Balancing Redox Equations — the Oxidation Number Method

What balancing a redox equation actually means

A skeletal equation carries the correct formulae of the reactants and products and nothing else. Balancing it means forcing three quantities to agree on the two sides: the atoms of every element, the total electrical charge, and — underneath both — the electrons handed from the reducing agent to the oxidising agent.

Two methods do this. The oxidation number method tracks the rise and fall of oxidation numbers and never writes an electron on the page. The half-reaction (ion-electron) method splits the change into two halves and cancels the electrons at the end. This section covers the first, the next section the second, and both give the same answer.

The oxidation number method is usually quicker on a molecular equation — one written with whole compounds such as $\mathrm{KMnO_4}$ and $\mathrm{HCl}$ — because it never forces you to invent ionic fragments. It rests on one arithmetic identity.

Key Point: Total increase in oxidation number = total decrease in oxidation number. Electrons are neither created nor destroyed in a chemical change, so every unit of oxidation number gained by one atom must be paid for by a unit lost by another.

That identity fixes the coefficients of the two species that change. Everything after it is book-keeping: spectator atoms, then oxygen, then hydrogen, then the charge check. That last check is not decoration. An equation can have every atom accounted for and still be wrong because the $\mathrm{H^+}$ count is off, and charge is the only thing that catches it.

The procedure, step by step

Work these steps in order every time. Balancing hydrogen before oxygen, or spectator atoms after adding water, produces a mess you then have to unpick.

Step 1 — Write the skeletal equation. Every reactant and product, with correct formulae. If a formula is wrong, no amount of balancing will save the equation.

Step 2 — Assign oxidation numbers to every atom and mark the ones that change. One element should be going up and one going down (in a disproportionation, the same element does both). If nothing changes, the reaction is not redox; if two things go down and nothing goes up, a formula is wrong.

Step 3 — Compute the total increase and the total decrease per formula unit. Change per atom, multiplied by the number of atoms of that element in the formula unit. This multiplication is where most marks are lost.

Step 4 — Equalise them. Multiply the oxidised and reduced species by whatever whole numbers make total increase equal to total decrease. The lowest common multiple of the two totals gives the smallest coefficients.

Step 5 — Balance every other atom except H and O. Metals, halides, sulphur in a spectator sulphate, and so on.

Step 6 — Balance oxygen with $\mathrm{H_2O}$, then hydrogen with $\mathrm{H^+}$ (acidic medium). Oxygen first, always: water carries hydrogen with it, so fixing hydrogen first means redoing it.

Step 7 — Convert to basic medium if the question says basic. Add as many $\mathrm{OH^-}$ ions to both sides as there are $\mathrm{H^+}$ ions, combine each $\mathrm{H^+ + OH^-}$ pair into one $\mathrm{H_2O}$, then cancel water that now appears on both sides.

Step 8 — Verify. Count every element on both sides, then add up the charges on both sides. Both must match.

Key Point (Definition): A balanced redox equation conserves atoms and charge at the same time. Conservation of charge is the visible signature of conservation of electrons, which is why an atom-balanced but charge-unbalanced equation is wrong.

Step 3 in detail, because this is where marks go

Take dichromate reduced to chromium(III):

$$\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}$$

Each chromium falls from $+6$ to $+3$, a change of $3$ per atom, and there are two chromium atoms in the ion. The total decrease per dichromate ion is $2 \times 3 = 6$, not $3$.

Take oxalate oxidised to carbon dioxide:

$$\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$$

Each carbon rises from $+3$ to $+4$, a change of $1$ per atom, over two carbons. The total increase per oxalate ion is $2$, not $1$.

The rule is short: change per atom $\times$ number of atoms in the formula unit. Write that product down rather than doing it in your head.

Once the two totals are in front of you the ratio follows. Iron gives $+1$ per $\mathrm{Fe^{2+}}$ and dichromate takes $6$, so six $\mathrm{Fe^{2+}}$ meet one $\mathrm{Cr_2O_7^{2-}}$. Permanganate takes $5$ and oxalate gives $2$, whose lowest common multiple is $10$, so two $\mathrm{MnO_4^-}$ meet five $\mathrm{C_2O_4^{2-}}$ — the $2:5$ ratio behind every permanganate titration.

[JEE Main] Coefficient questions are almost always testing this one multiplication. If a question asks for the coefficient of $\mathrm{H^+}$ or $\mathrm{H_2O}$, get the two totals right first; the rest is arithmetic that cannot go wrong.

Question 1: Iron(II) oxidised by dichromate in acid

Balance $\mathrm{Fe^{2+} + Cr_2O_7^{2-} \rightarrow Fe^{3+} + Cr^{3+}}$ in acidic medium.

Answer:

First I put oxidation numbers on the atoms that matter. Iron goes $+2 \rightarrow +3$; chromium goes $+6 \rightarrow +3$.

Increase: $+1$ per $\mathrm{Fe^{2+}}$. Decrease: $3$ per Cr atom, and two Cr atoms per ion, so $6$ per $\mathrm{Cr_2O_7^{2-}}$.

The smallest whole numbers making $1 \times a = 6 \times b$ are $a = 6$, $b = 1$. Six iron ions per dichromate ion:

$$\mathrm{6Fe^{2+} + Cr_2O_7^{2-} \rightarrow 6Fe^{3+} + 2Cr^{3+}}$$

Fe and Cr are now balanced. Oxygen: seven on the left, none on the right, so seven waters on the right.

$$\mathrm{6Fe^{2+} + Cr_2O_7^{2-} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O}$$

Fourteen hydrogens have appeared on the right, so I add fourteen $\mathrm{H^+}$ on the left.

$$\mathrm{6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O}$$

Atom check: Fe $6 = 6$, Cr $2 = 2$, O $7 = 7$, H $14 = 14$. Charge check: left $6(+2) + (-2) + 14(+1) = +12 - 2 + 14 = +24$. Right $6(+3) + 2(+3) = +18 + 6 = +24$. Equal.

Ans: $\mathrm{6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O}$ Watch out: Taking the decrease as $3$ instead of $6$ gives a $3:1$ ratio, and then the charges refuse to match no matter how you juggle the water.

Question 2: Permanganate and sulphite in acid

Balance $\mathrm{MnO_4^- + SO_3^{2-} \rightarrow Mn^{2+} + SO_4^{2-}}$ in acidic medium.

Answer:

Manganese: $+7 \rightarrow +2$, one Mn per ion, so the decrease is $5$. Sulphur: $+4 \rightarrow +6$, an increase of $2$ per ion.

The lowest common multiple of $5$ and $2$ is $10$, so two permanganate ions and five sulphite ions.

$$\mathrm{2MnO_4^- + 5SO_3^{2-} \rightarrow 2Mn^{2+} + 5SO_4^{2-}}$$

Mn and S are already balanced. Oxygen: left $8 + 15 = 23$, right $20$, so three waters on the right.

$$\mathrm{2MnO_4^- + 5SO_3^{2-} \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O}$$

Six hydrogens on the right, so six $\mathrm{H^+}$ on the left.

$$\mathrm{2MnO_4^- + 5SO_3^{2-} + 6H^+ \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O}$$

Atom check: Mn $2 = 2$, S $5 = 5$, O $23 = 23$, H $6 = 6$. Charge check: left $2(-1) + 5(-2) + 6(+1) = -2 - 10 + 6 = -6$. Right $2(+2) + 5(-2) = +4 - 10 = -6$. Equal.

Ans: $\mathrm{2MnO_4^- + 5SO_3^{2-} + 6H^+ \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O}$ Watch out: The $\mathrm{H^+}$ coefficient here is $6$, not $16$. Sixteen is what the permanganate half reaction alone needs; the sulphite half reaction releases ten $\mathrm{H^+}$ that cancel against it.

Question 3: Copper with dilute nitric acid

Balance $\mathrm{Cu + HNO_3(dilute) \rightarrow Cu(NO_3)_2 + NO + H_2O}$.

Answer:

Copper starts as the free element, $0$, and ends as $+2$ in $\mathrm{Cu(NO_3)_2}$. Increase $2$ per Cu.

Nitrogen in $\mathrm{HNO_3}$ is $+5$; in NO it is $+2$, a decrease of $3$ per N. The nitrogen inside $\mathrm{Cu(NO_3)_2}$ is still $+5$, so only the nitrogen that becomes NO counts in the arithmetic. This is the trap in every nitric acid balancing question.

Lowest common multiple of $2$ and $3$ is $6$: three Cu atoms oxidised, two N atoms reduced.

$$\mathrm{3Cu + HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + H_2O}$$

Now count nitrogen on the right: $3 \times 2 = 6$ inside the nitrate, plus $2$ in NO, giving $8$. So eight $\mathrm{HNO_3}$ on the left.

$$\mathrm{3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + H_2O}$$

Eight hydrogens on the left means four waters on the right.

$$\mathrm{3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O}$$

Atom check: Cu $3 = 3$, N $8 = 6 + 2 = 8$, H $8 = 8$, O left $24$, right $18 + 2 + 4 = 24$. Charge check: both sides are made of neutral species, so the charge is $0 = 0$. Writing it in net ionic form, $\mathrm{3Cu + 8H^+ + 2NO_3^- \rightarrow 3Cu^{2+} + 2NO + 4H_2O}$, the left carries $+8 - 2 = +6$ and the right carries $3(+2) = +6$. Equal.

Ans: $\mathrm{3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O}$ Watch out: Six of the eight nitric acid molecules are behaving as an acid, not as an oxidant. Only two are reduced. Counting all eight as reduced gives a decrease of $24$ and a hopeless equation.

Question 4: Copper with concentrated nitric acid

Balance $\mathrm{Cu + HNO_3(concentrated) \rightarrow Cu(NO_3)_2 + NO_2 + H_2O}$, and say why the product differs from the dilute case.

Answer:

Copper again rises $0 \rightarrow +2$, an increase of $2$. Nitrogen now falls only from $+5$ to $+4$ in $\mathrm{NO_2}$, a decrease of $1$ per N. To match an increase of $2$ I need two nitrogens reduced per copper atom.

$$\mathrm{Cu + HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + H_2O}$$

Nitrogen on the right: $2$ in the nitrate plus $2$ in $\mathrm{NO_2}$, giving $4$. So four $\mathrm{HNO_3}$ on the left, and four hydrogens means two waters.

$$\mathrm{Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O}$$

Atom check: Cu $1 = 1$, N $4 = 2 + 2 = 4$, H $4 = 4$, O left $12$, right $6 + 4 + 2 = 12$. Charge check: neutral on both sides. As a net ionic equation, $\mathrm{Cu + 4H^+ + 2NO_3^- \rightarrow Cu^{2+} + 2NO_2 + 2H_2O}$ gives left $+4 - 2 = +2$ and right $+2$. Equal.

Concentrated acid supplies nitrate in such excess that the reduction stops one step earlier, at $\mathrm{NO_2}$; dilute acid carries it further, to NO. The reducing agent is the same copper, and the depth of reduction is set by the concentration.

Ans: $\mathrm{Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O}$; the product is $\mathrm{NO_2}$ rather than NO because concentrated acid supplies nitrate in such excess that the reduction stops one step earlier. Watch out: Compare acid per copper, never the bare coefficients. The eight in the dilute equation is shared among three copper atoms, which is $8/3 = 2.67$ each, while the four here sits on a single copper, which is $4.00$. Setting $8$ against $4$ without first dividing by the copper coefficient reverses the conclusion.

The same oxidant, three different products

A balanced equation is only correct for the medium it was written for. Permanganate is the standard illustration, and all three of its fates appear in exam papers.

Written out as half changes with the medium's own species:

$$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$$

$$\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$$

$$\mathrm{MnO_4^- + e^- \rightarrow MnO_4^{2-}}$$

Check the middle one for charge: left $-1 - 3 = -4$, right $4(-1) = -4$. It balances, and it contains no $\mathrm{H^+}$, which is what a basic-medium equation must look like.

These electron counts are the n-factors of $\mathrm{KMnO_4}$: $5$ in acid, $3$ in neutral or weakly basic solution, $1$ in strong alkali. With a molar mass of $158\ \mathrm{g\ mol^{-1}}$ the equivalent masses are $158/5 = 31.6$, $158/3 = 52.67$ and $158$.

Key Point: A redox equation is not a property of the reactants alone. Change the medium and the products change, so the coefficients change. Read the words "in acidic medium" or "in alkaline medium" before you write anything.

[NEET] Question stems often hide the medium in a single word — "acidified", "in the presence of $\mathrm{KOH}$", "neutral". That word decides the answer.

Question 5: The permanganate-oxalate reaction

Balance $\mathrm{C_2O_4^{2-} + MnO_4^- \rightarrow CO_2 + Mn^{2+}}$ in acidic medium.

Answer:

Carbon in oxalate is $+3$; in $\mathrm{CO_2}$ it is $+4$. That is $+1$ per carbon over two carbons, so the increase is $2$ per oxalate ion. Manganese falls $+7 \rightarrow +2$, a decrease of $5$.

Lowest common multiple of $2$ and $5$ is $10$: five oxalate ions and two permanganate ions.

$$\mathrm{2MnO_4^- + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2}$$

Carbon is balanced at $10$. Oxygen: left $8 + 20 = 28$, right $20$, so eight waters on the right.

$$\mathrm{2MnO_4^- + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$$

Sixteen hydrogens on the right, so sixteen $\mathrm{H^+}$ on the left.

$$\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$$

Atom check: Mn $2 = 2$, C $10 = 10$, O $28 = 20 + 8 = 28$, H $16 = 16$. Charge check: left $2(-1) + 5(-2) + 16(+1) = -2 - 10 + 16 = +4$. Right $2(+2) = +4$. Equal.

Ans: $\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$, mole ratio $\mathrm{MnO_4^-} : \mathrm{C_2O_4^{2-}} = 2:5$. Watch out: Using $+1$ as the oxalate increase instead of $+2$ gives a $1:5$ ratio and destroys the titration arithmetic that depends on $2:5$.

Question 6: Dichromate and sulphur dioxide

Balance $\mathrm{Cr_2O_7^{2-} + SO_2 \rightarrow Cr^{3+} + SO_4^{2-}}$ in acidic medium.

Answer:

Chromium falls $+6 \rightarrow +3$ over two atoms, so the decrease is $6$ per dichromate ion. Sulphur rises $+4 \rightarrow +6$, an increase of $2$ per $\mathrm{SO_2}$.

To match $6$ I need three $\mathrm{SO_2}$.

$$\mathrm{Cr_2O_7^{2-} + 3SO_2 \rightarrow 2Cr^{3+} + 3SO_4^{2-}}$$

Cr is $2 = 2$, S is $3 = 3$. Oxygen: left $7 + 6 = 13$, right $12$. One water on the right.

$$\mathrm{Cr_2O_7^{2-} + 3SO_2 \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O}$$

Two hydrogens on the right, so two $\mathrm{H^+}$ on the left.

$$\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O}$$

Atom check: Cr $2 = 2$, S $3 = 3$, O $13 = 12 + 1 = 13$, H $2 = 2$. Charge check: left $(-2) + 3(0) + 2(+1) = 0$. Right $2(+3) + 3(-2) = +6 - 6 = 0$. Equal.

Ans: $\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O}$ Watch out: The $\mathrm{H^+}$ coefficient is only $2$, not $14$. Fourteen belongs to the dichromate half reaction on its own; here the sulphur dioxide brings six oxygens of its own, so far less water — and therefore far less acid — is needed.

Basic medium: balance as if acidic, then neutralise

There is no separate procedure to learn for alkaline solution. Balance exactly as you would in acid, ending with some number of $\mathrm{H^+}$ ions on one side, then convert.

The conversion, in three moves. Suppose you finish with $n$ hydrogen ions on the left.

1. Add $n$ hydroxide ions to both sides. Adding the same thing to both sides changes nothing chemically, which is why this is legal.
2. On the left, every $\mathrm{H^+}$ now sits next to an $\mathrm{OH^-}$. Combine each pair into one $\mathrm{H_2O}$, giving $n$ waters on the left.
3. Water may now appear on both sides. Cancel the smaller number from both, leaving water on one side only.

The right side keeps its $n$ hydroxide ions, and the finished equation contains no $\mathrm{H^+}$ at all.

Key Point: A basic-medium answer must not contain $\mathrm{H^+}$, and an acidic-medium answer must not contain $\mathrm{OH^-}$. If either appears, the neutralisation step was skipped or done on one side only.

The commonest error. Adding the hydroxide to one side only is not an identity operation — it shifts the charge on that side by $-n$ and quietly breaks the equation. The charge check catches it immediately, which is the whole reason for doing the charge check.

[Board] Marks are given for the neutralisation step being shown, not just for the final equation. Write the intermediate line with $\mathrm{H^+}$ and $\mathrm{OH^-}$ both visible before you collapse them into water.

Question 7: Permanganate and iodide in basic medium

Permanganate ion oxidises iodide ion in basic solution to iodine, while manganese ends up as $\mathrm{MnO_2}$. Balance the equation.

Answer:

Skeleton: $\mathrm{MnO_4^- + I^- \rightarrow MnO_2 + I_2}$.

Manganese: $+7 \rightarrow +4$, a decrease of $3$. Iodine: $-1 \rightarrow 0$, an increase of $1$ per atom, and $\mathrm{I_2}$ holds two of them, so $2$ per $\mathrm{I_2}$ produced.

Lowest common multiple of $3$ and $2$ is $6$. Two permanganate ions take six units; six iodide ions supply them and give three $\mathrm{I_2}$.

$$\mathrm{2MnO_4^- + 6I^- \rightarrow 2MnO_2 + 3I_2}$$

Now I pretend the medium is acidic. Oxygen: left $8$, right $4$, so four waters on the right and then eight $\mathrm{H^+}$ on the left.

$$\mathrm{2MnO_4^- + 6I^- + 8H^+ \rightarrow 2MnO_2 + 3I_2 + 4H_2O}$$

Charge here: left $-2 - 6 + 8 = 0$, right $0$. The acidic version is sound, so the conversion will be sound too.

The medium is basic, so I add eight $\mathrm{OH^-}$ to both sides and collapse the eight $\mathrm{H^+ + OH^-}$ pairs on the left into eight waters:

$$\mathrm{2MnO_4^- + 6I^- + 8H_2O \rightarrow 2MnO_2 + 3I_2 + 4H_2O + 8OH^-}$$

Water appears on both sides. Cancelling four from each:

$$\mathrm{2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-}$$

Atom check: Mn $2 = 2$, I $6 = 6$, O left $8 + 4 = 12$, right $4 + 8 = 12$, H left $8$, right $8$. Charge check: left $2(-1) + 6(-1) = -8$. Right $8(-1) = -8$. Equal.

Ans: $\mathrm{2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-}$ Watch out: Forgetting that $\mathrm{I_2}$ carries two iodine atoms makes the increase $1$ instead of $2$, and the iodide coefficient comes out as $3$ rather than $6$.

Question 8: Permanganate and bromide in basic medium

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Balance it.

Answer:

Skeleton: $\mathrm{MnO_4^- + Br^- \rightarrow MnO_2 + BrO_3^-}$.

Manganese: $+7 \rightarrow +4$, decrease $3$. Bromine: $-1 \rightarrow +5$, increase $6$. To match $6$ I need two manganese atoms:

$$\mathrm{2MnO_4^- + Br^- \rightarrow 2MnO_2 + BrO_3^-}$$

Balancing as if acidic: oxygen left $8$, right $4 + 3 = 7$, so one water on the right and two $\mathrm{H^+}$ on the left.

$$\mathrm{2MnO_4^- + Br^- + 2H^+ \rightarrow 2MnO_2 + BrO_3^- + H_2O}$$

Charge: left $-2 - 1 + 2 = -1$, right $-1$. Sound.

Adding two $\mathrm{OH^-}$ to both sides and collapsing the pair on the left gives two waters on the left and one on the right; cancelling one from each side:

$$\mathrm{2MnO_4^- + Br^- + H_2O \rightarrow 2MnO_2 + BrO_3^- + 2OH^-}$$

Atom check: Mn $2 = 2$, Br $1 = 1$, O left $8 + 1 = 9$, right $4 + 3 + 2 = 9$, H left $2$, right $2$. Charge check: left $2(-1) + (-1) = -3$. Right $(-1) + 2(-1) = -3$. Equal.

Ans: $\mathrm{2MnO_4^- + Br^- + H_2O \rightarrow 2MnO_2 + BrO_3^- + 2OH^-}$ Watch out: Bromide going to bromate is a six-electron change, the largest common one in this chapter. Treating it as a one-electron or five-electron change is the usual slip.

When the same element is oxidised and reduced

In a disproportionation one element leaves the reactant in two directions at once — some atoms go up, some go down. The method still works, with one adjustment: increase and decrease are both computed for the same starting species, and the ratio you extract is the ratio of the two products, not of two reactants.

The procedure:

1. Identify the intermediate oxidation state and the two products it splits into.
2. Compute the rise per atom to the higher product and the fall per atom to the lower product.
3. Equalise: if the fall is $f$ and the rise is $r$, then for every one atom that falls, $f/r$ atoms must rise. Convert to whole numbers.
4. Only then fix the coefficient of the reactant, so that its atom count covers both products.
5. Finish with the usual oxygen, hydrogen and charge balancing.

Step 4 is the one that trips people up. In $\mathrm{P_4}$ the coefficient is decided only after you know how many phosphorus atoms end up in each product, because all four come out of the same molecule.

Key Point: Disproportionation needs an element in an intermediate oxidation state — one with a higher state and a lower state both available. Fluorine has no positive oxidation state, so fluorine never disproportionates.

The standard cases worth recognising on sight:

• $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$, oxygen at $-1$ going to $-2$ and to $0$.
• $\mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O}$ in cold dilute alkali.
• $\mathrm{3Cl_2 + 6OH^- \rightarrow 5Cl^- + ClO_3^- + 3H_2O}$ in hot concentrated alkali.
• $\mathrm{2Cu^+ \rightarrow Cu^{2+} + Cu}$ in aqueous solution.
• $\mathrm{3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O}$, manganese at $+6$ splitting to $+7$ and $+4$.

Question 9: White phosphorus in hot alkali

Balance $\mathrm{P_4 + OH^- \rightarrow PH_3 + H_2PO_2^-}$ in basic medium.

Answer:

Phosphorus in $\mathrm{P_4}$ is a free element, so it is $0$. In $\mathrm{PH_3}$ hydrogen is $+1$, so phosphorus is $-3$. In $\mathrm{H_2PO_2^-}$ the two hydrogens are $+1$ each and the two oxygens $-2$ each, so $2 - 4 + x = -1$ gives $x = +1$. The same element goes both ways, which makes this a disproportionation.

Fall: $0 \rightarrow -3$, a decrease of $3$ per P atom going to $\mathrm{PH_3}$. Rise: $0 \rightarrow +1$, an increase of $1$ per P atom going to $\mathrm{H_2PO_2^-}$.

For every one atom that falls by $3$, three atoms must rise by $1$, so the product ratio is $\mathrm{PH_3} : \mathrm{H_2PO_2^-} = 1 : 3$.

$$\mathrm{P_4 + OH^- \rightarrow PH_3 + 3H_2PO_2^-}$$

Phosphorus check: the right side has $1 + 3 = 4$ atoms and the left has exactly four in one $\mathrm{P_4}$, so the coefficient of $\mathrm{P_4}$ is $1$ — a lucky fit here, but always worth checking.

Oxygen next. The right has $3 \times 2 = 6$ oxygens; the left has only what the hydroxide supplies. Charge fixes the hydroxide count: the right carries $3(-1) = -3$, so the left needs three $\mathrm{OH^-}$.

$$\mathrm{P_4 + 3OH^- \rightarrow PH_3 + 3H_2PO_2^-}$$

Three hydroxides bring three oxygens, and six are needed, so I add three waters on the left.

$$\mathrm{P_4 + 3OH^- + 3H_2O \rightarrow PH_3 + 3H_2PO_2^-}$$

Atom check: P $4 = 1 + 3 = 4$; O left $3 + 3 = 6$, right $3 \times 2 = 6$; H left $3 + 6 = 9$, right $3 + 3 \times 2 = 9$. Charge check: left $3(-1) = -3$. Right $3(-1) = -3$. Equal.

Ans: $\mathrm{P_4 + 3OH^- + 3H_2O \rightarrow PH_3 + 3H_2PO_2^-}$ Watch out: The two half changes here start from the same species, so you cannot look for "the oxidant" and "the reductant" as different reactants. $\mathrm{P_4}$ is both. Hydroxide is neither — its oxidation numbers do not change.

Question 10: Dichromate and sulphite in acid

Write the net ionic equation for the reaction of potassium dichromate(VI) with sodium sulphite in acid solution, giving chromium(III) ion and sulphate ion.

Answer:

Potassium and sodium are spectators, so I drop them and keep the ions that react: $\mathrm{Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_4^{2-}}$.

Chromium is $+6$ in dichromate and $+3$ in the product, a fall of $3$ per atom over two atoms, total $6$. Sulphur is $+4$ in sulphite and $+6$ in sulphate, a rise of $2$. Dichromate is the oxidant, sulphite the reductant, and matching a decrease of $6$ needs three sulphite ions.

$$\mathrm{Cr_2O_7^{2-} + 3SO_3^{2-} \rightarrow 2Cr^{3+} + 3SO_4^{2-}}$$

The charges are $-2 - 6 = -8$ on the left and $+6 - 6 = 0$ on the right. The medium is acidic, so I add eight $\mathrm{H^+}$ on the left to close the gap.

$$\mathrm{Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-}}$$

Eight hydrogens on the left must reappear as four waters on the right.

$$\mathrm{Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O}$$

Atom check: Cr $2 = 2$, S $3 = 3$, O left $7 + 9 = 16$, right $12 + 4 = 16$, H $8 = 8$. Charge check: left $(-2) + 3(-2) + 8(+1) = -2 - 6 + 8 = 0$. Right $2(+3) + 3(-2) = 0$. Equal.

Ans: $\mathrm{Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O}$ Watch out: This version uses the charge gap to fix the $\mathrm{H^+}$ count and water afterwards, rather than oxygen first. Both routes give the same equation; pick one and stay with it so you do not add hydrogen twice.

Question 11: Permanganate with hydrochloric acid

Balance $\mathrm{KMnO_4 + HCl \rightarrow KCl + MnCl_2 + Cl_2 + H_2O}$, and say what this tells you about running a permanganate titration.

Answer:

Manganese is $+7$ in $\mathrm{KMnO_4}$ and $+2$ in $\mathrm{MnCl_2}$, a fall of $5$. Chlorine rises from $-1$ to $0$, a rise of $1$ per atom.

I take two permanganates so that whole $\mathrm{Cl_2}$ molecules come out: a total fall of $10$, matched by ten chlorine atoms oxidised, which is five $\mathrm{Cl_2}$.

$$\mathrm{2KMnO_4 + HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + H_2O}$$

Chlorine on the right: $2$ in KCl, $4$ in $\mathrm{2MnCl_2}$, $10$ in $\mathrm{5Cl_2}$, total $16$. So sixteen HCl, and sixteen hydrogens give eight waters.

$$\mathrm{2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O}$$

Atom check: K $2 = 2$, Mn $2 = 2$, O $8 = 8$, H $16 = 16$, Cl $16 = 2 + 4 + 10 = 16$. Charge check: all species neutral, $0 = 0$.

Since permanganate oxidises chloride, hydrochloric acid can never acidify a permanganate titration — permanganate would be spent on the acid instead of on the substance being estimated, and the titre would read high. Dilute sulphuric acid is used instead, nitric acid being an oxidant itself.

Ans: $\mathrm{2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O}$; the acid for a permanganate titration must be dilute sulphuric acid. Watch out: Ten chlorine atoms are oxidised but only five $\mathrm{Cl_2}$ molecules appear. Writing $10\mathrm{Cl_2}$ is the standard slip.

Question 12: Permanganate and hydrogen peroxide in acid

Balance $\mathrm{MnO_4^- + H_2O_2 \rightarrow Mn^{2+} + O_2}$ in acidic medium.

Answer:

Manganese falls $+7 \rightarrow +2$, a decrease of $5$.

Oxygen in $\mathrm{H_2O_2}$ is $-1$, the peroxide value, not $-2$; in $\mathrm{O_2}$ it is $0$. That is a rise of $1$ per oxygen atom over two atoms, so the increase is $2$ per $\mathrm{H_2O_2}$, which is acting as the reducing agent.

Lowest common multiple of $5$ and $2$ is $10$: two permanganate and five peroxide.

$$\mathrm{2MnO_4^- + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2}$$

Oxygen: left $8 + 10 = 18$, right $10$, so eight waters on the right.

$$\mathrm{2MnO_4^- + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O}$$

Hydrogen: right has $16$, left has $10$, so six $\mathrm{H^+}$ on the left.

$$\mathrm{2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O}$$

Atom check: Mn $2 = 2$, O left $8 + 10 = 18$, right $10 + 8 = 18$, H left $10 + 6 = 16$, right $16$. Charge check: left $2(-1) + 6(+1) = +4$. Right $2(+2) = +4$. Equal.

Ans: $\mathrm{2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O}$ Watch out: Taking oxygen in $\mathrm{H_2O_2}$ as $-2$ makes the peroxide look unchanged and the equation unbalanceable. Hydrogen peroxide acts either way — as a reductant it goes to $\mathrm{O_2}$, as an oxidant to $\mathrm{H_2O}$ — with an n-factor of $2$ in both roles.

The verification habit

Every equation in this section ended with two counts. Make that automatic.

Atom count. Go element by element, left then right, and write each pair down. Doing it in your head is how a stray water molecule survives to the end of a paper.

Charge count. Multiply each species' charge by its coefficient and add. Neutral molecules contribute zero, so a molecular equation always gives $0 = 0$. In a net ionic equation it is the check that matters:

$$\mathrm{6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O}$$

Left: $6 \times (+2) = +12$; $1 \times (-2) = -2$; $14 \times (+1) = +14$. Sum $+24$. Right: $6 \times (+3) = +18$; $2 \times (+3) = +6$; water contributes $0$. Sum $+24$.

If the two sums differ by $k$, the $\mathrm{H^+}$ count is almost always wrong by $k$, since $\mathrm{H^+}$ is the only species added on charge grounds. That makes the check a diagnosis, not just a verdict.

When to use which method. Use the oxidation number method for molecular equations, for anything carrying a large excess of spectator ions such as the nitric acid and hydrochloric acid problems above, and whenever the oxidation numbers are visible at a glance. Use the half-reaction method when the question asks for the number of electrons transferred, or when you want the two halves for an electrochemical cell. The next section develops that method in full.

[JEE/NEET] For a purely numerical answer — "the coefficient of water is" — the oxidation number method is faster, because it never asks you to write electrons that you will only cancel again.