Balancing Redox

8 MCQs9-step worked example
Source: NCERT Redox ReactionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Balancing redox reactions is a procedural skill that NEET tests at the recall and direct-application level — yet aspirants lose marks not because they cannot balance, but because they misidentify oxidation states or skip the medium-dependent ion/molecule additions in the ion-electron method.

The core trap: In the ion-electron (half-reaction) method, forgetting to add H₂O and H⁺ (acidic) or OH⁻ (basic) to balance oxygen and hydrogen AFTER balancing atoms that change oxidation state. This leads to an unbalanced charge, and the final equation fails the charge-audit check.

Two standard methods (NCERT Class 11 Chemistry, Chapter 8, pages 14–16):

  1. Oxidation-number method — Assign oxidation numbers to every atom. Identify the atoms whose oxidation state changes. Equalize total increase and total decrease by multiplying with appropriate coefficients. Balance remaining atoms by inspection.

  2. Ion-electron (half-reaction) method — Split into oxidation and reduction half-reactions. Balance atoms other than O and H. Balance O using H₂O. Balance H using H⁺ (acidic) or OH⁻ (basic). Balance charge using electrons. Multiply half-reactions so electrons cancel. Combine.

Watch-out for NEET: The ion-electron method dominates NEET questions. When a question specifies "acidic medium" or "basic medium," that is your cue to use this method. Always perform the final check: atoms balanced AND charge balanced on both sides. If charge does not balance, you missed a step — go back to the H⁺/OH⁻ addition.

The oxidation-number method is faster for molecular equations (no medium specified). NEET occasionally frames questions as "identify the coefficient of X in the balanced equation" — these reward systematic stepwise work over guessing.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Direct ApplicationPractice

In the reaction MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ (acidic medium), what is the ratio of moles of MnO₄⁻ to C₂O₄²⁻ in the balanced equation?

MCQ 2Easy RecallPractice

What is the oxidation state of Cr in Cr₂O₇²⁻?

MCQ 3Easy RecallPractice

In the ion-electron method for balancing in acidic medium, oxygen atoms are balanced by adding:

MCQ 4Easy RecallPractice

In the balanced half-reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, how many electrons are gained per MnO₄⁻?

MCQ 5Direct ApplicationPractice

When balancing Fe²⁺ → Fe³⁺ as a half-reaction, what must be added to the left side to balance charge?

MCQ 6Direct ApplicationPractice

In basic medium, after balancing oxygen with H₂O and hydrogen with H⁺, the next step in the ion-electron method is:

MCQ 7Direct ApplicationPractice

For the unbalanced reaction in acidic medium: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺, the number of Fe²⁺ ions needed to balance one Cr₂O₇²⁻ is:

MCQ 8CalculationPractice

After combining balanced half-reactions, a student obtains: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂. To verify, the total charge on the left side should be:

Worked Example

  1. 1

    Given

    Unbalanced ionic equation in acidic medium. Species: Cr₂O₇²⁻ (Cr is +6) and SO₃²⁻ (S is +4).

  2. 2

    Required

    Balanced net ionic equation with atoms and charge balanced.

  3. 3

    Concept

    Ion-electron (half-reaction) method in acidic medium: split into half-reactions, balance atoms (other than O, H first), then O with H₂O, then H with H⁺, then charge with electrons. Equalize electrons and combine.

  4. 4

    Half-reactions identified

    - Reduction: Cr₂O₇²⁻ → Cr³⁺ - Oxidation: SO₃²⁻ → SO₄²⁻

  5. 5

    Balance each half-reaction

    **Reduction half-reaction:** 1. Balance Cr: Cr₂O₇²⁻ → 2Cr³⁺ 2. Balance O with H₂O: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O 3. Balance H with H⁺: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O 4. Balance charge with e⁻: Left charge = −2 + 14 = +12. Right charge = 2(+3) = +6. Add 6e⁻ to left. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O **Oxidation half-reaction:** 1. Balance S: already 1:1. 2. Balance O with H₂O: SO₃²⁻ + H₂O → SO₄²⁻ 3. Balance H with H⁺: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ 4. Balance charge with e⁻: Left charge = −2. Right charge = −2 + 2 = 0. Add 2e⁻ to right. SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

  6. 6

    Equalize electrons and combine

    Reduction needs 6e⁻; oxidation supplies 2e⁻. Multiply oxidation by 3: 3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻ Add to reduction half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ + 3SO₃²⁻ + 3H₂O → 2Cr³⁺ + 7H₂O + 3SO₄²⁻ + 6H⁺ + 6e⁻ Cancel 6e⁻ from both sides. Simplify H⁺: 14H⁺ − 6H⁺ = 8H⁺ (left). Simplify H₂O: 7H₂O − 3H₂O = 4H₂O (right).

  7. 7

    Final balanced equation

    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

  8. 8

    Verification

    - Cr: 2 = 2 ✓ - S: 3 = 3 ✓ - O: 7 + 9 = 16 left; 12 + 4 = 16 right ✓ - H: 8 left; 8 right ✓ - Charge left: −2 + 3(−2) + 8(+1) = −2 − 6 + 8 = 0 - Charge right: 2(+3) + 3(−2) + 0 = +6 − 6 = 0 ✓

  9. 9

    Common trap

    The most common error is forgetting to multiply the oxidation half-reaction by 3, leading to unequal electron transfer. Always verify: electrons in reduction = electrons in oxidation before adding. **Similar NEET-style question:** Balance in acidic medium: MnO₄⁻ + I⁻ → Mn²⁺ + I₂. Find the coefficient of I⁻. ---

Before solving, remember these

Key Fact

Balancing redox reactions (oxidation number method)

1) Assign oxidation numbers. 2) Identify atoms changing ON. 3) Balance change in ON by appropriate stoichiometric coefficients (loss = gain). 4) Balance other atoms; balance charge with H⁺/OH⁻ (acidic/basic medium).

-- NCERT, p. 14
Key Fact

Half-reaction (ion-electron) method

Split into oxidation and reduction half-reactions. Balance atoms (other than O, H), then O with H₂O, then H with H⁺ (acidic) or OH⁻ (basic). Balance charge with electrons. Multiply to equalise electrons; add halves.

-- NCERT, p. 16

Formulas

Cell EMF

Both as reduction potentials. E°_cell > 0 → spontaneous.

SymbolQuantitySI Unit
E°_cellstandard cell EMFV
E°_redreduction potentialV

Valid when

  • Standard conditions (1 M, 1 bar, 298 K)
  • Both half-reactions as reductions

Faraday's law of electrolysis

Mass deposited at electrode. M = molar mass; I = current; t = time; n = electrons per ion.

SymbolQuantitySI Unit
mmass depositedg
Mmolar massg/mol
IcurrentA
ttimes
nelectrons per ion-
FFaradayC/mol

Valid when

  • Steady current
  • Single product

ΔG from EMF

Connection between thermodynamics and electrochemistry. F = 96485 C/mol.

SymbolQuantitySI Unit
ΔGGibbs energy changeJ
nelectrons transferred-
FFaraday 96485C/mol
Ecell EMFV

Valid when

  • Single redox process

Molar conductivity

Molar conductivity from specific conductance. Increases with dilution as more ions are free.

SymbolQuantitySI Unit
Λ_mmolar conductivityS cm^2/mol
κspecific conductanceS/cm
Cmolaritymol/L

Valid when

  • Aqueous solution
  • Single electrolyte

Nernst equation

Cell potential at non-standard conditions. n = electrons transferred. At equilibrium E=0, Q=K.

SymbolQuantitySI Unit
Ecell potentialV
E°standardV
nelectrons-
Qreaction quotient-

Valid when

  • 298 K (else use RT/F)
  • Single redox process

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Nernst n electrons cumulative error

Category: Negative Marking

Multi-step Nernst problem: identify electrons, write Q correctly, plug into 0.0591/n. Each sub-step has factor errors.

When it triggers

Cell EMF problem at non-standard conditions.

How to avoid

Step-by-step: (1) write balanced redox; (2) count n electrons; (3) compute Q from concentrations; (4) plug into Nernst. Verify by checking limits: at standard conditions Q=1, log Q=0, E=E°.

Trap

KMnO4 oxidation product depends on medium

Category: Inorganic Exception

Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).

When it triggers

Question gives KMnO4 oxidation in unspecified or specific medium.

How to avoid

Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.

Past Year Questions

17 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2024Revised key

Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

1Both Statement I and Statement II are correct.
2Both Statement I and Statement II are incorrect.
3Statement I is correct but Statement II is incorrect.
4Statement I is incorrect but Statement II is correct.
NTA Answer: Option 1(revised_final)
NEET 2022

Which of the following statement is not correct about diborane?

1Both the Boron atoms are sp2 hybridised.
2There are two 3-centre-2-electron bonds.
3The four terminal B-H bonds are two centre two electron bonds.
4The four terminal Hydrogen atoms and the two Boron atoms lie in one plane.
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

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