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State Board · Class 11 · Chemistry

Lesson 6: Redox Reactions

7 concepts · 56 practice questions with answers · Free to read

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Ideas of oxidation and reduction: classical and electronic

Link to Std 9–10 : In Std 8–9 oxidation meant gaining oxygen; here electron transfer makes the idea work for every reaction.

From daily life

Iron rusts (gains oxygen) and a cut apple turns brown; both are oxidation.

Simple definition

Classical: gain of O / loss of H = oxidation; loss of O / gain of H = reduction. Electronic: loss of e⁻ = oxidation; gain of e⁻ = reduction (OIL RIG). An oxidising agent oxidises another and is itself reduced; a reducing agent does the opposite.

Oxidation and reductionZn + Cu²⁺ → Zn²⁺ + CuOXIDATIONZn → Zn²⁺ + 2e⁻loses e⁻ · O.N. 0 → +2 ↑REDUCTIONCu²⁺ + 2e⁻ → Cugains e⁻ · O.N. +2 → 0 ↓OIL RIG : Oxidation Is Loss, Reduction Is Gain (e⁻)Zn = reducing agent · Cu²⁺ = oxidising agentBoth half-reactions occur together (redox)

In the figure : Zn + Cu²⁺: two half-reactions together.

Rule 1: Classical concepts — Addition or removal of oxygen, hydrogen, or an electronegative / electropositive element.

  • Easy : 2Mg + O₂ → 2MgO: Mg gains O ⇒ oxidation.
  • Medium : H₂S + Cl₂ → 2HCl + S: H is removed from H₂S ⇒ oxidation.
  • Exam-type : 2HgO → 2Hg + O₂: HgO is reduced (O removed).

Rule 2: Electronic concept — Oxidation: M → Mⁿ⁺ + ne⁻; reduction: X + ne⁻ → Xⁿ⁻. They always occur together (redox).

  • Easy : Na → Na⁺ + e⁻ (oxidation).
  • Medium : Cl₂ + 2e⁻ → 2Cl⁻ (reduction).
  • Exam-type : Fe³⁺ + e⁻ → Fe²⁺ is reduction; Fe²⁺ → Fe³⁺ + e⁻ is oxidation.

Rule 3: Oxidising and reducing agents — An oxidising agent takes e⁻ (O₂, Cl₂, KMnO₄, K₂Cr₂O₇, H₂O₂); a reducing agent gives e⁻ (Na, Zn, H₂, I⁻, Fe²⁺).

  • Easy : Zn + Cu²⁺ → Zn²⁺ + Cu: Zn is the reducing agent, Cu²⁺ the oxidising agent.
  • Medium : Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂: Cl₂ is the oxidising agent.
  • Exam-type : The species that is itself oxidised is the reducing agent.

Derivation: half-reactions of Zn + CuSO₄

  • Step 1: Ionic equation: Zn + Cu²⁺ → Zn²⁺ + Cu (SO₄²⁻ is a spectator).
  • Step 2: Oxidation: Zn → Zn²⁺ + 2e⁻; reduction: Cu²⁺ + 2e⁻ → Cu.
  • Step 3: Electrons equal (2) ⇒ sum gives the original equation ✓.

CET level (MCQ)

  • CET 1) What is reduced in 2Mg + O₂ → 2MgO? 1) Mg 2) O₂ 3) MgO 4) none — Answer: 2). (O₂ gains e⁻.)
  • CET 2) A reducing agent is one that? 1) accepts e⁻ 2) donates e⁻ 3) loses H 4) gains O — Answer: 2).

Real-life use: Redox runs batteries, respiration, rusting and bleaching; KMnO₄ solution is a disinfectant.

Remember

OIL RIG: Oxidation Is Loss, Reduction Is Gain (e⁻); an oxidising agent is itself reduced.

Common mistake

Writing that an oxidising agent is oxidised; it is reduced.

CET trap

'Oxidising agent' and 'species oxidised' are opposites; CET asks both.

Shortcut

Memory clue: "OIL RIG" and "LEO says GER".

Spot the difference

ItemOxidationReduction
e⁻LossGain
O.N.IncreasesDecreases
OGainedLost

Practice questions

Easy
  1. 1Give the electronic definition of oxidation.
  2. 2Explain OIL RIG.
  3. 3What is an oxidising agent?
Medium
  1. 4Identify the oxidising and reducing agents in Zn + CuSO₄.
  2. 5What is oxidised in H₂S + Cl₂ → 2HCl + S?
  3. 6Why is Fe²⁺ → Fe³⁺ oxidation?
Challenging
  1. 7Write the half-reactions of Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.
  2. 8Why do oxidation and reduction occur together?
Show answers (try them yourself first)
  1. 1Loss of electrons.
  2. 2Oxidation Is Loss, Reduction Is Gain.
  3. 3One that oxidises another and is itself reduced.
  4. 4Zn is the reducing agent; Cu²⁺ the oxidising agent.
  5. 5H₂S (it loses H).
  6. 6It loses an electron.
  7. 72Br⁻ → Br₂ + 2e⁻; Cl₂ + 2e⁻ → 2Cl⁻.
  8. 8The electrons given must be taken by something.

Oxidation number: rules and calculation

From daily life

As scores are tallied with +/− signs, an O.N. records the 'imaginary charge' of an atom.

Simple definition

The oxidation number (O.N.) is the hypothetical charge an atom would have if all bonding electrons went to the more electronegative atom.

Oxidation numbersFinding O.N.: sum = total chargeH₂SO₄2(+1) + S + 4(−2) = 0S = +6KMnO₄(+1) + Mn + 4(−2) = 0Mn = +7Cr₂O₇²⁻2Cr + 7(−2) = −2Cr = +6Fe₃O₄3Fe + 4(−2) = 0Fe = +8/3H₂O₂2(+1) + 2O = 0O = −1H = +1 (−1 in hydrides) · O = −2 (−1 in peroxides) · F = −1 alwaysfree element = 0 · group 1 = +1 · group 2 = +2

In the figure : Finding O.N. in five compounds.

Rule 1: Basic rules — Free element 0; monatomic ion = its charge; sum is 0 in a compound; in a polyatomic ion sum = charge; group 1 = +1, group 2 = +2, F = −1.

  • Easy : O₂, S₈, Fe: 0.
  • Medium : NaCl: Na +1, Cl −1.
  • Exam-type : SO₄²⁻: S + 4(−2) = −2 ⇒ S = +6.

Rule 2: Exceptions for H and O — H is usually +1, −1 in metal hydrides; O is usually −2, −1 in peroxides, −½ in superoxides, +2 in OF₂.

  • Easy : H₂O₂: O = −1.
  • Medium : CaH₂: H = −1.
  • Exam-type : OF₂: O = +2.

Rule 3: Average / fractional O.N. — In some compounds identical atoms differ; the average may be fractional.

  • Easy : Fe₃O₄: 3x − 8 = 0 ⇒ x = +8/3.
  • Medium : Na₂S₄O₆: 2(+1) + 4x + 6(−2) = 0 ⇒ x = +2.5.
  • Exam-type : Na₂S₂O₃: 2 + 2x − 6 = 0 ⇒ x = +2.

Example: O.N. of Cr in K₂Cr₂O₇

  • Step 1: K = +1, O = −2; Cr = x.
  • Step 2: 2(+1) + 2x + 7(−2) = 0.
  • Step 3: 2x = 12 ⇒ x = +6.

CET level (MCQ)

  • CET 1) O.N. of Mn in KMnO₄? 1) +2 2) +4 3) +6 4) +7 — Answer: 4). (1 + x − 8 = 0.)
  • CET 2) O.N. of O in H₂O₂? 1) −2 2) −1 3) 0 4) +1 — Answer: 2).

Real-life use: O.N. tells in advance which atom will be reduced when making drugs or chemicals.

Remember

Sum = total charge; O −2, H +1; F always −1; peroxide O −1.

Common mistake

Taking O.N. in covalent compounds as the real charge; it is hypothetical.

CET trap

In CrO₅ Cr is +6 (four peroxide O at −1, one O at −2); do not assume −2 for all.

Shortcut

Memory clue: "sum = charge; F never changes".

Spot the difference

CompoundAtomO.N.
H₂SO₄S+6
Fe₃O₄Fe+8/3
H₂O₂O−1
CaH₂H−1

Practice questions

Easy
  1. 1What is O.N.?
  2. 2O.N. of S in S₈?
  3. 3O.N. of S in H₂SO₄?
Medium
  1. 4O.N. of Cr in Cr₂O₇²⁻?
  2. 5O.N. of H in CaH₂?
  3. 6O.N. of S in Na₂S₂O₃?
Challenging
  1. 7Average O.N. of Fe in Fe₃O₄?
  2. 8O.N. of N in NH₄⁺ and NO₃⁻?
Show answers (try them yourself first)
  1. 1A hypothetical charge.
  2. 20.
  3. 3+6.
  4. 4+6.
  5. 5−1.
  6. 6+2.
  7. 7+8/3.
  8. 8−3 and +5.

Stock notation, variable oxidation states and oxidising–reducing ability

From daily life

A teacher plays different roles in different classes; an element shows different states in different compounds.

Simple definition

Stock notation writes the O.N. as a Roman numeral in brackets, e.g. iron(III) chloride. Highest O.N. ⇒ only oxidising; lowest O.N. ⇒ only reducing; intermediate ⇒ both.

O.N. rangeRange of oxidation states-4-3-2-10+1+2+3+4+5+6+7NSClMnHighest O.N. ⇒ only an oxidising agent (KMnO₄, HNO₃, H₂SO₄)Lowest O.N. ⇒ only a reducing agent (H₂S, NH₃, I⁻)Intermediate ⇒ both (SO₂, H₂O₂, NO₂⁻)

In the figure : O.N. ranges of N, S, Cl and Mn.

Rule 1: Stock notation — Name = element(Roman numeral) + compound; Fe²⁺ = iron(II), Fe³⁺ = iron(III).

  • Easy : FeCl₃: iron(III) chloride.
  • Medium : Cu₂O: copper(I) oxide; CuO: copper(II) oxide.
  • Exam-type : MnO₂: manganese(IV) oxide.

Rule 2: Oxoacids of chlorine — HClO +1, HClO₂ +3, HClO₃ +5, HClO₄ +7. Cl in HClO₄ is only an oxidising agent; Cl⁻ (−1) only a reducing agent.

  • Easy : HClO₄: 1 + x − 8 = 0 ⇒ +7.
  • Medium : HClO₃: 1 + x − 6 = 0 ⇒ +5.
  • Exam-type : HClO: 1 + x − 2 = 0 ⇒ +1.

Rule 3: Species that play both roles — Intermediate O.N.: SO₂ (+4), H₂O₂ (−1), NO₂⁻ (+3), Fe²⁺ (+2) can act both ways.

  • Easy : H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂: H₂O₂ oxidises.
  • Medium : 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → …: H₂O₂ reduces.
  • Exam-type : SO₂ + 2H₂S → 3S + 2H₂O: SO₂ oxidises.

Method: deciding the role

  • Step 1: Write the atom's O.N. and its possible range.
  • Step 2: At an end of the range ⇒ one role; in the middle ⇒ both.
  • Step 3: Compare with the partner; the one with lower E° gets oxidised.

CET level (MCQ)

  • CET 1) Which can act only as an oxidising agent? 1) H₂S 2) SO₂ 3) H₂SO₄ 4) Na₂S — Answer: 3). (S = +6, highest.)
  • CET 2) Stock name of Cu₂O? 1) copper(I) oxide 2) copper(II) oxide 3) copper(III) 4) dicopper — Answer: 1).

Real-life use: H₂O₂ is a disinfectant and bleach (oxidising); SO₂ preserves wine (reducing).

Remember

Highest O.N. ⇒ oxidiser; lowest ⇒ reducer; intermediate ⇒ both.

Common mistake

Treating the highest O.N. as a reducing agent; it is an oxidiser.

CET trap

N ranges −3 to +5; NH₃ (−3) only reduces, HNO₃ (+5) only oxidises.

Shortcut

Memory clue: "end = one role, middle = two roles".

Spot the difference

O.N.RoleExample
HighestOxidisingKMnO₄, HNO₃
LowestReducingH₂S, I⁻
MiddleBothSO₂, H₂O₂

Practice questions

Easy
  1. 1What is Stock notation?
  2. 2Write the Stock name of FeCl₂.
  3. 3O.N. of Cl in HClO₃?
Medium
  1. 4Why is KMnO₄ only an oxidiser?
  2. 5Why is H₂S only a reducer?
  3. 6Why can H₂O₂ play both roles?
Challenging
  1. 7Write Cl's O.N. in all its oxoacids.
  2. 8Identify the roles in SO₂ + 2H₂S → 3S + 2H₂O.
Show answers (try them yourself first)
  1. 1O.N. shown as a Roman numeral in brackets.
  2. 2Iron(II) chloride.
  3. 3+5.
  4. 4Mn is +7, its highest; it cannot be oxidised further.
  5. 5S is −2, its lowest.
  6. 6O at −1 lies between −2 and 0.
  7. 7+1, +3, +5, +7.
  8. 8SO₂ is the oxidiser, H₂S the reducer.

Redox in terms of O.N. and types of redox reactions

From daily life

Marks rise or fall as pupils change classes; if O.N. rise or fall in a reaction, it is redox.

Simple definition

O.N. increase = oxidation; O.N. decrease = reduction. Total increase = total decrease. Types: combination, decomposition, displacement, disproportionation (one atom goes both up and down), comproportionation (two states merge into one).

Redox typesTypes of redox reactionsCombination2Mg + O₂ → 2MgO0 → +2 , 0 → −2Decomposition2KClO₃ → 2KCl + 3O₂Cl +5 → −1 , O −2 → 0DisplacementZn + CuSO₄ → ZnSO₄ + CuZn 0 → +2 , Cu +2 → 0DisproportionationCl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂OCl 0 → −1 and +1Comproportionation2H₂S + SO₂ → 3S + 2H₂OS −2 and +4 → 0

In the figure : Five types of redox reactions.

Rule 1: Recognising redox — If any atom's O.N. changes it is redox; if none changes it is not.

  • Easy : AgNO₃ + NaCl → AgCl + NaNO₃: no O.N. change ⇒ not redox.
  • Medium : 2KClO₃ → 2KCl + 3O₂: Cl +5 → −1, O −2 → 0 ⇒ redox.
  • Exam-type : NaOH + HCl → NaCl + H₂O: not redox.

Rule 2: Disproportionation — One atom is oxidised and reduced at once. It needs an intermediate O.N.

  • Easy : Cl₂ + 2NaOH → NaCl + NaClO + H₂O: Cl 0 → −1 and +1.
  • Medium : 2H₂O₂ → 2H₂O + O₂: O −1 → −2 and 0.
  • Exam-type : 3I₂ + 6OH⁻ → 5I⁻ + IO₃⁻ + 3H₂O.

Rule 3: Displacement reactions — Metal displacement: a more reactive metal displaces a less reactive one from solution. Non-metal displacement: Cl₂ displaces Br⁻ (halogen order).

  • Easy : Zn + CuSO₄ → ZnSO₄ + Cu.
  • Medium : Cl₂ + 2KBr → 2KCl + Br₂.
  • Exam-type : Cu + ZnSO₄ → no reaction (Cu is less reactive).

Method: checking redox with an O.N. table

  • Step 1: Write each atom's O.N. on both sides.
  • Step 2: Find atoms that changed: increase = oxidised, decrease = reduced.
  • Step 3: Total increase = total decrease ✓ (otherwise an error).

CET level (MCQ)

  • CET 1) Which is not redox? 1) Zn + CuSO₄ 2) AgNO₃ + NaCl 3) 2KClO₃ → 2KCl + 3O₂ 4) Cl₂ + 2KBr — Answer: 2).
  • CET 2) Which is disproportionation? 1) 2H₂O₂ → 2H₂O + O₂ 2) H₂ + Cl₂ → 2HCl 3) CaCO₃ → CaO + CO₂ 4) Zn + 2HCl — Answer: 1).

Real-life use: Bleaching powder arises from disproportionation; respiration and combustion are redox.

Remember

O.N. increase = oxidation; disproportionation = one atom both up and down; acid–base reactions are not redox.

Common mistake

Assuming every decomposition/combination is redox; CaCO₃ → CaO + CO₂ is not.

CET trap

H₂O₂ decomposition shows both roles; disproportionation needs an intermediate state.

Shortcut

Memory clue: "disproportion = one atom, two directions; comproportion = two atoms, one destination".

Spot the difference

TypeO.N.Example
Disproportion0 → −1, +1Cl₂ + OH⁻
Comproportion−2, +4 → 0H₂S + SO₂
Displacement0 → +2, +2 → 0Zn + CuSO₄

Practice questions

Easy
  1. 1Define oxidation in terms of O.N.
  2. 2Give an example of disproportionation.
  3. 3Why is AgNO₃ + NaCl not redox?
Medium
  1. 4Write the O.N. changes in 2KClO₃ → 2KCl + 3O₂.
  2. 5Why does Cu + ZnSO₄ not react?
  3. 6Write the half-reactions of Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.
Challenging
  1. 7Which type is 2H₂S + SO₂ → 3S + 2H₂O? Give the reason.
  2. 8What is oxidised and reduced in H₂O₂ decomposition?
Show answers (try them yourself first)
  1. 1Increase in O.N.
  2. 2Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.
  3. 3No O.N. changes.
  4. 4Cl +5 → −1; O −2 → 0.
  5. 5Cu is less reactive and cannot reduce Zn²⁺.
  6. 6Cl₂ → Cl⁻ (reduction); Cl₂ → ClO⁻ (oxidation).
  7. 7Comproportionation: S −2 and +4 → 0.
  8. 8O (−1) is reduced to −2 and oxidised to 0.

Balancing redox equations by the oxidation number method

From daily life

To pair pupils equally you multiply one group by 5 and another by 1; similarly the electrons given and taken are equalised.

Simple definition

Oxidation number method: (1) write O.N. (2) find the O.N. changes (3) multiply to equalise the change (4) balance other atoms: H⁺/H₂O in acid, OH⁻/H₂O in base (5) check atoms and charge.

Oxidation number methodOxidation-number method: five steps1Write O.N.2Find change3Equalise4Balance O, H5Check chargeMnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂OMn: +7 → +2 (gains 5 e⁻) · Fe: +2 → +3 (loses 1 e⁻)multipliers: MnO₄⁻ × 1, Fe²⁺ × 5MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂Ocharge: −1 + 10 + 8 = +17 ; right side 2 + 15 = +17 ✓

In the figure : Five steps: the MnO₄⁻ + Fe²⁺ example.

Rule 1: Acidic medium — Balance charge with H⁺ and then H and O with H₂O.

  • Easy : MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
  • Medium : Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
  • Exam-type : 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O.

Rule 2: Basic medium — Balance charge with OH⁻, then H and O with H₂O.

  • Easy : 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
  • Medium : Charge check: left −2 − 6 = −8; right −8 ✓.
  • Exam-type : Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O ✓.

Rule 3: Balancing molecular equations — Fix the redox part first and balance spectator ions (K⁺, SO₄²⁻) at the end.

  • Easy : 2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O.
  • Medium : Cu + 4HNO₃ (conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O.
  • Exam-type : Zn + 2HCl → ZnCl₂ + H₂.

Worked: balancing 3Cu + 8HNO₃ (dilute)

  • Step 1: Cu 0 → +2 (loses 2e⁻); N +5 → +2 (gains 3e⁻).
  • Step 2: Equalise e⁻: Cu × 3, NO × 2.
  • Step 3: N for 6 NO₃⁻ in Cu(NO₃)₂ + 2 NO = 8 ⇒ 8 HNO₃; 4 H₂O for H.
  • Step 4: Check O: 24 = 18 + 2 + 4 ✓.

CET level (MCQ)

  • CET 1) In MnO₄⁻ + Fe²⁺ (acid) the coefficient of Fe²⁺ is? 1) 1 2) 2 3) 5 4) 8 — Answer: 3). (Mn changes by 5.)
  • CET 2) Total electron change for Cr₂O₇²⁻ → Cr³⁺? 1) 3 2) 6 3) 2 4) 4 — Answer: 2). (2 Cr × 3.)

Real-life use: The balanced equation is used when titrating Fe²⁺ with KMnO₄.

Remember

Total increase = total decrease; acid: H⁺, H₂O; base: OH⁻, H₂O; finally check atoms and charge.

Common mistake

Forgetting the charge check in ionic equations; atoms alone are not enough.

CET trap

Cr₂O₇²⁻ has two Cr atoms so the change is 2 × 3 = 6, not 3.

Shortcut

Memory clue: "cross the changes" (Mn 5 ⇒ Fe ×5, Fe 1 ⇒ Mn ×1).

Spot the difference

MediumBalance chargeExample
AcidicH⁺MnO₄⁻ → Mn²⁺
BasicOH⁻MnO₄⁻ → MnO₂

Practice questions

Easy
  1. 1State the steps of the oxidation number method.
  2. 2What balances charge in acidic medium?
  3. 3What is the O.N. change per Cr in Cr₂O₇²⁻?
Medium
  1. 4Balance MnO₄⁻ + Fe²⁺ in acid.
  2. 5Balance Cr₂O₇²⁻ + Fe²⁺ in acid.
  3. 6Balance Cu + dilute HNO₃.
Challenging
  1. 7Balance MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic medium.
  2. 8Why is the charge check needed?
Show answers (try them yourself first)
  1. 1Write O.N.; changes; multipliers; other atoms; check.
  2. 2H⁺.
  3. 3+6 → +3 = 3 each.
  4. 4MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
  5. 5Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
  6. 63Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O.
  7. 72MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
  8. 8The total charge must be equal on both sides of an ionic equation.

Ion–electron (half-reaction) method

From daily life

Keep separate scoreboards for two players and add them at the end; likewise two half-reactions are balanced separately and added.

Simple definition

Half-reaction method: write oxidation and reduction separately, balance each (atoms, then charge with e⁻), equalise e⁻ and add.

Half reactionsIon–electron (half-reaction) methodReduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OH₂O for O, H⁺ for H, e⁻ for chargeOxidation: Fe²⁺ → Fe³⁺ + e⁻  (× 5)MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂OAcidic medium: H⁺ ; basic: OH⁻ (finally H⁺ + OH⁻ → H₂O)

In the figure : Half-reactions of MnO₄⁻ / Fe²⁺.

Rule 1: How to balance a half-reaction — (1) main atoms (2) H₂O for O (3) H⁺ for H (4) e⁻ for charge.

  • Easy : MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
  • Medium : Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
  • Exam-type : 2I⁻ → I₂ + 2e⁻.

Rule 2: Basic medium — Balance as in acid, then add one OH⁻ per H⁺ on both sides ⇒ H₂O.

  • Easy : MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.
  • Medium : ×2 and ×3: 2MnO₄⁻ + 4H₂O + 6I⁻ → 2MnO₂ + 8OH⁻ + 3I₂.
  • Exam-type : Cl₂ + 2e⁻ → 2Cl⁻; Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻.

Rule 3: Adding the two half-reactions — Make e⁻ equal, cancel them, and simplify leftover H⁺/H₂O.

  • Easy : MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
  • Medium : Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O (charge +6 = +6 ✓).
  • Exam-type : Zn + Cu²⁺ → Zn²⁺ + Cu.

Worked: MnO₄⁻/Fe²⁺ in acid

  • Step 1: Reduction: MnO₄⁻ → Mn²⁺; 4H₂O for O; 8H⁺ for H; 5e⁻ for charge.
  • Step 2: Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (× 5).
  • Step 3: Add and cancel e⁻ ⇒ final equation; charge −1 + 8 + 10 = +17 = 2 + 15 ✓.

CET level (MCQ)

  • CET 1) Electrons in MnO₄⁻ → Mn²⁺ (acid)? 1) 1 2) 3 3) 5 4) 7 — Answer: 3).
  • CET 2) H⁺ needed for Cr₂O₇²⁻ → 2Cr³⁺? 1) 7 2) 14 3) 6 4) 3 — Answer: 2).

Real-life use: In batteries and electroplating each electrode hosts a half-reaction; the same method applies.

Remember

O ⇒ H₂O; H ⇒ H⁺; charge ⇒ e⁻; basic: H⁺ + OH⁻ → H₂O.

Common mistake

Wrong electron side: right for oxidation, left for reduction.

CET trap

In basic medium keep OH⁻ only; no H⁺ in the final equation.

Shortcut

Memory clue: "O → H₂O, H → H⁺, charge → e⁻".

Spot the difference

ItemOxidationReduction
e⁻ sideRightLeft
O.N.↑↓

Practice questions

Easy
  1. 1State the steps of the half-reaction method.
  2. 2Write the reduction half-reaction MnO₄⁻ → Mn²⁺.
  3. 3On which side do the electrons appear in Fe²⁺ → Fe³⁺?
Medium
  1. 4Balance Cr₂O₇²⁻ + I⁻ in acid.
  2. 5Write the reduction half-reaction for MnO₄⁻ + I⁻ in basic medium.
  3. 6How do you convert an acid half-reaction to basic?
Challenging
  1. 7Write and check the half-reactions of Zn + Cu²⁺.
  2. 8Check the charge for MnO₄⁻/Fe²⁺.
Show answers (try them yourself first)
  1. 1Separate half-reactions; atoms; charge by e⁻; equal e⁻; add.
  2. 2MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
  3. 3On the right.
  4. 4Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O.
  5. 5MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.
  6. 6Add one OH⁻ per H⁺ on both sides.
  7. 7Zn → Zn²⁺ + 2e⁻; Cu²⁺ + 2e⁻ → Cu.
  8. 8−1 + 8 + 10 = +17; 2 + 15 = +17 ✓.

Electrode potential, electrochemical series and feasibility of redox reactions

From daily life

The stronger of two players is the one who wins more often; electrode potential measures the urge to gain electrons.

Simple definition

The standard reduction potential E° is measured at 1 M, 1 atm, 298 K against the standard hydrogen electrode (E° = 0). E°cell = E°cathode − E°anode. E°cell > 0 ⇒ the reaction is feasible.

Electrochemical seriesLi⁺/Li-3.05K⁺/K-2.93Na⁺/Na-2.71Mg²⁺/Mg-2.36Al³⁺/Al-1.66Zn²⁺/Zn-0.76Fe²⁺/Fe-0.44H⁺/H₂0.00Cu²⁺/Cu+0.34Ag⁺/Ag+0.80Cl₂/Cl⁻+1.36F₂/F⁻+2.87Zn–Cu cellE°cell = E°cat − E°an= 0.34 − (−0.76)= +1.10 VE° > 0 ⇒ feasibleΔG° = −nFE°= −212 kJlow E° = strong reducing agent · high E° = strong oxidising agent

In the figure : Electrochemical series and the Zn–Cu cell.

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Electrode potential, electrochemical series and feasibility of redox reactions

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Rule 1: Using the series — Lower E° ⇒ stronger reducing agent (oxidised easily); higher E° ⇒ stronger oxidising agent. A metal higher in the series displaces one below.

  • Easy : Zn (−0.76) displaces Cu²⁺ (+0.34).
  • Medium : Fe + 2H⁺ → Fe²⁺ + H₂: E° = 0 − (−0.44) = +0.44 V feasible.
  • Exam-type : Cu + 2H⁺: E° = −0.34 V ⇒ not feasible (Cu gives no H₂ with acid).

Rule 2: E°cell and ΔG° — ΔG° = −nFE°; F = 96500 C/mol. For K: log K = nE° ÷ 0.0592 (298 K).

  • Easy : Zn–Cu: E° = 0.34 + 0.76 = 1.10 V.
  • Medium : ΔG° = −2 × 96500 × 1.10 = −212 kJ/mol.
  • Exam-type : log K = 2 × 1.10 ÷ 0.0592 = 37.2.

Rule 3: Feasibility for halogens and other pairs — The oxidiser with higher E° oxidises the species with lower E°.

  • Easy : Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂: 1.36 − 1.09 = +0.27 V feasible.
  • Medium : Cl₂ + 2F⁻ → F₂: 1.36 − 2.87 = −1.51 V not feasible.
  • Exam-type : Cu + 2Ag⁺ → Cu²⁺ + 2Ag: 0.80 − 0.34 = +0.46 V feasible.

Derivation: ΔG° = −nFE°

  • Step 1: n moles of electrons flow ⇒ charge = nF.
  • Step 2: Maximum electrical work = charge × potential difference = nF × E°.
  • Step 3: The fall in free energy equals this work ⇒ ΔG° = −nFE°.

Formulas, units and dimensional formulae

QuantityFormulaUnit
E°cellE°cat − E°anV
ΔG°−nFE°J mol⁻¹
F96500C mol⁻¹
log KnE° ÷ 0.0592—

CET level (MCQ)

  • CET 1) E° of the Zn–Cu cell? (E° Zn²⁺/Zn = −0.76, Cu²⁺/Cu = +0.34) 1) 0.42 2) 1.10 3) −1.10 4) 0.76 — Answer: 2). (0.34 + 0.76.)
  • CET 2) Strongest reducing agent? 1) Li 2) Zn 3) Cu 4) Ag — Answer: 1). (lowest E°.)

Real-life use: Iron is galvanised with zinc because Zn has the lower E° and corrodes first; the basis of batteries and fuel cells.

Remember

Low E° = reducer; E°cell = E°cathode − E°anode; > 0 ⇒ feasible; ΔG° = −nFE°.

Common mistake

Multiplying E° by coefficients (E° is intensive and does not scale with n).

CET trap

Oxidation at the anode, reduction at the cathode; subtract cathode − anode, not the reverse.

Shortcut

Memory clue: "AN OX, RED CAT" (Anode Oxidation, Reduction Cathode).

Spot the difference

ItemAnodeCathode
ReactionOxidationReduction
Zn–CuZnCu
e⁻ReleasedConsumed

Practice questions

Easy
  1. 1What is E° measured against?
  2. 2Write the formula for E°cell.
  3. 3What role does a species with low E° play?
Medium
  1. 4Find E° of the Zn–Cu cell.
  2. 5Why does Cu + 2H⁺ not react?
  3. 6Is Cl₂ + 2Br⁻ feasible? Find E°.
Challenging
  1. 7Find ΔG° of the Zn–Cu cell.
  2. 8Why is iron coated with zinc?
Show answers (try them yourself first)
  1. 1The standard hydrogen electrode (E° = 0).
  2. 2E°cell = E°cathode − E°anode.
  3. 3A reducing agent.
  4. 40.34 − (−0.76) = 1.10 V.
  5. 5E° = 0 − 0.34 = −0.34 V < 0.
  6. 61.36 − 1.09 = +0.27 V ⇒ feasible.
  7. 7−2 × 96500 × 1.10 ≈ −212 kJ/mol.
  8. 8Zn oxidises first and protects Fe.

Quick revision of the whole lesson

ConceptKey formula / fact
OIL RIGOxidation Is Loss, Reduction Is Gain (e⁻)
O.N. rulessum = charge; H +1, O −2, F −1
Rolehighest O.N. ⇒ oxidiser; lowest ⇒ reducer
Typescombination, decomposition, displacement, disproportionation, comproportionation
Balancingequalise change → H⁺/OH⁻ → H₂O → check charge
Half-reactionO ⇒ H₂O, H ⇒ H⁺, charge ⇒ e⁻
E°cellE°cat − E°an ; ΔG° = −nFE°
Serieslow E° = strong reducer
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Classical and electronic ideas; agents.
  • O.N. rules and fractional O.N.
  • Stock notation; O.N. range and roles.
  • Types of redox; disproportionation.
  • O.N. balancing: acid/base.
  • Half-reaction method.
  • E°, series, feasibility and ΔG° = −nFE°.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.