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 Class 11 Chemistry, Ch. 7, p. 246Balancing Redox
8 MCQs9-step worked example
Source: NCERT Redox Reactions and ElectrochemistryPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 27 Sep 2026
Balancing Redox, explained for NEET
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 7, pages 246–247):
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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.
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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.
Can you answer these Balancing Redox 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?
Show answer and why every option is right or wrong
Answer: C. Mn goes from +7 to +2 (gain of 5e⁻). Each C₂O₄²⁻ has 2 carbons going from +3 to +4 (loss of 2e⁻ total per oxalate ion). To equalize: 2 × 5e⁻ = 5 × 2e⁻. So MnO₄⁻ : C₂O₄²⁻ = 2 : 5. (NCERT Class 11 Chemistry, Chapter 7, page 247.)
Why A is wrong: A uses 1 instead of 2 for MnO₄⁻, ignoring that 5 electrons per Mn must match 2 electrons per oxalate, requiring a multiplier of 2.
Why B is wrong: B reverses the ratio — this error comes from swapping which species gains and which loses electrons.
Why D is wrong: D takes MnO₄⁻ as gaining only 3e⁻ (reduction to MnO₂, as in neutral medium) instead of 5e⁻ (to Mn²⁺ in acid): 3 × 2 = 2 × 3 gives 2 : 3.
MCQ 2Easy RecallPractice
What is the oxidation state of Cr in Cr₂O₇²⁻?
Show answer and why every option is right or wrong
Answer: A. Let x be the oxidation state of Cr. 2x + 7(−2) = −2. 2x = +12. x = +6. (NCERT Class 11 Chemistry, Chapter 7.)
Why B is wrong: B confuses Cr₂O₇²⁻ with MnO₄⁻ where Mn is +7. Cr in dichromate is +6.
Why C is wrong: C is the oxidation state of Cr in Cr₂O₃ (chromic oxide), not in dichromate.
Why D is wrong: D forgets to divide by 2 — it gives the total contribution of both Cr atoms combined, not the per-atom oxidation state.
MCQ 3Easy RecallPractice
In the ion-electron method for balancing in acidic medium, oxygen atoms are balanced by adding:
Show answer and why every option is right or wrong
Answer: A. In acidic medium, add H₂O to balance oxygen, then H⁺ to balance hydrogen. OH⁻ is used in basic medium. (NCERT Class 11 Chemistry, Chapter 7, page 246.)
Why B is wrong: B applies the basic-medium rule. In acidic medium, O is balanced with H₂O, not OH⁻.
Why C is wrong: C is not a standard balancing species in aqueous half-reaction methods — O²⁻ does not exist free in solution.
Why D is wrong: D (H₂O₂) is a reactant/oxidant, not a balancing species in the ion-electron method.
MCQ 4Easy RecallPractice
In the balanced half-reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, how many electrons are gained per MnO₄⁻?
Show answer and why every option is right or wrong
Answer: B. Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺, a decrease of 5 units, requiring 5 electrons. (NCERT Class 11 Chemistry, Chapter 7, page 247.)
Why A is wrong: A corresponds to a +7 → +5 change (as in MnO₄⁻ → MnO₄³⁻), which is not the product here.
Why C is wrong: C corresponds to a +7 → +4 change (MnO₂ product in neutral medium), not the acidic-medium product Mn²⁺.
Why D is wrong: D equals the oxidation state of Mn itself (+7), not the change in oxidation state. Electrons gained = change, not absolute value.
MCQ 5Direct ApplicationPractice
When balancing Fe²⁺ → Fe³⁺ as a half-reaction, what must be added, and on which side, to balance charge?
Show answer and why every option is right or wrong
Answer: D. Fe²⁺ → Fe³⁺ is oxidation (loss of electron). The equation is: Fe²⁺ → Fe³⁺ + e⁻. The electron appears on the right (product) side. Left side charge: +2. Right side: +3 + (−1) = +2. Balanced.
Why A is wrong: A places the electron on the wrong side — that would represent reduction (gain), not oxidation.
Why B is wrong: B adds H⁺ to balance charge, but H⁺ is used only to balance hydrogen atoms, not charge. Electrons balance charge in half-reactions.
Why C is wrong: C uses 2 electrons, which would apply to Fe → Fe²⁺ (metallic iron losing 2e⁻), not Fe²⁺ → Fe³⁺.
MCQ 6CalculationPractice
The reduction of MnO₄⁻ to MnO₂ is first balanced as if in acidic medium: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O. To convert this to basic medium, an equal number of OH⁻ ions are added to both sides to neutralize the H⁺ (forming H₂O), and any H₂O common to both sides is then cancelled. Which of the following is the correctly converted and simplified basic-medium half-reaction?
Show answer and why every option is right or wrong
Answer: A. Adding 4OH⁻ to both sides converts the 4H⁺ on the left into 4H₂O; combining this with the 2H₂O already on the right and cancelling the 2H₂O common to both sides leaves a net 2H₂O on the left and 4OH⁻ on the right: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (NCERT Class 11 Chemistry, Chapter 7, page 249).
Why B is wrong: B keeps all 4H₂O formed from neutralizing 4H⁺ without subtracting the 2H₂O already present on the product side, so the common water is never cancelled.
Why C is wrong: C supplies only 2 OH⁻, matching the original H₂O count instead of the 4 H⁺ that actually need neutralizing, undercounting the hydroxide required.
Why D is wrong: D swaps the sides — after conversion and cancellation, the surviving water is on the reactant side and hydroxide is on the product side, not the reverse.
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:
Show answer and why every option is right or wrong
Answer: C. Each Cr goes from +6 to +3 (gain of 3e⁻); 2 Cr atoms gain 6e⁻ total. Each Fe²⁺ → Fe³⁺ loses 1e⁻. To supply 6e⁻, need 6 Fe²⁺. (NCERT Class 11 Chemistry, Chapter 7, page 247.)
Why A is wrong: A accounts for only one Cr atom (3e⁻) instead of both Cr atoms in Cr₂O₇²⁻ (6e⁻ total). This is the most common trap — forgetting the subscript 2.
Why B is wrong: B confuses the subscript in Cr₂O₇²⁻ (2 Cr atoms) with the electron requirement per Fe²⁺.
Why D is wrong: D uses the number of oxygen atoms in dichromate (7) rather than the electron-transfer calculation.
MCQ 8CalculationPractice
After combining balanced half-reactions, a student obtains: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂. To verify it, the total charge on the left side should be:
Show answer and why every option is right or wrong
Answer: B. B is correct. Left side: 2(−1) + 16(+1) + 5(−2) = −2 + 16 − 10 = +4. Right side: 2(+2) = +4, since H₂O and CO₂ are neutral. The charges match, which is the check that the electrons were balanced correctly when the half-reactions were combined (NCERT Class 11 Chemistry, Chapter 7, page 247).
Why A is wrong: A is wrong because −4 comes from counting 8H⁺ instead of 16H⁺ — carrying the 8 over from 8H₂O on the other side: −2 + 8 − 10 = −4. The number of H⁺ is set by the oxygen balance, and it is 16.
Why C is wrong: C is wrong because +6 comes from leaving the permanganate ions out of the count: 16 − 10 = +6. The two MnO₄⁻ ions carry −2 between them.
Why D is wrong: D is wrong because it assumes a balanced equation must be neutral on each side. That is true of molecular equations, but an ionic equation only has to carry the SAME charge on both sides, and here that charge is +4.
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How do you solve a Balancing Redox question? A worked example
- 1
Given
Unbalanced ionic equation in acidic medium. Species: Cr₂O₇²⁻ (Cr is +6) and SO₃²⁻ (S is +4).
- 2
Required
Balanced net ionic equation with atoms and charge balanced.
- 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
Half-reactions identified
• Reduction: Cr₂O₇²⁻ → Cr³⁺• Oxidation: SO₃²⁻ → SO₄²⁻
- 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
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
Final balanced equation
Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
- 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
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⁻.
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What to remember before solving Balancing Redox questions
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 Class 11 Chemistry, Ch. 7, p. 248More in Redox Reactions and Electrochemistry: 4 exam traps and mistakes · 5 formulas · 2 question patterns from its other lessons.
Balancing Redox questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys.
NEET 2023
13, 8, 1
21, 8, 3
38, 1, 3
41, 3, 8
NTA Answer: Option 4(final)
All 21 past-paper questions from Redox Reactions and Electrochemistry →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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