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Redox Reactions · TOPIC LESSON

Redox Balancing and Stoichiometry

Track electrons, atoms and charge together; the reaction medium determines the permitted bookkeeping.

JEE Main · JEE Advanced · NEET UG43 min study guide9 indexed sections

BY THE END, YOU CAN

  • Assign oxidation states with explicit exceptions and distinguish them from formal charges.
  • Identify oxidants, reductants, disproportionation and comproportionation.
  • Balance aqueous redox equations in acidic and basic media by half-reactions.
  • Derive reaction-specific electron factors rather than memorizing universal equivalent masses.
  • Use balanced equations in titration, limiting-reagent and purity calculations.

BEFORE YOU START

  • Atomic composition and ionic charge.
  • Moles, molarity and stoichiometric coefficients.
  • Acidic versus basic aqueous media.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

Available in this lesson: JEE Advanced 2012. This is a limited lesson sample, not a complete exam-question bank.

1 sourced sample question available across the listed exams and years.

Matched topics: Oxidation States.

JEE Advanced · 2012 · Paper 1 · Chemistry · Q29

Paraphrased exam question

Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.

Arrange HNO₃, NO, NH₄Cl and N₂ by decreasing nitrogen oxidation state.

Reveal answer & worked solution

Answer: HNO₃ > NO > N₂ > NH₄Cl (original option B; independently derived).

  1. Using H=+1 and O=−2, neutrality gives nitrogen +5 in HNO₃ and +2 in NO.
  2. Elemental N₂ has oxidation state 0. In NH₄⁺, x+4=+1 gives nitrogen −3.
  3. Ordering +5,+2,0,−3 gives the stated sequence.

Source: IIT-JEE 2012, Paper 1, Chemistry Q29; independently derived solution; key finality unverified. Solution written for this lesson.

01 / UNDERSTAND & APPLY

1. Redox is a change in oxidation state

Oxidation is an increase in oxidation state and reduction is a decrease. In a simple ionic electron-transfer event these descriptions correspond directly to electron loss and gain. In a covalent reaction the oxidation-state change is a formal accounting method; it need not mean that isolated integer-charge ions physically existed as intermediates.

The oxidizing agent is reduced because it accepts electron equivalents from something else. The reducing agent is oxidized because it supplies those equivalents. Naming agents by what happens to themselves reverses the labels: a species that becomes more oxidized is the reducing agent, not the oxidizing agent.

Electron conservation requires total electron equivalents lost to equal total electron equivalents gained. A closed overall reaction cannot have leftover electrons on one side. Half-reactions may contain electrons because they deliberately represent only one part of the coupled process, as in an electrochemical electrode reaction.

Not every reaction with ions is redox. Precipitation of AgCl from Ag⁺ and Cl⁻ leaves silver at +1 and chlorine at −1. Neutralization of H⁺ with OH⁻ leaves hydrogen at +1 and oxygen at −2. Identify actual oxidation-state changes before applying a redox balancing method; ordinary mass and charge balancing may be enough.

02 / UNDERSTAND & APPLY

2. Assign oxidation states without confusing three kinds of charge

An atom in a free element has oxidation state zero, whether the element appears as isolated atoms or a molecular form such as O₂, P₄ or S₈. A monatomic ion has oxidation state equal to its ionic charge. In a neutral species the sum of all oxidation states is zero; in a polyatomic ion the sum equals the ion’s net charge.

Fluorine is −1 in its compounds. Oxygen is commonly −2, but in a peroxide such as H₂O₂ it is −1, and in a superoxide ion O₂⁻ its average oxidation state is −1/2. Oxygen is +2 in OF₂ because fluorine is assigned −1. These exceptions follow the bonding situation rather than an arbitrary refusal to use the sum rule.

Hydrogen is commonly +1 when bonded to more electronegative nonmetals and −1 in ionic metal hydrides such as NaH. Alkali metals are usually +1 and alkaline-earth metals +2 in their ordinary compounds. Apply well-supported assignments first, then solve for the remaining unknown while checking that the chemical species has been identified correctly.

Formal charge divides covalent bonding electrons equally between bonded atoms; oxidation state assigns them to the more electronegative partner in the ionic approximation. Partial charge describes actual uneven electron density and is not generally an integer. For example, the oxygen formal charge in a usual water Lewis structure is zero, its oxidation state is −2, and its actual partial charge is neither of those bookkeeping numbers.

An average oxidation state can be fractional without every atom having that fractional local assignment. Fe₃O₄ has average iron oxidation state +8/3, consistent with a mixed-valence description involving Fe(II) and Fe(III). Structural information is needed before an average is turned into a statement about distinct atomic sites. The sum rule alone does not determine which site has which value.

Σ(number of atoms × oxidation state)=net species charge

This is an accounting constraint, not a claim about measured partial charges.

ORIGINAL WORKED EXAMPLE

Find the central-atom oxidation state

Find manganese in MnO₄⁻, oxygen in H₂O₂, and the average iron oxidation state in Fe₃O₄.

  1. For MnO₄⁻, Mn+4(−2)=−1, giving Mn=+7.
  2. For H₂O₂, 2(+1)+2O=0, giving O=−1, consistent with peroxide.
  3. For Fe₃O₄, 3Fe+4(−2)=0, giving average Fe=+8/3.
  4. The fractional iron average does not require each individual Fe site to carry a fractional oxidation-state assignment.

Answer Mn +7; peroxide O −1; average Fe +8/3.

03 / UNDERSTAND & APPLY

3. Balance an acidic half-reaction in a fixed order

First write the chemically specified reactant and product of the changing element. Balance atoms other than hydrogen and oxygen. Balance oxygen by adding H₂O, then hydrogen by adding H⁺. Finally balance electrical charge by adding electrons. Electrons appear on the reactant side for reduction and on the product side for oxidation.

The method balances a proposed transformation; it does not predict the product from nothing. Permanganate can form different manganese products under different conditions, so the product and medium must be known or justified independently. A formally balanced equation for a chemically inappropriate product is still an inappropriate description of the reaction.

After writing both half-reactions, multiply them by the smallest integers that equalize electron counts, add them, and cancel species that occur on both sides. Reduce all remaining coefficients by any common integer. Never cancel two chemically different species merely because they contain the same element.

Perform two independent final checks. Count atoms of every element, including hydrogen and oxygen, and compare total charge on each side. Atom balance alone does not guarantee charge balance. A redox equation can look plausible while missing an H⁺ or an electron factor that changes its stoichiometric ratio.

ORIGINAL WORKED EXAMPLE

Acidic permanganate and iron(II)

Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic aqueous medium.

  1. Manganese reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxygen gives four waters, hydrogen gives eight protons, and charge requires five electrons.
  2. Iron oxidation: Fe²⁺ → Fe³⁺ + e⁻. Multiply this half-reaction by five.
  3. Add and cancel electrons: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
  4. Left charge is −1+8+10=+17; right charge is +2+15=+17. All atoms also balance.

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

04 / UNDERSTAND & APPLY

4. Combine different electron counts before using mole ratios

An oxidation-state change is counted for each atom that changes, not merely for each written formula. Dichromate contains two chromium atoms, each decreasing from +6 to +3 when it forms Cr³⁺. It therefore accepts six electrons per dichromate ion, not three. Oxalate contains two carbons, each increasing from +3 to +4 on conversion to CO₂, so it releases two electrons per oxalate.

The least common multiple of half-reaction electron counts sets the scaling factors. If one half-reaction accepts five electrons and the other releases two, scale them by two and five respectively. The resulting ten-electron exchange is the electron count for that balanced overall reaction as written.

Stoichiometric ratios come from the finished equation, after medium-dependent H⁺, OH⁻ and water terms have been reconciled. Spectator counterions can then be restored if a molecular equation is requested. A net ionic equation is often the clearer route because it keeps spectators from obscuring the actual electron balance.

An electron factor can simplify arithmetic, but it is only a compact representation of the same balanced chemistry. Equating oxidant electron equivalents to reductant electron equivalents is valid when the stated products form quantitatively and no competing reaction consumes a reagent. It does not replace checking reaction identity or completion.

ORIGINAL WORKED EXAMPLE

Dichromate oxidizes iodide

Balance Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ in acidic medium.

  1. Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
  2. Oxidation: 2I⁻ → I₂ + 2e⁻; multiply by three to produce six electrons.
  3. Adding yields Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O.
  4. Charge check: −2+14−6=+6 on the left and 2(+3)=+6 on the right.

Answer Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O.

05 / UNDERSTAND & APPLY

5. In basic medium, eliminate the temporary protons

A convenient method is to balance the equation first using H⁺ and water as though it were acidic, then add enough OH⁻ to both sides to neutralize every H⁺. Replace each H⁺+OH⁻ pair by H₂O and cancel excess water. The final basic-medium equation should not contain an uncanceled H⁺ term.

Adding OH⁻ to only one side changes the charge and invents a different equation. The same amount must be added to each side before simplification. This is an algebraic transformation of the balance, not a claim that free protons accumulate in the basic solution.

In basic or neutral conditions where permanganate forms MnO₂, manganese decreases from +7 to +4 and accepts three electrons. The acidic Mn²⁺ product instead requires five. In strongly alkaline conditions a manganate product MnO₄²⁻ can correspond to a one-electron reduction. These examples show why a universal permanganate electron factor is unsafe.

If the final equation consumes or produces hydroxide, keep that coefficient. It may be needed in a material balance or to assess whether the assumed medium can be maintained. A reagent ratio derived for a sufficiently basic solution should not automatically be extrapolated to a mixture whose base is exhausted during the process.

ORIGINAL WORKED EXAMPLE

Permanganate and sulfite in basic medium

Balance MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ under conditions giving MnO₂.

  1. Reduction half: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.
  2. Oxidation half: SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻.
  3. Multiply the first by two and the second by three, then cancel six electrons, six hydroxides and three waters.
  4. Result: 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻. Both sides have charge −8.

Answer 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻.

06 / UNDERSTAND & APPLY

6. One element can be both oxidized and reduced

In disproportionation, atoms of the same element in one initial oxidation state form products with both a higher and a lower oxidation state. Chlorine in Cl₂ can form chloride at −1 and hypochlorite at +1 under suitable cold dilute alkaline conditions. Different chlorine atoms undergo the two changes; one atom is not simultaneously assigned two final states.

Comproportionation is the reverse pattern: two initial oxidation states combine to form an intermediate oxidation state. For example, iodide and iodate can form iodine in acid. The number of atoms coming from each starting state is determined by electron balance, not by assuming equal amounts of the two reactants.

Hydrogen peroxide can act as an oxidant when oxygen is reduced from −1 to −2, or as a reductant when oxygen is oxidized from −1 to zero. Its decomposition into water and oxygen is itself a disproportionation. Knowing that a reagent is often called an oxidizer does not prevent it from being reduced or oxidized differently in another reaction.

The ability to write a balanced disproportionation equation does not alone establish that it proceeds rapidly or favorably under all conditions. Medium, thermodynamics, catalyst and kinetic barriers matter. Use the problem’s specified products and conditions, then use oxidation-state accounting to explain the atom ratios.

ORIGINAL WORKED EXAMPLE

Cold alkaline chlorine disproportionation

Balance Cl₂ → Cl⁻ + ClO⁻ in basic medium and identify both roles of chlorine.

  1. One chlorine atom goes from 0 to −1 and gains one electron. The other goes from 0 to +1 and loses one electron, giving a 1:1 product ratio.
  2. Balance oxygen and hydrogen with hydroxide and water: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.
  3. There are two chlorine atoms, two oxygen atoms and two hydrogen atoms on each side; both sides have charge −2.
  4. The chlorine becoming chloride is reduced; the chlorine becoming hypochlorite is oxidized.

Answer Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O; Cl₂ supplies both oxidizing and reducing roles.

07 / UNDERSTAND & APPLY

7. Titration turns an electron ratio into an amount

At the equivalence point of a quantitative redox titration, oxidant electron equivalents consumed equal reductant electron equivalents supplied. If the reaction-specific factors are z_ox and z_red, then z_ox C_ox V_ox=z_red C_red V_red. This equation assumes compatible volume units and that the relevant reaction has reached its intended stoichiometric endpoint.

The observed endpoint is an experimental signal, which may differ slightly from the theoretical equivalence point. An indicator response, reagent self-color or instrumental signal must be appropriate to the chemistry. The present numerical exercises assume negligible endpoint error rather than claiming that every visible color change is exact equivalence.

Equivalent mass is molar mass divided by the electron factor for the specified reaction. It is not a fixed material constant independent of reaction. Dichromate reducing to Cr³⁺ in acid has a factor of six per formula unit, while peroxide has factor two per H₂O₂ for either complete reduction to water or oxidation to oxygen.

Purity and aliquot calculations add ordinary material accounting around the redox ratio. Determine analyte moles in the titrated aliquot, scale to the full prepared solution, convert to analyte mass, and divide by the original sample mass. State that impurities are inert toward the titrant if the calculation relies on that assumption.

ORIGINAL WORKED EXAMPLE

Find iron(II) concentration

A 25.00 mL Fe²⁺ aliquot requires 18.00 mL of 0.02000 M MnO₄⁻ in sufficient acid, with Mn²⁺ as product and no interfering reductants. Find [Fe²⁺].

  1. Permanganate amount is 0.01800×0.02000=0.0003600 mol.
  2. The balanced reaction consumes five Fe²⁺ per permanganate, so Fe²⁺ amount is 0.001800 mol.
  3. Divide by aliquot volume 0.02500 L to obtain 0.07200 mol L⁻¹.
  4. The factor five depends on the Mn²⁺ product in acid; it would not apply to MnO₂ formation.

Answer [Fe²⁺]=0.07200 M.

ORIGINAL WORKED EXAMPLE

A peroxide purity calculation

A 0.5000 g sample containing H₂O₂ and titrant-inert material is diluted to 100.0 mL. A 20.00 mL aliquot consumes 12.00 mL of 0.01000 M acidic permanganate, producing Mn²⁺ and O₂. Find H₂O₂ mass percent using 34.0 g mol⁻¹.

  1. Electron balance gives 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O.
  2. Permanganate in the aliquot titration is 0.0001200 mol; peroxide amount is (5/2)×0.0001200=0.0003000 mol.
  3. The full sample contains five aliquots, so peroxide amount is 0.001500 mol and mass is 0.0510 g.
  4. Mass percent is 100×0.0510/0.5000=10.2%.

Answer 10.2% H₂O₂ by mass.

08 / UNDERSTAND & APPLY

8. Limiting reagents and final checks prevent false ratios

For multiple measured reactants, compare available moles divided by their balanced coefficients. The smallest extent limits the reaction. Comparing raw mole amounts is unreliable because an oxidant accepting six electrons may consume several moles of a one-electron reductant. Acid or base can also be limiting if it appears as a consumed reagent and is not stated to be in excess.

When a reagent remains after reaction, its amount is initial amount minus stoichiometric consumption. To report concentration, divide by the final solution volume under an explicit volume model. Do not divide by the initial volume of whichever reagent happened to be in the flask first.

For reactions involving several changing elements, account for the net electron change of the entire reacting species. A simple difference in one central atom’s oxidation state can fail if another atom in the same reagent also changes. The safest check remains a fully balanced equation with both mass and charge conserved.

Keep computational precision until the last step. A titration ratio is exact only with respect to the balanced integer coefficients; measured concentration, delivered volume and sample mass carry experimental precision. Report a result consistent with those data and avoid treating extra calculator digits as extra chemical certainty.

  • Calling the species that is oxidized the oxidizing agent.
  • Treating oxidation state as actual partial charge or always as formal charge.
  • Using oxygen −2 in a peroxide without checking.
  • Forgetting two chromium atoms in dichromate.
  • Using a five-electron permanganate factor for every medium and product.
  • Leaving H⁺ in a final basic-medium equation.
  • Checking atoms but not total charge.

09 / UNDERSTAND & APPLY

9. Eight solved checks

Identify the specified products and medium before calculating. Each result can be checked by oxidation-state change, atom count or charge conservation.

ORIGINAL PRACTICE · 1

Find sulfur oxidation state in SO₄²⁻.

Reveal answer & reasoning

+6.

Let it be s. Then s+4(−2)=−2, so s=+6.

ORIGINAL PRACTICE · 2

In Zn + Cu²⁺ → Zn²⁺ + Cu, identify the oxidizing agent.

Reveal answer & reasoning

Cu²⁺.

Copper decreases from +2 to 0 and is reduced. It therefore oxidizes zinc and is the oxidizing agent.

ORIGINAL PRACTICE · 3

How many electrons are accepted per Cr₂O₇²⁻ when both chromium atoms become Cr³⁺?

Reveal answer & reasoning

Six electrons.

Each chromium decreases from +6 to +3, accepting three electrons; there are two chromium atoms.

ORIGINAL PRACTICE · 4

Balance Fe³⁺ + Sn²⁺ → Fe²⁺ + Sn⁴⁺.

Reveal answer & reasoning

2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺.

Tin loses two electrons while each iron gains one. Two iron ions are required, and both sides have charge +8.

ORIGINAL PRACTICE · 5

For MnO₄⁻ → MnO₂, what is the electron factor per permanganate?

Reveal answer & reasoning

Three.

Manganese falls from +7 to +4. This differs from reduction to Mn²⁺, which needs five electrons.

ORIGINAL PRACTICE · 6

A reaction consumes one mole dichromate per six moles Fe²⁺. How much Fe²⁺ reacts with 0.00200 mol dichromate in sufficient acid?

Reveal answer & reasoning

0.0120 mol Fe²⁺.

Multiply the dichromate amount by the balanced ratio six. This assumes the intended complete reaction and enough Fe²⁺.

ORIGINAL PRACTICE · 7

Balance peroxide disproportionation into water and oxygen.

Reveal answer & reasoning

2H₂O₂ → 2H₂O + O₂.

Four oxygen atoms begin at −1. Two end at −2 in water and two at 0 in oxygen; electron gain and loss match. Hydrogen and oxygen atom counts also balance.

ORIGINAL PRACTICE · 8

For MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺, start with 0.00100 mol permanganate and 0.00300 mol Fe²⁺ in excess acid. Which limits and how much permanganate remains?

Reveal answer & reasoning

Fe²⁺ limits; 0.000400 mol permanganate remains.

The iron amount consumes 0.00300/5=0.000600 mol permanganate. Subtracting from 0.00100 gives 0.000400 mol; the original permanganate would have required 0.00500 mol iron to be exhausted.

KEEP THE BIG IDEAS

Quick revision

  • Oxidation state is a formal electron-accounting number, distinct from formal and partial charge.
  • Oxidants are reduced and reductants are oxidized.
  • Balance atoms and total charge independently.
  • The medium and products determine electron factors.
  • Half-reactions must cancel electrons in the overall equation.
  • Use the finished equation for titration, limiting-reagent and purity arithmetic.
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