Redox reactions for NEET

Redox is a short chapter with a long reach. Oxidation numbers turn up in the d-block, coordination compounds, electrochemistry and titrations, so a slip here costs marks in four other places. The good news is that the chapter is almost entirely method: a set of rules, a list of exceptions and one balancing procedure.

In the NMC syllabus, redox and electrochemistry sit together as one unit. NEET usually asks this part as an oxidation number with a twist (a peroxide or an average value), "which is a disproportionation reaction", "which can act only as an oxidising agent", or the coefficient or mole ratio in a balanced equation.

Oxidation and reduction, three ways

The electronic view is the one to use. In Zn + Cu²⁺ → Zn²⁺ + Cu, zinc loses two electrons (oxidised) and Cu²⁺ gains them (reduced). The two always happen together, which is why they are called redox reactions.

• The oxidising agent takes electrons, so it is itself reduced (Cu²⁺ here).
• The reducing agent gives electrons, so it is itself oxidised (Zn here).

Rules for oxidation numbers

Apply these in order. When two rules clash, the earlier one wins.

1. An element in its free state is 0: Na, O₂, P₄, S₈.
2. A monatomic ion carries its charge: Na⁺ is +1, Cl⁻ is −1.
3. Fluorine is always −1 in compounds.
4. Alkali metals are +1 and alkaline earth metals +2 in compounds.
5. Hydrogen is +1 with non-metals and −1 in metal hydrides such as NaH and CaH₂.
6. Oxygen is −2, except in peroxides (−1, as in H₂O₂ and Na₂O₂), superoxides (−½, as in KO₂) and OF₂ (+2).
7. The oxidation numbers add up to 0 for a neutral compound and to the charge for an ion.

Most NEET questions are built on the exceptions:

Oxidising agents, reducing agents and the types of redox reaction

An element's oxidation state tells you what it can do:

• Highest state: can only be reduced, so it acts only as an oxidising agent. Examples: KMnO₄ (Mn +7), HNO₃ (N +5), K₂Cr₂O₇ (Cr +6).
• Lowest state: can only be oxidised, so it acts only as a reducing agent. Examples: H₂S (S −2), NH₃ (N −3), I⁻.
• Intermediate state: can act as both. Examples: H₂O₂ (O −1), SO₂ (S +4), HNO₂ (N +3).

Types of redox reaction:

• Combination: C + O₂ → CO₂.
• Decomposition: 2KClO₃ → 2KCl + 3O₂. Not every decomposition is redox: CaCO₃ → CaO + CO₂ involves no change in oxidation number.
• Displacement: a more reactive metal displaces a less reactive one, as in Zn + CuSO₄ → ZnSO₄ + Cu. Non-metals do the same: Cl₂ + 2KBr → 2KCl + Br₂.
• Disproportionation: one element in one oxidation state is oxidised and reduced at the same time.

Disproportionation examples to know:

• 2H₂O₂ → 2H₂O + O₂ (O goes from −1 to −2 and to 0)
• Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O (Cl goes from 0 to −1 and to +1)
• P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻ (P goes from 0 to −3 and to +1)
• 2Cu⁺ → Cu²⁺ + Cu in water

A species in its highest or lowest state cannot disproportionate. ClO₄⁻ (Cl +7) cannot, and neither can F₂, because fluorine has no positive oxidation states.

Balancing redox equations

Ion–electron (half-reaction) method

1. Split the reaction into oxidation and reduction half-reactions.
2. Balance atoms other than O and H.
3. Balance O by adding H₂O.
4. Balance H by adding H⁺ (in acidic medium).
5. Balance charge by adding electrons.
6. Multiply the half-reactions so the electrons cancel, and add them.
7. In basic medium, add OH⁻ to both sides to neutralise each H⁺, then cancel water.

Permanganate and iron(II), acidic:

• Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
• Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 5)
• Overall: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
• Check the charge: −1 + 10 + 8 = +17 on the left; +2 + 15 = +17 on the right.

Dichromate and iron(II), acidic: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, so Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Oxidation number method

Find the change in oxidation number of the atoms oxidised and reduced, and cross-multiply so the total increase equals the total decrease. Then balance O with H₂O and H with H⁺ or OH⁻. For permanganate and iodide in basic medium: Mn falls from +7 to +4 (3 units), I rises from −1 to +5 (6 units), so two MnO₄⁻ are needed per I⁻:

2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

Worked NEET-style questions

Example 1: an average oxidation number

Find the oxidation number of S in Na₂S₄O₆.

• 2(+1) + 4x + 6(−2) = 0, so 4x = 10 and x = +2.5.
• This is an average. Structurally, the two bridging S atoms are 0 and the two outer S atoms are +5: (0 + 0 + 5 + 5)/4 = 2.5.

Example 2: titration from a balanced equation

What volume of 0.02 M KMnO₄ is needed to oxidise 0.01 mol of Fe²⁺ in acid?

• From the balanced equation, 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺.
• Moles of MnO₄⁻ = 0.01/5 = 0.002 mol.
• Volume = 0.002/0.02 = 0.1 L = 100 mL.

Example 3: permanganate and oxalate

Balance the oxidation of C₂O₄²⁻ by MnO₄⁻ in acid and give the mole ratio.

• Carbon goes from +3 to +4, so each C₂O₄²⁻ loses 2 electrons. Each MnO₄⁻ gains 5.
• The lowest common multiple is 10: 2MnO₄⁻ and 5C₂O₄²⁻.
• 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Charge: −2 − 10 + 16 = +4 on the left, +4 on the right.
• Mole ratio MnO₄⁻ : C₂O₄²⁻ = 2 : 5.

Example 4: spotting disproportionation

Which of these is a disproportionation reaction? (a) 2KClO₃ → 2KCl + 3O₂ (b) 3ClO⁻ → 2Cl⁻ + ClO₃⁻ (c) Zn + 2HCl → ZnCl₂ + H₂ (d) CaCO₃ → CaO + CO₂

• In (a), Cl is reduced and O is oxidised: two different elements.
• In (b), Cl goes from +1 to −1 and to +5: the same element both ways.
• Answer: (b).

Practice MCQs

1. The oxidation number of S in Na₂S₂O₃ is: (a) +2 (b) +4 (c) +6 (d) −2
2. The oxidation number of Fe in Fe₃O₄ is: (a) +2 (b) +3 (c) +8/3 (d) +4
3. Which can act only as an oxidising agent? (a) SO₂ (b) H₂O₂ (c) HNO₃ (d) HNO₂
4. The reaction 2Cu⁺ → Cu²⁺ + Cu is an example of: (a) combination (b) displacement (c) disproportionation (d) not a redox reaction
5. The number of electrons gained in Cr₂O₇²⁻ → 2Cr³⁺ is: (a) 3 (b) 5 (c) 6 (d) 12
6. How many moles of KMnO₄ oxidise 1 mol of C₂O₄²⁻ in acid? (a) 0.2 (b) 0.4 (c) 2.5 (d) 5
7. The oxidation number of O in OF₂ is: (a) −2 (b) −1 (c) +1 (d) +2
8. The oxidation number of Mn in K₂MnO₄ is: (a) +4 (b) +6 (c) +7 (d) +5

Answers

1. (a) 2 + 2x − 6 = 0, so x = +2 (average).
2. (c) 3x − 8 = 0, so x = 8/3.
3. (c) N is at +5, its highest state.
4. (c) Copper(I) is both oxidised and reduced.
5. (c) Each Cr goes from +6 to +3: 2 × 3 = 6 electrons.
6. (b) The ratio is 2 : 5, so 2/5 = 0.4 mol.
7. (d) F is −1, so O is +2.
8. (b) 2 + x − 8 = 0, so x = +6.

What to do next

• Learn the ten tricky oxidation numbers in the table, then write the reason for each from memory.
• Balance five equations by the ion–electron method in acid and five in base, checking the charge every time.
• Write the three reduction products of MnO₄⁻ with their media and electron counts.
• Move on to electrochemistry, where half-reactions become electrode potentials, and use d- and f-block elements for KMnO₄ and K₂Cr₂O₇ chemistry.