

3NO₂ + H₂O → 2HNO₃ + NO

N in NO₂: x + 2(−2) = 0 ⇒ +4N in HNO₃: +1 + x + 3(−2) = 0 ⇒ +5N in NO: x − 2 = 0 ⇒ +2

H (+1) and O (−2) don't change — only N is both oxidised and reduced.

(ii) Why disproportionation? [2]

Because simultaneous reduction and oxidation has occurred. Nitrogen in NO₂ was oxidised from +4 to +5 in HNO₃ by losing an electron, and also reduced from +4 in NO₂ to +2 in NO by gaining 2 electrons.

Change in oxidation number = final − initial: +5 − 4 = +1 ; +2 − 4 = −2.

(b)structured · chlorine in water

Cl₂(g) + H₂O → HCl + HClO

Cl₂ ⇒ 0 (element)HCl: +1 + x = 0 ⇒ −1HClO: +1 + x − 2 = 0 ⇒ +1

(ii) Type of reaction: disproportionation — Cl goes 0 → −1 (reduced) and 0 → +1 (oxidised) at the same time. It is redox, but “more specifically” disproportionation.

(iii) Why chlorine is used in water purification [1]:

It lowers the pH of the water and kills harmful micro-organisms as it forms HCl and HClO.

(c) “HCl + HClO added to cold dilute NaOH; a product behaves as bleach — equation?” — crossed out on the board: “will learn later” (Group 17 chapter). Skipped for now.

MCQ 1MCQ · greatest change in N oxidation number

Ammonia → nitric acid on commercial scale. In which reaction is the change in oxidation number of N greatest?

Rule: change = final − initial; compare magnitudes (a −2 is a change of 2). H = +1, O = −2 first; then N.

A ✓ 4NH₃ + 5O₂ → 4NO + 6H₂O : N −3 → +2 = 5 B ✗ 3NO₂ + H₂O → 2HNO₃ + NO : +4 → +5 (+1) and +4 → +2 (−2) ⇒ 1, 2 C ✗ 2NO + O₂ → 2NO₂ : +2 → +4 = 2 (not 6 — the +2 is N's, −2 is O's) D ✗ 4NH₃ + 6NO → 5N₂ + 6H₂O : −3 → 0 (+3) and +2 → 0 (−2) ⇒ 3, 2

Answer: A (biggest magnitude, 5)

MCQ 2MCQ · urea + NO₂ · balance without redox

2CON₂H₄ + xNO₂ → 2CO₂ + yH₂O + zN₂ — find x, y, z. (No hint to use redox, no known agent ⇒ balance normally.)

C: 2 = 2 ✓. O and N are in many places ⇒ leave. H is in only one place each side ⇒ start there.

H: 2 × 4 = 8 = 2yy = 4 (kills option A)O: 2 + 2x = 2(2) + y = 8x = 3 (kills B)N: 4 + x = 7 = 2zz = 3.5

Answer: C (3, 4, 3½)

MCQ 5MCQ · ammonia oxidation

wNH₃ + xO₂ → yNO + zH₂O

N: w = y (all options agree ⇒ no help)H: 3w = 2zz = 3w/2Test each: 4 → 6 ✓ ; 5 → 7.5 ✗ ; 6 → 9 ✗

Options B, C, D fail the z = 3w/2 test ⇒ only A left (no further work — speed matters). Full: O: 2x = y + z = 4 + 6 = 10 ⇒ x = 5 ✓ (board: 2x = w + 3w/2).

A ✓ 4, 5, 4, 6 — z = 6 = 3(4)/2B ✗ 4, 6, 4, 5 — z = 5 ≠ 6C ✗ 5, 6, 5, 4 — z = 4 ≠ 7.5D ✗ 6, 5, 6, 4 — z = 4 ≠ 9

Answer: A

MCQ 10MCQ · HIO disproportionation (use ox. numbers)

mHIO → nI₂ + pHIO₃ + qH₂O — find n, p. Examiner tells you to use oxidation numbers.

I: HIO: +1 + x − 2 = 0 ⇒ +1I in HIO₃: +1 + x + 3(−2) = 0 ⇒ +5I₂ = 0+1 → 0 : gain 1 e⁻ per I ⇒ 2HIO + 2e⁻ → I₂ × 2+1 → +5 : lose 4 e⁻ ⇒ HIO → HIO₃ + 4e⁻4HIO + 4e⁻ + HIO → 2I₂ + HIO₃ + 4e⁻5HIO → 2I₂ + HIO₃ (+ 2H₂O)

Only I matters — ignore H and O ("don't bother"); they balance last: H: 5 = 1 + 4 (2H₂O), O: 5 = 3 + 2 ✓.

Answer: B (n = 2, p = 1) — full: 5HIO → 2I₂ + HIO₃ + 2H₂O

MCQ 22MCQ · combine given half-equations

IO₃⁻ + 2Cl⁻ + 6H⁺ + 4e⁻ → ICl₂⁻ + 3H₂O and I₂ + 4Cl⁻ → 2ICl₂⁻ + 2e⁻. Ratio IO₃⁻ : I₂ in overall equation?

e⁻: 4 in the first, 2 in the second2nd × 2 : 2I₂ + 8Cl⁻ → 4ICl₂⁻ + 4e⁻Add: IO₃⁻ + 2I₂ + 10Cl⁻ + 6H⁺ → 5ICl₂⁻ + 3H₂Oratio IO₃⁻ : I₂ = 1 : 2

Answer: B (1 : 2)

Easy one: make e⁻ equal and read off the coefficients.

MCQ 37MCQ · K₂MnO₄ in water (redox path)

aMnO₄²⁻ + bH₂O → cMnO₄⁻ + dMnO₂ + eOH⁻ — find a and c. MnO₄ is a known oxidising-agent ion ⇒ use redox balancing.

Mn: MnO₄²⁻ x + 4(−2) = −2 ⇒ +6 ; MnO₄⁻ ⇒ +7 ; MnO₂ ⇒ +4+6 → +7 : lose 1 e⁻ ⇒ MnO₄²⁻ → MnO₄⁻ + e⁻ × 2+6 → +4 : gain 2 e⁻ ⇒ MnO₄²⁻ + 2e⁻ + 2H₂O → MnO₂ + 4OH⁻2MnO₄²⁻ + MnO₄²⁻ + 2e⁻ + 2H₂O → 2MnO₄⁻ + 2e⁻ + MnO₂ + 4OH⁻

Class mistake (even sir's old video): everyone said A — they forgot to ADD the MnO₄²⁻ from both half-equations (it's a reactant in both): 2 + 1 = 3, not 2.

3MnO₄²⁻ + 2H₂O → 2MnO₄⁻ + MnO₂ + 4OH⁻ ⇒ a = 3, c = 2 ⇒ Answer: B

Check: O 12 + 2 = 14 = 8 + 2 + 4 ✓ ; H 4 = 4 ✓ ; charge −6 = −2 − 4 ✓. (Board had A circled — that's the slip; correct is B.)

Balance by counting: H: 2b = e ⇒ e = 2b ; O: 12 + b = 8 + 2 + 2b ⇒ b = 2, e = 4.

MCQ 31MCQ · AsCl₃ + NaBH₄ (algebra)

pAsCl₃ + qNaBH₄ → rAsH₃ + sNaCl + tBCl₃ — find p, q, r, s, t.

Elements in one place each side: As ⇒ p = r; Na ⇒ q = s (Cl is in two places — skip); B ⇒ q = t. Check the options against these relations:

A ✗ 2, 3, 2, 3, 1 — q ≠ tB ✗ 3, 3, 3, 3, 2 — q ≠ tC ✓ 4, 3, 4, 3, 3 — p = r, q = s = tD ✗ 4, 4, 4, 4, 3 — q ≠ t

Board algebra: H: 4q = 3r = 3pq = 3p/4Cl: 3p = s + 3t = 3p/4 + 3tt = 3p/4 (= q ✓)

Answer: C (4, 3, 4, 3, 3)

MCQ 36MCQ · Cu + HNO₃ (longer, algebra)

pCu + qHNO₃ → rCu(NO₃)₂ + sH₂O + tNO — find p, q, r, s, t.

Cu: p = r. H (one place each side): q = 2s. N: q = 2r + t. O: 3q = 6r + s + t. Every option passes the H and N relations (the examiner set the trap) — so bring in O and simplify:

O: 3q = 6r + s + t , with q = 2s6s = 6r + s + tt = 5s − 6rTest: A 5(2)−6(1) = 4 ✗ ; B 5(3)−6(2) = 3 ✗ ; C 5(4)−6(2) = 8 ✗D: 5(4) − 6(3) = 2 ✓ (t = 2)

Answer: D (3, 8, 3, 4, 2) — 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO

Lesson: when several options survive the easy checks, simplify the harder relation (here t = 5s − 6r) and test again.

13. Quick Recap — Full Chapter Cheat-Sheet

✓ Definitions

Oxidation = loss of e⁻ / ↑ ox. no. | Reduction = gain of e⁻ / ↓ ox. no. (O/H versions only sometimes work)
Oxidation number = number of electrons an atom has gained control of or lost control of (+ lost, − gained; ≠ charge)
Reducing agent: reduces others, itself oxidised. Oxidising agent: oxidises others, itself reduced.
Disproportionation = simultaneous oxidation and reduction of the same element.
Change in ox. no. = final − initial; neutralisation is never redox.

✓ Oxidation-number rules

Element = 0 · simple ion = charge · Grp 1 +1 · Grp 2 +2 · Grp 13 +3 · Grp 18 = 0 (always)
F = −1 always · O = −2 (not in H₂O₂ or with F) · H = +1 (nearly) · Grp 17 usually −1 · Grp 16 usually −2 · Grp 15 usually −3 · Grp 14 usually +4
Neutral compound Σ = 0 · ion Σ = charge (solve for x)
Ions ★★: NO₃⁻ · CO₃²⁻ · SO₄²⁻ · OH⁻ · NH₄⁺ · HCO₃⁻ · PO₄³⁻ — work forgotten ones out from H₂SO₄, H₃PO₄, HNO₃, NaHCO₃, NH₄Cl
Formula: Roman numeral = charge → simplify → swap → brackets for polyatomic (Fe₂(SO₄)₃, Mg(NO₃)₂)

✓ Half-equations & balancing

Oxidation: e⁻ on the right · reduction: e⁻ on the left · make e⁻ equal (LCM, multiply everything) · add · cancel
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O · 10Cl⁻ + 2MnO₄⁻ + 16H⁺ → 5Cl₂ + 2Mn²⁺ + 8H₂O
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O · 2Cu⁺ → Cu²⁺ + Cu · 3ClO⁻ → ClO₃⁻ + 2Cl⁻ · 2Br⁻ + Cl₂ → Br₂ + 2Cl⁻ · 4Na + O₂ → 4Na⁺ + 2O²⁻
Oxidising agents (MnO₄⁻, Cr₂O₇²⁻) must be acidified · use redox balancing only if the examiner says so or a known agent (MnO₄⁻, Cr₂O₇²⁻, Fe²⁺) is present · never on non-redox
Otherwise: balance the element that appears in one place each side first; eliminate MCQ options quickly

✓ Exam wording

Name the species (element) and give the reason in words: “as its oxidation number decreased from … to …” (no arrows)
“In terms of electron transfer” ⇒ say who gained / lost electrons and how many (+ the numbers)
“Type of reaction” ⇒ check ox. nos. — redox vs disproportionation vs neither (neutralisation)