Solved Examples

What This Set Is

Everything the chapter taught you meets here. Forty worked questions, numbered straight through, arranged so that each group leans on the one before it. The first few are one-line oxidation numbers. The last few are full back-titrations and cell EMFs.

Work them with a pen. Cover the answer, try the question, then read the working and see where your route differed from mine. A question you only read is a question you have not done.

Three habits are worth building while you work through this set.

Write the oxidation number over every atom that could change, before you do anything else. Half the mistakes in this chapter are made in the first ten seconds, by guessing which atom is the interesting one.

Check charge as well as atoms. An equation can have every atom matched and still be wrong. Add up the charge on the left, add it up on the right, and make the two agree before you call it balanced.

Say out loud which species is the oxidant. The oxidising agent is the one that is reduced. Students lose more marks to that single swap than to any calculation error in the chapter.

Questions	Topic	What it tests
1-5	Classical and electronic definitions	Reading a reaction three ways and naming the oxidant and the reductant
6-13	Oxidation number assignment	The rules in priority order, the peroxide-linkage traps and the fractional cases
14-16	Stock notation and naming	Turning an oxidation number into a Roman numeral, and a name back into a formula
17-21	Types of redox reaction	Combination, decomposition, displacement, disproportionation and the non-redox impostor
22-26	Oxidation number method	Equalising rise and fall, then finishing with water, hydrogen ion or hydroxide
27-31	Half-reaction method	Splitting, balancing each half separately, matching electrons and adding back
32-37	Titration arithmetic	n-factor, equivalent mass, mole-ratio chains and a back titration
38-40	Electrode potentials	Cell EMF, spontaneity and why one acid is banned from a permanganate flask

Key Point: The three definitions must agree on every example. If the classical reading says a species is oxidised, the electron count and the oxidation number must say the same thing. Whenever they disagree, one of them has been read wrongly.

[JEE/NEET] Oxidation number assignment and balancing carry the most marks per minute in this chapter. Questions 6 to 31 are the ones to repeat until they are automatic.

Classical and Electronic Definitions

Question 1: Magnesium burning in carbon dioxide

A strip of burning magnesium keeps burning when it is lowered into a jar of carbon dioxide, leaving white powder and black specks.

$$\mathrm{2Mg(s) + CO_2(g) \rightarrow 2MgO(s) + C(s)}$$

Read this reaction by all three definitions and name the oxidant and the reductant.

Answer:

Classical first. Magnesium picks up oxygen, so magnesium is oxidised. Carbon dioxide loses its oxygen and comes out as black carbon, so carbon dioxide is reduced.

Electronic next. Magnesium metal ends up as $\mathrm{Mg^{2+}}$ in magnesium oxide, so each magnesium atom hands over $2$ electrons. Those electrons go to carbon.

Oxidation number last. Magnesium goes from $0$ to $+2$, a rise. Carbon goes from $+4$ in $\mathrm{CO_2}$ to $0$ in free carbon, a fall of $4$. Oxygen stays at $-2$ throughout.

All three readings agree. Two magnesium atoms rise by $2$ each, giving $4$; one carbon falls by $4$. The electron count matches.

The species that is reduced is the oxidising agent, so carbon dioxide is the oxidant. The species that is oxidised is the reducing agent, so magnesium is the reductant.

Ans: $\mathrm{Mg}$ is oxidised and is the reducing agent; $\mathrm{CO_2}$ is reduced and is the oxidising agent. Watch out: Carbon dioxide is normally thought of as a fire extinguisher. Here it is the oxidant, because magnesium is hungry enough for oxygen to strip it from carbon.

Question 2: Hydrogen sulphide and chlorine

$$\mathrm{H_2S(g) + Cl_2(g) \rightarrow 2HCl(g) + S(s)}$$

Which reactant is oxidised? Answer once by the hydrogen rule and once by oxidation numbers.

Answer:

By the hydrogen rule: $\mathrm{H_2S}$ loses its hydrogen and comes out as sulphur, and removal of hydrogen is oxidation. Chlorine takes that hydrogen up as $\mathrm{HCl}$, and addition of hydrogen is reduction.

By oxidation numbers: sulphur in $\mathrm{H_2S}$ is $-2$, since hydrogen is $+1$ and the molecule is neutral. Free sulphur is $0$. So sulphur climbs by $2$. Chlorine starts at $0$ as the free element and ends at $-1$ in $\mathrm{HCl}$, a fall of $1$ per atom, or $2$ for the molecule.

Rise of $2$ against fall of $2$. The two readings agree.

Ans: $\mathrm{H_2S}$ is oxidised and is the reductant; $\mathrm{Cl_2}$ is reduced and is the oxidant.

Question 3: Tin(II) with iron(III)

$$\mathrm{Sn^{2+}(aq) + 2Fe^{3+}(aq) \rightarrow Sn^{4+}(aq) + 2Fe^{2+}(aq)}$$

Split this into half reactions, check the charge, and name the oxidant and the reductant.

Answer:

Tin goes from $+2$ to $+4$, so it loses electrons. That is the oxidation half:

$$\mathrm{Sn^{2+} \rightarrow Sn^{4+} + 2e^-}$$

Iron goes from $+3$ to $+2$, so it gains one electron per ion. Two ions are involved, so I write:

$$\mathrm{2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}}$$

Two electrons released, two electrons absorbed. Adding the halves cancels them and gives back the overall equation.

Charge check on the overall equation. Left: $(+2) + 2(+3) = +8$. Right: $(+4) + 2(+2) = +8$. They match.

Neither half happens by itself. A half reaction is only a bookkeeping device that lets me count electrons; the electrons never appear in the real flask on their own.

Ans: $\mathrm{Sn^{2+}}$ is oxidised and is the reducing agent; $\mathrm{Fe^{3+}}$ is reduced and is the oxidising agent.

Question 4: Naming the agents in three reactions

Name the oxidising and the reducing agent in each.

(i) $\mathrm{2K(s) + Cl_2(g) \rightarrow 2KCl(s)}$

(ii) $\mathrm{CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(g)}$

(iii) $\mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_2(g)}$

Answer:

(i) Potassium goes $0 \rightarrow +1$, a rise, so potassium is oxidised. Chlorine goes $0 \rightarrow -1$, a fall, so chlorine is reduced. Each potassium hands one electron to chlorine.

(ii) Copper goes $+2 \rightarrow 0$, a fall. Hydrogen goes $0 \rightarrow +1$, a rise. Classical reading agrees: copper oxide loses oxygen and hydrogen gains it.

(iii) Iron goes $+3 \rightarrow 0$ over two atoms, so $6$ electrons are gained in total. Carbon goes $+2$ in $\mathrm{CO}$ to $+4$ in $\mathrm{CO_2}$, losing $2$ each, and there are three of them, so $6$ electrons are lost. The counts match.

In every case the reduced species is the oxidant.

Ans: (i) oxidant $\mathrm{Cl_2}$, reductant $\mathrm{K}$; (ii) oxidant $\mathrm{CuO}$, reductant $\mathrm{H_2}$; (iii) oxidant $\mathrm{Fe_2O_3}$, reductant $\mathrm{CO}$. Watch out: In (iii) carbon monoxide is the reductant even though it contains oxygen. What matters is the change in oxidation number, not whether the formula looks reduced.

Question 5: Hydrogen peroxide, both ways round

Decide what hydrogen peroxide is doing in each reaction.

(a) $\mathrm{H_2O_2 + 2KI + H_2SO_4 \rightarrow I_2 + K_2SO_4 + 2H_2O}$

(b) $\mathrm{2KMnO_4 + 5H_2O_2 + 3H_2SO_4 \rightarrow 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O}$

Answer:

The oxygen in hydrogen peroxide is at $-1$, because it holds a peroxide linkage. That is an in-between value, so it can go either way.

(a) Iodide goes from $-1$ to $0$ in $\mathrm{I_2}$, so iodide is oxidised. The oxygen of $\mathrm{H_2O_2}$ ends up in water at $-2$, a fall from $-1$. So hydrogen peroxide is reduced here, which makes it the oxidant. Electron count: two iodides lose $1$ each, two oxygens gain $1$ each. Balanced.

(b) Manganese falls from $+7$ to $+2$, gaining $5$ electrons per manganese, so $10$ for the two. The oxygen of $\mathrm{H_2O_2}$ leaves as free $\mathrm{O_2}$ at $0$, a rise from $-1$. Each $\mathrm{H_2O_2}$ gives up $2$ electrons, and five of them give $10$. So here hydrogen peroxide is oxidised, which makes it the reductant.

Ans: In (a) $\mathrm{H_2O_2}$ acts as the oxidising agent; in (b) it acts as the reducing agent. Watch out: Both roles are possible only because oxygen at $-1$ sits between $-2$ and $0$. Read the other reactant before deciding which role peroxide has taken.

Oxidation Numbers, Traps and Fractions

Question 6: The two nitrogens of ammonium nitrate

Find the oxidation number of nitrogen in $\mathrm{NH_4NO_3}$ and in $\mathrm{NH_4NO_2}$.

Answer:

If I treat all the nitrogen as one kind, I get an average. For $\mathrm{NH_4NO_3}$ there are two nitrogens, four hydrogens at $+1$ and three oxygens at $-2$:

$$2x + 4(+1) + 3(-2) = 0 \quad \Rightarrow \quad 2x = +2 \quad \Rightarrow \quad x = +1$$

That $+1$ is only an average. The salt is really $\mathrm{NH_4^+}$ and $\mathrm{NO_3^-}$, so I do each ion on its own. In $\mathrm{NH_4^+}$: $x + 4(+1) = +1$, so $x = -3$. In $\mathrm{NO_3^-}$: $x + 3(-2) = -1$, so $x = +5$. The mean of $-3$ and $+5$ is $+1$, which matches.

For $\mathrm{NH_4NO_2}$: $2x + 4(+1) + 2(-2) = 0$ gives $2x = 0$, so the average is $0$. Splitting the ions, ammonium nitrogen is $-3$ and nitrite nitrogen is $+3$, since $x + 2(-2) = -1$. The mean of $-3$ and $+3$ is $0$, which matches.

Ans: $\mathrm{NH_4NO_3}$: average $+1$, actual $-3$ and $+5$. $\mathrm{NH_4NO_2}$: average $0$, actual $-3$ and $+3$. Watch out: An average of $0$ in $\mathrm{NH_4NO_2}$ does not mean nitrogen is uncombined. When a salt has the element in two different ions, always split the ions.

Question 7: Chlorine in bleaching powder

Bleaching powder is written $\mathrm{CaOCl_2}$. Find the oxidation number of chlorine in it.

Answer:

Straight arithmetic first. Calcium is $+2$, oxygen is $-2$, and there are two chlorines:

$$(+2) + (-2) + 2x = 0 \quad \Rightarrow \quad x = 0$$

An average of $0$ for a chlorine that is clearly combined should make me suspicious. The real structure is $\mathrm{Ca(OCl)Cl}$: one chlorine sits in the hypochlorite ion and the other is a plain chloride.

In $\mathrm{OCl^-}$: $x + (-2) = -1$, so that chlorine is $+1$. The chloride is $-1$. The mean of $+1$ and $-1$ is $0$, which is the average I calculated.

Ans: average $0$; actually one chlorine at $+1$ and one at $-1$. Watch out: Zero here does not mean free chlorine. It is the mean of two chlorines in genuinely different environments, which is exactly why bleaching powder both bleaches and gives chloride.

Question 8: Two sulphur oxoanions that look alike

Find the oxidation number of sulphur in the peroxodisulphate ion $\mathrm{S_2O_8^{2-}}$ and in the dithionate ion $\mathrm{S_2O_6^{2-}}$.

Answer:

I start with $\mathrm{S_2O_8^{2-}}$ the naive way, taking every oxygen at $-2$:

$$2x + 8(-2) = -2 \quad \Rightarrow \quad 2x = +14 \quad \Rightarrow \quad x = +7$$

Sulphur cannot exceed $+6$, since it has only six valence electrons to lose. That impossible answer tells me an oxygen assumption is wrong. This ion carries one peroxide linkage, $\mathrm{-O-O-}$, and those two oxygens are $-1$, not $-2$. Redoing it with six oxygens at $-2$ and two at $-1$:

$$2x + 6(-2) + 2(-1) = -2 \quad \Rightarrow \quad 2x - 14 = -2 \quad \Rightarrow \quad 2x = +12 \quad \Rightarrow \quad x = +6$$

Now $\mathrm{S_2O_6^{2-}}$. Dithionate has a sulphur-sulphur bond, not a peroxide linkage, so every oxygen really is $-2$:

$$2x + 6(-2) = -2 \quad \Rightarrow \quad 2x = +10 \quad \Rightarrow \quad x = +5$$

A bond between two like atoms contributes nothing to either atom, so the $\mathrm{S-S}$ bond does not disturb the count.

Ans: $+6$ in $\mathrm{S_2O_8^{2-}}$; $+5$ in $\mathrm{S_2O_6^{2-}}$. Watch out: The signal is a value above the group maximum. Getting $+7$ for sulphur is a message from the arithmetic that a peroxide linkage is present.

Question 9: Manganese in hausmannite

Find the average oxidation number of manganese in $\mathrm{Mn_3O_4}$, then give the real values.

Answer:

Average first, oxygen at $-2$:

$$3x + 4(-2) = 0 \quad \Rightarrow \quad 3x = +8 \quad \Rightarrow \quad x = +\frac{8}{3}$$

No atom can lose two-thirds of an electron, so $+8/3$ is a bookkeeping mean, not a real charge. The structure is $\mathrm{MnO \cdot Mn_2O_3}$, so one manganese is $+2$ and two are $+3$.

Checking: $(+2) + 2(+3) = +8$ against $4(-2) = -8$, and the mean of $+2, +3, +3$ is $+8/3$.

Ans: average $+8/3$; actual states $+2$ for one manganese and $+3$ for two. Watch out: If the question says "oxidation states" in the plural, give $+2$ and $+3$. If it says "oxidation number", give $+8/3$.

Question 10: Nitrogen in the azides

Find the oxidation number of nitrogen in hydrazoic acid $\mathrm{HN_3}$ and in sodium azide $\mathrm{NaN_3}$.

Answer:

Hydrogen is $+1$, so for $\mathrm{HN_3}$:

$$(+1) + 3x = 0 \quad \Rightarrow \quad 3x = -1 \quad \Rightarrow \quad x = -\frac{1}{3}$$

Sodium is $+1$ too, so $\mathrm{NaN_3}$ gives exactly the same equation and the same answer.

The three nitrogens in the azide ion are not identical in bonding, but there is no simple whole-number split of the kind that works for $\mathrm{Fe_3O_4}$, so $-1/3$ is the answer that is quoted.

Ans: $-1/3$ in both. Watch out: A fraction is a perfectly legal oxidation number. Rounding it to $0$ or forcing it to $-1$ is the error to avoid.

Question 11: Carbon in acetic acid

Find the average oxidation number of carbon in $\mathrm{CH_3COOH}$, then assign each carbon separately.

Answer:

Average, with hydrogen at $+1$ and oxygen at $-2$:

$$2x + 4(+1) + 2(-2) = 0 \quad \Rightarrow \quad 2x = 0 \quad \Rightarrow \quad x = 0$$

Now the individual atoms. For each bond I hand both shared electrons to the more electronegative atom, and a carbon-carbon bond splits evenly and counts as nothing.

The methyl carbon is bonded to three hydrogens and one carbon. Each $\mathrm{C-H}$ bond gives it $-1$ and the $\mathrm{C-C}$ bond gives $0$, so it is $-3$.

The carboxyl carbon is bonded to one carbon ($0$), to a doubly bonded oxygen ($+2$) and to the oxygen of the $\mathrm{-OH}$ group ($+1$). That is $+3$.

Mean of $-3$ and $+3$ is $0$, matching the average.

Ans: average $0$; methyl carbon $-3$, carboxyl carbon $+3$. Watch out: An average of zero does not mean carbon is unoxidised. Organic redox questions almost always want the individual carbon, not the mean.

Question 12: Iodine in three of its own compounds

Find the oxidation number of iodine in $\mathrm{I_3^-}$, in $\mathrm{ICl}$ and in $\mathrm{ICl_3}$.

Answer:

For the triiodide ion, all three atoms are iodine, so:

$$3x = -1 \quad \Rightarrow \quad x = -\frac{1}{3}$$

Structurally the ion is an $\mathrm{I_2}$ molecule holding on to an $\mathrm{I^-}$, so two iodines are really $0$ and one is $-1$. The mean of $0, 0, -1$ is $-1/3$, as calculated.

For $\mathrm{ICl}$ I compare electronegativities. Chlorine is more electronegative than iodine, so chlorine takes $-1$ and iodine is $+1$.

For $\mathrm{ICl_3}$, three chlorines at $-1$ give $-3$, so iodine is $+3$.

Ans: $-1/3$ in $\mathrm{I_3^-}$; $+1$ in $\mathrm{ICl}$; $+3$ in $\mathrm{ICl_3}$. Watch out: In an interhalogen the less electronegative halogen carries the positive value. Iodine is positive here even though halogens are usually pictured as $-1$.

Question 13: The three oxoacids of phosphorus

Find the oxidation number of phosphorus in $\mathrm{H_3PO_2}$, $\mathrm{H_3PO_3}$ and $\mathrm{H_3PO_4}$.

Answer:

Hydrogen $+1$, oxygen $-2$ in all three, because none of them holds a peroxide linkage.

$\mathrm{H_3PO_2}$: $3(+1) + x + 2(-2) = 0$, so $x + 3 - 4 = 0$ and $x = +1$.

$\mathrm{H_3PO_3}$: $3(+1) + x + 3(-2) = 0$, so $x = +3$.

$\mathrm{H_3PO_4}$: $3(+1) + x + 4(-2) = 0$, so $x = +5$.

The rise of $2$ from one acid to the next is just an extra oxygen each time.

Ans: $+1$, $+3$ and $+5$. Watch out: All three have three hydrogens, but they are not all tribasic. Basicity depends on how many hydrogens sit on oxygen, which is a different question from oxidation number, so do not read one off the other.

Stock Notation and Naming

Question 14: Six formulae into Stock names

Name $\mathrm{Hg_2Cl_2}$, $\mathrm{HgCl_2}$, $\mathrm{MnO}$, $\mathrm{MnO_2}$, $\mathrm{CrO_3}$ and $\mathrm{Tl_2O}$ in Stock notation.

Answer:

Stock notation puts the oxidation number of the metal as a Roman numeral in brackets after its name, so I work out that number first.

$\mathrm{Hg_2Cl_2}$: two chlorides give $-2$, shared over two mercuries, so each is $+1$.

$\mathrm{HgCl_2}$: two chlorides on one mercury, so mercury is $+2$.

$\mathrm{MnO}$: one oxygen at $-2$, so manganese is $+2$.

$\mathrm{MnO_2}$: two oxygens give $-4$, so manganese is $+4$.

$\mathrm{CrO_3}$: three oxygens give $-6$, so chromium is $+6$.

$\mathrm{Tl_2O}$: $2x + (-2) = 0$, so each thallium is $+1$.

Ans: mercury(I) chloride, mercury(II) chloride, manganese(II) oxide, manganese(IV) oxide, chromium(VI) oxide, thallium(I) oxide. Watch out: The Roman numeral belongs to one atom, never to the whole formula. Calling $\mathrm{Hg_2Cl_2}$ mercury(II) chloride adds the two mercuries together and names a completely different compound.

Question 15: Six Stock names into formulae

Write the formula for iron(III) sulphate, tin(IV) chloride, copper(I) iodide, chromium(III) oxide, nickel(II) nitrate and manganese(VII) oxide.

Answer:

I read the Roman numeral as the charge on the metal, then balance it against the anion.

Iron(III) with sulphate $\mathrm{SO_4^{2-}}$: two irons give $+6$, three sulphates give $-6$, so $\mathrm{Fe_2(SO_4)_3}$.

Tin(IV) with chloride: four chlorides, so $\mathrm{SnCl_4}$.

Copper(I) with iodide: one each, so $\mathrm{CuI}$.

Chromium(III) with oxide $\mathrm{O^{2-}}$: two chromiums give $+6$ against three oxides at $-6$, so $\mathrm{Cr_2O_3}$.

Nickel(II) with nitrate $\mathrm{NO_3^-}$: two nitrates, so $\mathrm{Ni(NO_3)_2}$.

Manganese(VII) with oxide: two manganese give $+14$, seven oxides give $-14$, so $\mathrm{Mn_2O_7}$.

Ans: $\mathrm{Fe_2(SO_4)_3}$, $\mathrm{SnCl_4}$, $\mathrm{CuI}$, $\mathrm{Cr_2O_3}$, $\mathrm{Ni(NO_3)_2}$, $\mathrm{Mn_2O_7}$. Watch out: Brackets matter when the anion has more than one atom. Dropping them turns nickel(II) nitrate into a string of subscripts that means something else entirely, and the same goes for iron(III) sulphate.

Question 16: The compound Stock notation cannot name in one word

Red lead is $\mathrm{Pb_3O_4}$. Find the oxidation number of lead, then say how the compound is named and why a single Roman numeral will not do.

Answer:

Average first:

$$3x + 4(-2) = 0 \quad \Rightarrow \quad 3x = +8 \quad \Rightarrow \quad x = +\frac{8}{3}$$

Roman numerals only come in whole numbers, so $+8/3$ cannot go into a Stock name at all.

The structure sorts it out. Red lead is $\mathrm{2PbO \cdot PbO_2}$, so two lead atoms are $+2$ and one is $+4$. Checking the mean: $(2 \times 2 + 4)/3 = 8/3$, which agrees.

Because the atoms sit in two genuinely different states, the name has to carry both: dilead(II) lead(IV) oxide.

Ans: average $+8/3$; two lead at $+2$ and one at $+4$; named dilead(II) lead(IV) oxide. Watch out: Stock notation only works when the element has one whole-number oxidation state in the compound. A fractional average is a signal that the formula unit contains atoms in more than one state.

Sorting Redox Reactions into Types

Question 17: Five reactions, five labels

Classify each as combination, decomposition, displacement, disproportionation or not a redox reaction at all.

(a) $\mathrm{2Na(s) + H_2(g) \rightarrow 2NaH(s)}$

(b) $\mathrm{2NaN_3(s) \rightarrow 2Na(s) + 3N_2(g)}$

(c) $\mathrm{Cl_2(g) + 2NaBr(aq) \rightarrow 2NaCl(aq) + Br_2(l)}$

(d) $\mathrm{2Cu^+(aq) \rightarrow Cu^{2+}(aq) + Cu(s)}$

(e) $\mathrm{BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)}$

Answer:

(a) Two elements combine into one compound, so it is combination. It is redox because sodium goes $0 \rightarrow +1$ and hydrogen goes $0 \rightarrow -1$. Hydrogen is $-1$ here because it is bonded to a metal.

(b) One compound splits into two products, so it is decomposition. Sodium falls from $+1$ to $0$ and nitrogen rises from $-1/3$ to $0$. This is the airbag reaction.

(c) Chlorine takes bromine's place, and both are non-metals, so it is non-metal displacement. Chlorine goes $0 \rightarrow -1$ and bromide goes $-1 \rightarrow 0$. It runs because chlorine is the stronger oxidant, $+1.36\ \mathrm{V}$ against $+1.09\ \mathrm{V}$.

(d) One species, copper(I), gives both products. Copper goes up to $+2$ and down to $0$ in the same reaction, so it is disproportionation.

(e) No oxidation number changes anywhere. Barium stays $+2$, sulphate stays intact, sodium stays $+1$, chloride stays $-1$. It is a precipitation, not a redox reaction.

Ans: (a) combination; (b) decomposition; (c) displacement; (d) disproportionation; (e) not redox. Watch out: A reaction that forms a solid and looks dramatic is not automatically redox. Check the oxidation numbers before choosing a label.

Question 18: The reaction that looks like disproportionation and is not

$$\mathrm{2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)}$$

Classify this reaction, and explain why it is not disproportionation.

Answer:

Oxidation numbers first. In $\mathrm{KClO_3}$, potassium is $+1$ and oxygen is $-2$, so chlorine is $+5$. On the right, chlorine in $\mathrm{KCl}$ is $-1$ and the oxygen leaves as free $\mathrm{O_2}$ at $0$.

Chlorine falls from $+5$ to $-1$, gaining $6$ electrons per atom, so $12$ for the two.

Oxygen rises from $-2$ to $0$, losing $2$ electrons per atom, and there are six of them, so $12$ lost. The counts match.

One compound breaks into two products, so the type is decomposition, and it is redox because oxidation numbers change.

It is not disproportionation because disproportionation needs one and the same element to be oxidised and reduced. Here chlorine is reduced and oxygen is oxidised. Two different elements, so two different roles.

Ans: Redox decomposition, with chlorine reduced and oxygen oxidised; not disproportionation. Watch out: Disproportionation is about one element, not one compound. A single reactant giving both an oxidised and a reduced product is not enough on its own.

Question 19: Nitrogen dioxide in water

$$\mathrm{2NO_2(g) + H_2O(l) \rightarrow HNO_3(aq) + HNO_2(aq)}$$

Classify this and justify the label with numbers.

Answer:

Nitrogen in $\mathrm{NO_2}$: $x + 2(-2) = 0$, so $x = +4$.

Nitrogen in $\mathrm{HNO_3}$: $(+1) + x + 3(-2) = 0$, so $x = +5$.

Nitrogen in $\mathrm{HNO_2}$: $(+1) + x + 2(-2) = 0$, so $x = +3$.

One nitrogen has gone up to $+5$ and the other down to $+3$, both starting from $+4$. Same element, both directions, one reaction, so it is disproportionation.

Electron count: one nitrogen loses $1$, one gains $1$. Balanced.

Atom check: two nitrogens on each side. Hydrogen: $2$ on the left in the water, and $1 + 1 = 2$ on the right. Oxygen: left $4 + 1 = 5$; right $3 + 2 = 5$. Everything matches.

It is possible only because $+4$ is an intermediate state for nitrogen, with $+5$ available above and $+3$ below.

Ans: Disproportionation; nitrogen goes from $+4$ to $+5$ and to $+3$.

Question 20: Running disproportionation backwards

$$\mathrm{SO_2(g) + 2H_2S(g) \rightarrow 3S(s) + 2H_2O(l)}$$

What is special about the way sulphur behaves here?

Answer:

Sulphur in $\mathrm{SO_2}$ is $+4$, since each oxygen is $-2$. Sulphur in $\mathrm{H_2S}$ is $-2$, since each hydrogen is $+1$. On the right, all three sulphurs come out as the free element at $0$.

So the sulphur that started high comes down, and the sulphur that started low goes up, and they meet in the middle.

Electron count: one sulphur falls from $+4$ to $0$, gaining $4$. Two sulphurs climb from $-2$ to $0$, losing $2$ each, giving $4$ lost. Balanced.

Atom check: sulphur $1 + 2 = 3$ on the left and $3$ on the right; hydrogen $4$ on each side; oxygen $2$ on each side.

This is the opposite of disproportionation. Two different oxidation states of one element converge on a single intermediate state, which is called comproportionation.

Ans: The same element in two different states, $+4$ and $-2$, converges to one state, $0$. It is comproportionation, the reverse of disproportionation. Watch out: $\mathrm{SO_2}$ is the oxidant here and $\mathrm{H_2S}$ is the reductant, even though both contain sulphur. The roles are decided by which direction each sulphur moves.

Question 21: Two displacements, two different kinds

Classify each and check the electron count.

(a) $\mathrm{Cr_2O_3(s) + 2Al(s) \rightarrow Al_2O_3(s) + 2Cr(s)}$

(b) $\mathrm{2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)}$

Answer:

(a) Aluminium goes $0 \rightarrow +3$, losing $3$ electrons each, so $6$ for two atoms. Chromium goes $+3 \rightarrow 0$, gaining $3$ each, so $6$ gained. Balanced.

One metal has pushed another metal out of its compound, so this is metal displacement, the thermite type. It runs because aluminium is the stronger reductant: $E^\circ$ for $\mathrm{Al^{3+}/Al}$ is $-1.66\ \mathrm{V}$ against $-0.74\ \mathrm{V}$ for $\mathrm{Cr^{3+}/Cr}$, so

$$E^\circ_{\mathrm{cell}} = -0.74 - (-1.66) = +0.92\ \mathrm{V}$$

which is positive, so the reaction is spontaneous as written.

(b) Sodium goes $0 \rightarrow +1$, losing $1$ each, so $2$ in all. Hydrogen falls from $+1$ to $0$, and two of the four hydrogens do so, gaining $1$ each, so $2$ gained. Balanced.

A metal has displaced hydrogen, which is a non-metal, so this is non-metal displacement.

Ans: (a) metal displacement, $E^\circ_{\mathrm{cell}} = +0.92\ \mathrm{V}$; (b) non-metal displacement of hydrogen. Watch out: In (b) only half the hydrogen changes. The two hydrogens that end up in $\mathrm{NaOH}$ are still $+1$; assuming all four are reduced gives a fall of $4$ against a rise of $2$.

Balancing by the Oxidation Number Method

The routine never changes. Find the atoms whose oxidation number moves, work out the total rise and the total fall, multiply the two species so that rise equals fall, then tidy up the remaining atoms and finish with water and either hydrogen ion or hydroxide ion.

Question 22: Dichromate with hydrogen sulphide in acid

Balance in acidic medium.

$$\mathrm{Cr_2O_7^{2-} + H_2S \rightarrow Cr^{3+} + S}$$

Answer:

First I mark the changes. Chromium in $\mathrm{Cr_2O_7^{2-}}$ is $+6$, since $2x + 7(-2) = -2$. It ends at $+3$, so each chromium falls by $3$, and there are two of them, so the fall is $6$ per dichromate ion.

Sulphur in $\mathrm{H_2S}$ is $-2$ and ends at $0$ as free sulphur, a rise of $2$ per sulphur.

To make rise equal fall I need three sulphurs against one dichromate: $3 \times 2 = 6$.

$$\mathrm{Cr_2O_7^{2-} + 3H_2S \rightarrow 2Cr^{3+} + 3S}$$

Now the leftovers. Seven oxygens on the left have nowhere to go except water, so I write $7\mathrm{H_2O}$ on the right. That needs $14$ hydrogens on the left, and only $6$ come from the three $\mathrm{H_2S}$, so I add $8\mathrm{H^+}$:

$$\mathrm{Cr_2O_7^{2-} + 3H_2S + 8H^+ \rightarrow 2Cr^{3+} + 3S + 7H_2O}$$

Atom check: $\mathrm{Cr}$ $2$ and $2$; $\mathrm{S}$ $3$ and $3$; $\mathrm{O}$ $7$ and $7$; $\mathrm{H}$ $6 + 8 = 14$ on the left and $14$ on the right.

Charge check: left $(-2) + 0 + (+8) = +6$; right $2(+3) = +6$. They match.

Ans: $\mathrm{Cr_2O_7^{2-} + 3H_2S + 8H^+ \rightarrow 2Cr^{3+} + 3S + 7H_2O}$

Question 23: Dichromate with concentrated hydrochloric acid

Balance this molecular equation by the oxidation number method.

$$\mathrm{K_2Cr_2O_7 + HCl \rightarrow KCl + CrCl_3 + Cl_2 + H_2O}$$

Answer:

Chromium goes from $+6$ to $+3$, gaining $3$ each. Two chromiums per formula unit, so the fall is $6$.

Chlorine is trickier. It starts at $-1$ in $\mathrm{HCl}$, but only some of it changes. The chlorine in $\mathrm{KCl}$ and $\mathrm{CrCl_3}$ is still $-1$; only the chlorine that leaves as $\mathrm{Cl_2}$ has risen to $0$. Each of those loses $1$.

To match a fall of $6$ I need six chlorines oxidised, which is $3\mathrm{Cl_2}$:

$$\mathrm{K_2Cr_2O_7 + HCl \rightarrow 2KCl + 2CrCl_3 + 3Cl_2 + H_2O}$$

Now I count the chlorine on the right: $2$ in $\mathrm{KCl}$, $6$ in $2\mathrm{CrCl_3}$ and $6$ in $3\mathrm{Cl_2}$, giving $14$. So I need $14\mathrm{HCl}$, which brings $14$ hydrogens and therefore $7\mathrm{H_2O}$:

$$\mathrm{K_2Cr_2O_7 + 14HCl \rightarrow 2KCl + 2CrCl_3 + 3Cl_2 + 7H_2O}$$

Atom check: $\mathrm{K}$ $2$ and $2$; $\mathrm{Cr}$ $2$ and $2$; $\mathrm{O}$ $7$ and $7$; $\mathrm{H}$ $14$ and $14$; $\mathrm{Cl}$ $14$ and $2 + 6 + 6 = 14$.

Both sides are neutral molecules, so the charge balances at zero.

Ans: $\mathrm{K_2Cr_2O_7 + 14HCl \rightarrow 2KCl + 2CrCl_3 + 3Cl_2 + 7H_2O}$ Watch out: Only $6$ of the $14$ chlorines are oxidised. Counting all $14$ as oxidised gives a rise of $14$ against a fall of $6$ and wrecks the whole balance.

Question 24: Iodine oxidised by concentrated nitric acid

Balance in acidic medium.

$$\mathrm{I_2 + HNO_3 \rightarrow HIO_3 + NO_2 + H_2O}$$

Answer:

Iodine starts as the free element at $0$. In $\mathrm{HIO_3}$ it is $(+1) + x + 3(-2) = 0$, so $x = +5$. That is a rise of $5$ per atom, and $\mathrm{I_2}$ has two atoms, so the rise is $10$ per molecule.

Nitrogen in $\mathrm{HNO_3}$ is $+5$ and in $\mathrm{NO_2}$ it is $+4$, a fall of only $1$ per atom.

So I need ten nitrogens for one iodine molecule:

$$\mathrm{I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + H_2O}$$

Hydrogen: $10$ on the left, $2$ in the two $\mathrm{HIO_3}$, so $8$ must go into water, giving $4\mathrm{H_2O}$.

$$\mathrm{I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O}$$

Atom check: $\mathrm{I}$ $2$ and $2$; $\mathrm{N}$ $10$ and $10$; $\mathrm{H}$ $10$ and $2 + 8 = 10$; $\mathrm{O}$ $30$ on the left and $6 + 20 + 4 = 30$ on the right.

Everything is neutral, so charge is fine.

Ans: $\mathrm{I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O}$ Watch out: The rise is $10$, not $5$, because $\mathrm{I_2}$ carries two iodine atoms. Forgetting the subscript halves every coefficient on the right.

Question 25: Chromite oxidised by hypochlorite in alkali

Balance in basic medium.

$$\mathrm{CrO_2^- + ClO^- \rightarrow CrO_4^{2-} + Cl^-}$$

Answer:

Chromium in $\mathrm{CrO_2^-}$: $x + 2(-2) = -1$, so $x = +3$. In $\mathrm{CrO_4^{2-}}$: $x + 4(-2) = -2$, so $x = +6$. That is a rise of $3$.

Chlorine in $\mathrm{ClO^-}$: $x + (-2) = -1$, so $x = +1$. It ends at $-1$ as chloride, a fall of $2$.

The lowest common multiple of $3$ and $2$ is $6$, so I take two chromiums and three chlorines:

$$\mathrm{2CrO_2^- + 3ClO^- \rightarrow 2CrO_4^{2-} + 3Cl^-}$$

Now I balance charge, because the medium is basic and hydroxide is what is available. Left: $2(-1) + 3(-1) = -5$. Right: $2(-2) + 3(-1) = -7$. The left needs to be more negative by $2$, so I add $2\mathrm{OH^-}$ to the left:

$$\mathrm{2CrO_2^- + 3ClO^- + 2OH^- \rightarrow 2CrO_4^{2-} + 3Cl^-}$$

That brings $2$ hydrogens in, which have to leave as water, so I put $1\mathrm{H_2O}$ on the right:

$$\mathrm{2CrO_2^- + 3ClO^- + 2OH^- \rightarrow 2CrO_4^{2-} + 3Cl^- + H_2O}$$

Atom check: $\mathrm{Cr}$ $2$ and $2$; $\mathrm{Cl}$ $3$ and $3$; $\mathrm{H}$ $2$ and $2$; $\mathrm{O}$ $4 + 3 + 2 = 9$ on the left and $8 + 1 = 9$ on the right.

Charge check: left $-2 - 3 - 2 = -7$; right $-4 - 3 = -7$.

Ans: $\mathrm{2CrO_2^- + 3ClO^- + 2OH^- \rightarrow 2CrO_4^{2-} + 3Cl^- + H_2O}$ Watch out: No $\mathrm{H^+}$ may survive in a basic answer. If you balanced as if the medium were acidic, neutralise every $\mathrm{H^+}$ with $\mathrm{OH^-}$ on both sides before you write the final line.

Question 26: Zinc reducing nitrate to ammonia in alkali

Balance in basic medium.

$$\mathrm{Zn + NO_3^- \rightarrow ZnO_2^{2-} + NH_3}$$

Answer:

Zinc goes from $0$ to $+2$ in the zincate ion, since $x + 2(-2) = -2$. That is a rise of $2$.

Nitrogen goes from $+5$ in nitrate to $-3$ in ammonia, a fall of $8$.

Four zincs against one nitrate makes rise equal fall: $4 \times 2 = 8$.

$$\mathrm{4Zn + NO_3^- \rightarrow 4ZnO_2^{2-} + NH_3}$$

Charge next. Left is $-1$; right is $4(-2) = -8$. I need $7$ more negative charges on the left, so $7\mathrm{OH^-}$ go there:

$$\mathrm{4Zn + NO_3^- + 7OH^- \rightarrow 4ZnO_2^{2-} + NH_3}$$

Hydrogen: $7$ on the left, $3$ in ammonia, so $4$ remain and go into $2\mathrm{H_2O}$:

$$\mathrm{4Zn + NO_3^- + 7OH^- \rightarrow 4ZnO_2^{2-} + NH_3 + 2H_2O}$$

Atom check: $\mathrm{Zn}$ $4$ and $4$; $\mathrm{N}$ $1$ and $1$; $\mathrm{O}$ $3 + 7 = 10$ on the left and $8 + 2 = 10$ on the right; $\mathrm{H}$ $7$ on the left and $3 + 4 = 7$ on the right.

Charge check: left $-1 - 7 = -8$; right $4(-2) = -8$.

Ans: $\mathrm{4Zn + NO_3^- + 7OH^- \rightarrow 4ZnO_2^{2-} + NH_3 + 2H_2O}$ Watch out: A fall of $8$ for one nitrogen is the largest single change in this chapter. Writing $+5$ to $-3$ as a fall of $2$ is the standard slip, because the sign flip is easy to miss.

Balancing by the Half-Reaction Method

Question 27: Chlorate oxidising iodide in acid

Balance by the ion-electron method.

$$\mathrm{ClO_3^- + I^- \rightarrow Cl^- + I_2}$$

Answer:

I split it into two skeleton halves.

Reduction half: $\mathrm{ClO_3^- \rightarrow Cl^-}$. Chlorine is already balanced. Three oxygens on the left need $3\mathrm{H_2O}$ on the right, and those six hydrogens need $6\mathrm{H^+}$ on the left. Charge is then $-1 + 6 = +5$ on the left and $-1$ on the right, so I add $6\mathrm{e^-}$ to the left:

$$\mathrm{ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O}$$

Oxidation half: $\mathrm{2I^- \rightarrow I_2}$. Charge is $-2$ on the left and $0$ on the right, so two electrons leave:

$$\mathrm{2I^- \rightarrow I_2 + 2e^-}$$

The reduction half needs $6$ electrons and the oxidation half supplies $2$, so I multiply the oxidation half by $3$ and add:

$$\mathrm{ClO_3^- + 6I^- + 6H^+ \rightarrow Cl^- + 3I_2 + 3H_2O}$$

Atom check: $\mathrm{Cl}$ $1$ and $1$; $\mathrm{I}$ $6$ and $6$; $\mathrm{O}$ $3$ and $3$; $\mathrm{H}$ $6$ and $6$.

Charge check: left $-1 - 6 + 6 = -1$; right $-1$.

Ans: $\mathrm{ClO_3^- + 6I^- + 6H^+ \rightarrow Cl^- + 3I_2 + 3H_2O}$

Question 28: Hydrogen peroxide oxidising iron(II) in acid

Balance by the ion-electron method.

$$\mathrm{H_2O_2 + Fe^{2+} \rightarrow Fe^{3+} + H_2O}$$

Answer:

Oxidation half is short: $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$. Charge is $+2$ on the left and $+3 - 1 = +2$ on the right.

Reduction half: $\mathrm{H_2O_2 \rightarrow 2H_2O}$. The oxygens are balanced at two each side. Hydrogen is $2$ on the left and $4$ on the right, so I add $2\mathrm{H^+}$ to the left. Charge is then $+2$ on the left and $0$ on the right, so $2\mathrm{e^-}$ join the left:

$$\mathrm{H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O}$$

The reduction half takes $2$ electrons, so the iron half is doubled and added:

$$\mathrm{H_2O_2 + 2Fe^{2+} + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O}$$

Atom check: $\mathrm{Fe}$ $2$ and $2$; $\mathrm{O}$ $2$ and $2$; $\mathrm{H}$ $2 + 2 = 4$ and $4$.

Charge check: left $0 + 2(+2) + 2(+1) = +6$; right $2(+3) = +6$.

Here the oxygen of the peroxide falls from $-1$ to $-2$, so hydrogen peroxide is the oxidant, exactly as in Question 5(a).

Ans: $\mathrm{H_2O_2 + 2Fe^{2+} + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O}$

Question 29: Thiosulphate and iodine

Balance the reaction behind every iodometric titration.

$$\mathrm{S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + I^-}$$

Answer:

Oxidation half: $\mathrm{2S_2O_3^{2-} \rightarrow S_4O_6^{2-}}$. I need two thiosulphate ions to make one tetrathionate, since the product has four sulphurs. Oxygen is $6$ on each side, which is already right. Charge is $-4$ on the left and $-2$ on the right, so two electrons leave:

$$\mathrm{2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^-}$$

Reduction half: $\mathrm{I_2 + 2e^- \rightarrow 2I^-}$. Charge is $-2$ on each side.

Both halves involve $2$ electrons, so I add them straight away:

$$\mathrm{2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-}$$

Charge check: left $-4$; right $-2 - 2 = -4$.

The oxidation numbers make the same point: sulphur goes from $+2$ in thiosulphate to an average of $+5/2$ in tetrathionate, a rise of $1/2$ per sulphur over four sulphurs, which is a total of $2$ electrons.

Ans: $\mathrm{2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-}$ Watch out: The mole ratio here is $\mathrm{I_2 : S_2O_3^{2-} = 1 : 2}$, and every iodometry calculation rests on it. Reading it as $1:1$ halves the final answer.

Question 30: Permanganate and sulphite in alkali

Balance in basic medium.

$$\mathrm{MnO_4^- + SO_3^{2-} \rightarrow MnO_2 + SO_4^{2-}}$$

Answer:

Reduction half: $\mathrm{MnO_4^- \rightarrow MnO_2}$. Manganese falls from $+7$ to $+4$, so three electrons are involved. In basic medium I balance oxygen by putting water on the side short of oxygen and hydroxide on the other. The left has four oxygens and the right has two, so I add $2\mathrm{H_2O}$ on the left and $4\mathrm{OH^-}$ on the right. Charge is then $-1$ on the left and $-4$ on the right, so $3\mathrm{e^-}$ join the left:

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

Oxidation half: $\mathrm{SO_3^{2-} \rightarrow SO_4^{2-}}$. Sulphur climbs from $+4$ to $+6$, so two electrons leave. The right needs one more oxygen, so $2\mathrm{OH^-}$ go on the left and $\mathrm{H_2O}$ on the right. Charge: left $-2 - 2 = -4$; right $-2$, so two electrons leave the right side:

$$\mathrm{SO_3^{2-} + 2OH^- \rightarrow SO_4^{2-} + H_2O + 2e^-}$$

To match electrons I take the reduction half twice and the oxidation half three times, giving $6$ each way. Adding:

$$\mathrm{2MnO_4^- + 3SO_3^{2-} + 4H_2O + 6OH^- \rightarrow 2MnO_2 + 3SO_4^{2-} + 8OH^- + 3H_2O}$$

Now I cancel what appears on both sides. Three waters cancel, leaving $\mathrm{H_2O}$ on the left, and six hydroxides cancel, leaving $2\mathrm{OH^-}$ on the right:

$$\mathrm{2MnO_4^- + 3SO_3^{2-} + H_2O \rightarrow 2MnO_2 + 3SO_4^{2-} + 2OH^-}$$

Atom check: $\mathrm{Mn}$ $2$ and $2$; $\mathrm{S}$ $3$ and $3$; $\mathrm{H}$ $2$ and $2$; $\mathrm{O}$ $8 + 9 + 1 = 18$ on the left and $4 + 12 + 2 = 18$ on the right.

Charge check: left $-2 - 6 = -8$; right $-6 - 2 = -8$.

Ans: $\mathrm{2MnO_4^- + 3SO_3^{2-} + H_2O \rightarrow 2MnO_2 + 3SO_4^{2-} + 2OH^-}$ Watch out: Permanganate stops at $\mathrm{MnO_2}$ here, so the n-factor is $3$, not $5$. The medium decides the product and the product decides the electron count.

Question 31: Stannite reducing bismuth hydroxide

Balance in basic medium.

$$\mathrm{SnO_2^{2-} + Bi(OH)_3 \rightarrow SnO_3^{2-} + Bi}$$

Answer:

Oxidation half: $\mathrm{SnO_2^{2-} \rightarrow SnO_3^{2-}}$. Tin rises from $+2$ to $+4$, so two electrons leave. The right needs one extra oxygen, so I add $2\mathrm{OH^-}$ on the left and $\mathrm{H_2O}$ on the right. Charge: left $-2 - 2 = -4$; right $-2$, so two electrons come off the right:

$$\mathrm{SnO_2^{2-} + 2OH^- \rightarrow SnO_3^{2-} + H_2O + 2e^-}$$

Reduction half: $\mathrm{Bi(OH)_3 \rightarrow Bi}$. Bismuth falls from $+3$ to $0$, so three electrons are needed, and the three hydroxide groups leave as $3\mathrm{OH^-}$. Charge: left $0 - 3 = -3$; right $-3$:

$$\mathrm{Bi(OH)_3 + 3e^- \rightarrow Bi + 3OH^-}$$

Six is the lowest common multiple of $2$ and $3$, so the tin half is taken three times and the bismuth half twice:

$$\mathrm{3SnO_2^{2-} + 6OH^- + 2Bi(OH)_3 \rightarrow 3SnO_3^{2-} + 3H_2O + 2Bi + 6OH^-}$$

Six hydroxides appear on both sides and cancel:

$$\mathrm{3SnO_2^{2-} + 2Bi(OH)_3 \rightarrow 3SnO_3^{2-} + 2Bi + 3H_2O}$$

Atom check: $\mathrm{Sn}$ $3$ and $3$; $\mathrm{Bi}$ $2$ and $2$; $\mathrm{H}$ $6$ and $6$; $\mathrm{O}$ $6 + 6 = 12$ on the left and $9 + 3 = 12$ on the right.

Charge check: left $3(-2) + 0 = -6$; right $3(-2) = -6$.

Ans: $\mathrm{3SnO_2^{2-} + 2Bi(OH)_3 \rightarrow 3SnO_3^{2-} + 2Bi + 3H_2O}$ Watch out: Hydroxide that ends up on both sides must be cancelled. Leaving $6\mathrm{OH^-}$ on each side is not wrong chemically, but it is an unfinished answer.