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Chemistry · Redox Reactions

Balancing Redox Reactions

A redox equation is balanced only when both mass and charge are conserved, which an ordinary by-inspection balance cannot guarantee. NCERT §7.3.2 sets out two systematic routes — the oxidation-number method and the ion-electron (half-reaction) method — and notes that the choice between them rests with the individual using them. For NEET, the recurring task is to recover the integer coefficients of permanganate, dichromate and oxalate equations, so mastery of both routes in acidic and basic media is decisive.

Why redox balancing needs a method

In a redox reaction electrons are transferred from the reducing agent to the oxidising agent. A correctly balanced equation must therefore satisfy two conservation laws at once: every atom must be conserved (mass balance) and the net electrical charge must be identical on both sides (charge balance). Trial-and-error balancing, which works for simple combination or precipitation reactions, becomes unreliable when polyatomic oxoanions such as $\ce{MnO4-}$ or $\ce{Cr2O7^2-}$ change their oxidation state, because the spectator hydrogen and oxygen atoms must be supplied by the medium itself.

NCERT §7.3.2 introduces two disciplined methods. The first, the oxidation-number method, equates the total increase in oxidation number of the reductant with the total decrease in oxidation number of the oxidant. The second, the ion-electron (half-reaction) method, splits the change into an oxidation half and a reduction half, balances each separately with explicit electrons, and recombines them. Both are exact; the difference is bookkeeping style.

Figure 1 · Balancing Workflow
Skeletal ionic equation Split into two half-reactions Oxidation half (loses e⁻) Reduction half (gains e⁻) In each half: balance atoms → balance O with H₂O → balance H with H⁺ → balance charge with e⁻ (basic medium: then add OH⁻ for each H⁺) Equalise e⁻, add halves, cancel e⁻
Schematic of the ion-electron method following NCERT §7.3.2. Each half-reaction is balanced for atoms, then oxygen with water, then hydrogen with protons, then charge with electrons, before the halves are equalised and added.

The oxidation-number method

This method tracks the change in oxidation number directly. Following NCERT and NIOS §13.3.1, the working proceeds through a fixed sequence: write the skeletal equation without coefficients, assign oxidation numbers to every atom, identify the atoms whose oxidation number changes, compute the total increase and the total decrease, and then scale the formulae so that the increase equals the decrease. Only after the electron bookkeeping is settled do you balance the remaining atoms and, finally, the medium species.

Consider the oxidation of sulphite by dichromate worked in NCERT. The skeletal equation is $\ce{Cr2O7^2- + SO3^2- -> Cr^3+ + SO4^2-}$. Chromium falls from $+6$ to $+3$, a decrease of $3$ per atom (so $6$ for two Cr atoms), while sulphur rises from $+4$ to $+6$, an increase of $2$ per atom. Equalising the change places a $2$ before $\ce{Cr^3+}$ and a $3$ before $\ce{SO4^2-}$ and $\ce{SO3^2-}$:

$$\ce{Cr2O7^2- + 3SO3^2- -> 2Cr^3+ + 3SO4^2-}$$

Because the reaction occurs in acidic medium and the ionic charges are unequal, $\ce{8H+}$ is added on the left to equalise charge, and $\ce{4H2O}$ on the right to balance hydrogen, giving the complete equation $\ce{Cr2O7^2- + 3SO3^2- + 8H+ -> 2Cr^3+ + 3SO4^2- + 4H2O}$. This is exactly the NEET 2023 result, where the coefficients $a,b,c$ work out to $1,3,8$.

NEET Trap

Assigning oxidation numbers per atom, not per ion

When two atoms of the same element change state — as with the two chromium atoms in $\ce{Cr2O7^2-}$ — the total change is the per-atom change multiplied by the number of atoms. Forgetting the factor of two for chromium is the single most common reason a dichromate balance comes out wrong.

Total decrease for Cr $= 3 \times 2 = 6$; this must equal the total increase contributed by the reductant.

The ion-electron (half-reaction) method

In the half-reaction method the two half equations are balanced separately and then added together to give the balanced equation, as NCERT states. The structure is rigid, which is precisely why it is reliable under examination pressure. The reduction halves of the two NEET workhorses can be memorised outright:

$$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$ $$\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$$

Each half-reaction is processed through seven NCERT steps: produce the unbalanced ionic form, separate into halves, balance atoms other than O and H, balance O with $\ce{H2O}$ and H with $\ce{H+}$, balance charge with electrons, equalise the electrons across the two halves, add the halves and cancel electrons, and finally verify atoms and charge. The merit of the method is that the electron count is made explicit at step five, so the equalisation in step six is mechanical.

e⁻
Build the foundation first

Every step here depends on correctly assigning oxidation states. Revise the rules in Oxidation Number before attempting full balances.

Acidic vs basic media: the step sequence

The two methods diverge only in how the medium supplies oxygen and hydrogen. In acidic medium oxygen is balanced with water and hydrogen with protons. In basic medium NCERT prescribes a two-stage device: first balance exactly as in acidic medium, then for each $\ce{H+}$ add an equal number of $\ce{OH-}$ to both sides, combining any $\ce{H+}$ and $\ce{OH-}$ on the same side into water. The table below sets the two sequences side by side.

StepAcidic mediumBasic medium
1Write the skeletal ionic equationWrite the skeletal ionic equation
2Split into oxidation and reduction halvesSplit into oxidation and reduction halves
3Balance atoms other than O and HBalance atoms other than O and H
4Balance O by adding H2OBalance O by adding H2O
5Balance H by adding H+Balance H by adding H+ (provisionally)
6Balance charge by adding e-For each H+, add equal OH- to both sides; combine into H2O
7Equalise e-, add halves, cancelBalance charge with e-, equalise, add halves, cancel
8Verify atoms and net chargeVerify atoms and net charge

The same logic carries over to the oxidation-number method: NIOS §13.3.1 instructs that in acidic medium you add $\ce{H2O}$ to the oxygen-deficient side and $\ce{H+}$ to the hydrogen-deficient side, whereas in basic medium you add $\ce{OH-}$ to the side deficient in negative charge and then balance with water. The medium never changes which atoms are oxidised or reduced; it only changes the spectator species used to close the mass and charge balance.

Worked problem: acidic medium

Balance the oxidation of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ by permanganate in acidic medium, $\ce{MnO4- + Fe^2+ -> Mn^2+ + Fe^3+}$, using both methods. This is the classic NEET-style coefficient question.

Worked Example · Half-Reaction Method

Step 1 — Split into halves.

Oxidation: $\ce{Fe^2+ -> Fe^3+}$. Reduction: $\ce{MnO4- -> Mn^2+}$.

Step 2 — Balance O and H in the reduction half (acidic).

$\ce{MnO4- + 8H+ -> Mn^2+ + 4H2O}$ (4 waters for 4 oxygens; 8 protons for 8 hydrogens).

Step 3 — Balance charge with electrons.

$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ and $\ce{Fe^2+ -> Fe^3+ + e-}$.

Step 4 — Equalise electrons and add.

Multiply the iron half by $5$: $\ce{5Fe^2+ -> 5Fe^3+ + 5e-}$. Adding the halves and cancelling electrons gives the balanced equation:

$$\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$$

Worked Example · Oxidation-Number Method

Same reaction, oxidation-number bookkeeping.

Mn falls from $+7$ to $+2$: decrease of $5$. Fe rises from $+2$ to $+3$: increase of $1$. To equalise, scale iron by $5$, giving $\ce{MnO4- + 5Fe^2+ -> Mn^2+ + 5Fe^3+}$.

Charge on the left is $-1 + 10 = +9$; on the right $+2 + 15 = +17$. Add $\ce{8H+}$ on the left to equalise charge, then $\ce{4H2O}$ on the right to balance hydrogen:

$$\ce{MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$$

Both methods deliver the identical coefficients $1,5,8,1,5,4$, confirming the balance.

The same permanganate reduction half drives the NEET 2018 oxalate problem, $\ce{MnO4- + C2O4^2- + H+ -> Mn^2+ + CO2 + H2O}$. There the oxidation half is $\ce{C2O4^2- -> 2CO2 + 2e-}$; multiplying it by $5$ and the permanganate half by $2$ yields $\ce{2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O}$, with reactant coefficients $2,5,16$.

Worked problem: basic medium

Balance the reaction in which permanganate oxidises bromide to bromate in basic medium, $\ce{MnO4- + Br- -> MnO2 + BrO3-}$, again by both methods.

Worked Example · Half-Reaction Method (Basic)

Step 1 — Halves.

Reduction: $\ce{MnO4- -> MnO2}$ (Mn: $+7 \to +4$). Oxidation: $\ce{Br- -> BrO3-}$ (Br: $-1 \to +5$).

Step 2 — Balance as if acidic.

Reduction: $\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}$. Oxidation: $\ce{Br- + 3H2O -> BrO3- + 6H+ + 6e-}$.

Step 3 — Equalise electrons.

Multiply the reduction half by $2$: $\ce{2MnO4- + 8H+ + 6e- -> 2MnO2 + 4H2O}$. Adding to the bromide half and cancelling $6e^-$ and the protons gives $\ce{2MnO4- + Br- + 2H+ -> 2MnO2 + BrO3- + H2O}$.

Step 4 — Convert to basic medium.

Add $\ce{2OH-}$ to both sides for the $\ce{2H+}$; the protons and hydroxide combine to $\ce{2H2O}$ on the left, one of which cancels the water on the right:

$$\ce{2MnO4- + Br- + H2O -> 2MnO2 + BrO3- + 2OH-}$$

Worked Example · Oxidation-Number Method (Basic)

Mn decreases by $3$ ($+7 \to +4$); Br increases by $6$ ($-1 \to +5$). Scaling permanganate by $2$ equalises the change: $\ce{2MnO4- + Br- -> 2MnO2 + BrO3-}$.

As the medium is basic and the left carries $-3$ charge against $-1$ on the right, add $\ce{2OH-}$ on the right to equalise charge, then balance the remaining hydrogen by adding one $\ce{H2O}$ on the left:

$$\ce{2MnO4- + Br- + H2O -> 2MnO2 + BrO3- + 2OH-}$$

This matches the NCERT Problem 7.9 answer exactly, and agrees with the half-reaction route.

Figure 2 · Electron Bookkeeping
Oxidation half Br⁻ → BrO₃⁻ + 6 e⁻ Reduction half (×2) 2 MnO₄⁻ + 6 e⁻ → 2 MnO₂ 6 e⁻ 6 electrons lost = 6 electrons gained → cancel
The electron count is the hinge of every balance: the six electrons released by bromide are exactly absorbed by two permanganate units, so they cancel when the halves are summed.

Common errors and final check

NCERT closes every worked balance with a verification step: confirm that each type of atom appears in equal numbers on both sides and that the net charge is the same left and right. This single check catches the great majority of mistakes — a missed water molecule shows up as an oxygen imbalance, and a wrong electron count shows up as a charge imbalance.

NEET Trap

Adding OH⁻ before the acidic balance is complete

In basic medium the discipline is to finish the acidic-style balance with $\ce{H2O}$ and $\ce{H+}$ first, and only then neutralise every $\ce{H+}$ with an equal $\ce{OH-}$ on both sides. Introducing hydroxide too early scrambles the oxygen count and is a frequent source of an off-by-one water molecule.

Rule: number of $\ce{OH-}$ added on each side = number of $\ce{H+}$ present, then merge co-located $\ce{H+}$ and $\ce{OH-}$ into $\ce{H2O}$.

Symptom after balancingLikely causeFix
Charge unequal on the two sidesWrong number of electrons in a halfRecount the oxidation-number change per species
Oxygen count offMissing or extra H2OOne H2O per oxygen on the deficient side
Hydrogen count offForgot H+ after adding waterAdd H+ to the hydrogen-deficient side
Non-integer coefficientsHalves not equalised before addingMultiply each half to a common electron count

Once verified, a balanced redox equation becomes the foundation for electrode and cell calculations, where the same half-reactions are reinterpreted as half-cell processes carrying measurable potentials.

Quick Recap

Balancing redox reactions in one screen

  • Two valid routes: oxidation-number method (equate total ON increase and decrease) and ion-electron method (balance separate halves, then add).
  • Acidic medium: balance O with $\ce{H2O}$, H with $\ce{H+}$, charge with $\ce{e-}$.
  • Basic medium: balance as if acidic, then add one $\ce{OH-}$ per $\ce{H+}$ on both sides and merge into $\ce{H2O}$.
  • Memorise $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ and $\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$.
  • Always finish with the NCERT check: equal atom counts and equal net charge on both sides.

NEET PYQ Snapshot — Balancing Redox Reactions

Coefficient-recovery questions are the dominant NEET format for this subtopic; the ion-electron method resolves each in seconds.

NEET 2018

For the redox reaction $\ce{MnO4- + C2O4^2- + H+ -> Mn^2+ + CO2 + H2O}$, the correct coefficients of the reactants $\ce{MnO4-}$, $\ce{C2O4^2-}$, $\ce{H+}$ for the balanced equation are:

  • (1) 16, 5, 2
  • (2) 2, 5, 16
  • (3) 2, 16, 5
  • (4) 5, 16, 2
Answer: (2) — 2, 5, 16

Oxidation half $\ce{C2O4^2- -> 2CO2 + 2e-}$ ($\times 5$); reduction half $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ ($\times 2$). Adding gives $\ce{2MnO4- + 5C2O4^2- + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O}$.

NEET 2023

On balancing $\ce{aCr2O7^2- + bSO3^2- + cH+ -> 2aCr^3+ + bSO4^2- + H2O}$, the coefficients $a$, $b$, $c$ are respectively:

  • (1) 8, 1, 3
  • (2) 1, 3, 8
  • (3) 3, 8, 1
  • (4) 1, 8, 3
Answer: (2) — 1, 3, 8

Reduction $\ce{Cr2O7^2- + 6e- -> 2Cr^3+}$; oxidation $\ce{SO3^2- -> SO4^2- + 2e-}$ ($\times 3$). Balancing O with $\ce{4H2O}$ and H with $\ce{8H+}$ gives $a=1$, $b=3$, $c=8$.

NEET 2022

In the neutral or faintly alkaline medium, $\ce{KMnO4}$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from:

  • (1) +6 to +4
  • (2) +7 to +3
  • (3) +6 to +5
  • (4) +7 to +4
Answer: (4) — +7 to +4

The balanced change is $\ce{2MnO4- + H2O + I- -> 2MnO2 + 2OH- + IO3-}$; manganese goes from $+7$ in $\ce{MnO4-}$ to $+4$ in $\ce{MnO2}$, the same reduction half used to balance basic-medium permanganate reactions.

NEET 2022

Given the half cells $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ ($E^\circ_{\ce{MnO4-}/\ce{Mn^2+}} = +1.510$ V) and $\ce{O2 + 2H+ + 2e- -> H2O}$ ($E^\circ = +1.223$ V), will $\ce{MnO4-}$ liberate $\ce{O2}$ from water in acid?

  • (1) No, because E°cell = −0.287 V
  • (2) Yes, because E°cell = +2.733 V
  • (3) No, because E°cell = −2.733 V
  • (4) Yes, because E°cell = +0.287 V
Answer: (4) — Yes, E°cell = +0.287 V

Combining $2\times$ the permanganate half with $5\times$ the reversed oxygen half gives $\ce{2MnO4- + 6H+ -> 2Mn^2+ + O2 + 3H2O}$. $E^\circ_{cell} = 1.510 - 1.223 = +0.287$ V $> 0$, so the reaction is spontaneous — the same half-reaction balancing underpins the cell analysis.

FAQs — Balancing Redox Reactions

The recurring conceptual hurdles students meet when balancing redox equations for NEET.

What is the difference between the oxidation-number method and the ion-electron method?

The oxidation-number method balances the equation by equating the total increase in oxidation number of the reducing agent with the total decrease in oxidation number of the oxidising agent, then balancing the remaining atoms and charge. The ion-electron (half-reaction) method splits the reaction into separate oxidation and reduction half-reactions, balances each one for atoms, oxygen, hydrogen and charge with explicit electrons, equalises the electrons, and then adds the halves so the electrons cancel. Both are valid; NCERT states the choice rests with the individual using them.

How do you balance oxygen and hydrogen in acidic medium?

In acidic medium you balance oxygen by adding one H2O molecule for each oxygen atom needed on the deficient side, then balance hydrogen by adding H+ ions to the side short of hydrogen. For example, in reducing Cr2O7^2- to Cr3+ you add 7H2O on the right to balance oxygen and 14H+ on the left to balance hydrogen.

How does the basic medium differ from the acidic medium when balancing?

In a basic medium you first balance the half-reaction exactly as in acidic medium using H2O and H+. Then, for every H+ ion present, add an equal number of OH- ions to both sides of the equation. Wherever H+ and OH- appear on the same side, combine them into H2O, and cancel any excess water. This converts the acidic-style balance into one valid for alkaline conditions.

Why must the number of electrons be equal in the two half-reactions before adding them?

Electrons are neither created nor destroyed in a redox reaction; every electron lost in oxidation must be gained in reduction. The electrons can cancel completely only when the oxidation and reduction halves carry the same number of electrons, so each half is multiplied by a suitable integer to make the electron counts equal before the halves are added.

How do I verify that a balanced redox equation is correct?

Carry out the final check NCERT prescribes: confirm that each type of atom occurs in equal numbers on both sides, and that the total electrical charge is identical on the left and the right. If both the atom count and the net charge match, the equation is fully balanced.

Which method is faster for NEET multiple-choice questions on coefficients?

For permanganate and dichromate problems where the half-reactions are standard, the ion-electron method is usually fastest because the reduction halves MnO4- + 8H+ + 5e- to Mn2+ + 4H2O and Cr2O7^2- + 14H+ + 6e- to 2Cr3+ + 7H2O can be recalled directly. NEET solutions such as the 2018 MnO4-/C2O4^2- and 2023 Cr2O7^2-/SO3^2- problems are worked using the ion-electron method.

Related Topics
Redox Reactions Return to the full chapter hub and all subtopics. Oxidation Number The rules every balance depends on. Types of Redox Reactions Combination, displacement and disproportionation. Redox and Electrode Processes Half-reactions reinterpreted as half-cells. Electrochemistry Cell potentials built on balanced redox halves.