Chapter 7 · Redox Reactions

7.1 Oxidation, Reduction and Oxidation Number

From "adding oxygen" to "losing electrons": what oxidation and reduction really are, why a zinc strip turns copper sulphate colourless, and the bookkeeping of oxidation numbers that lets you spot a redox change in any reaction — ionic or covalent.

A cut apple turns brown, an iron gate rusts, a cell phone battery runs down, a bleach removes a stain, and every cell in your body burns glucose to stay alive. These look like five unrelated events, but chemically they are one kind of reaction. In each, one substance gives up electrons and another takes them: something is oxidised and something is reduced. Reactions of this kind — redox reactions — power metallurgy, batteries, fuel cells, photosynthesis, respiration and much of industrial chemistry.

This section follows the idea of oxidation as it grew: first "adding oxygen", then "losing electrons", and finally a general bookkeeping device, the oxidation number, that works even for covalent molecules in which no electron is completely transferred. Along the way you will see metals compete for electrons — the first step towards the batteries of Section 7.3 — and learn to read oxidation and reduction straight off a balanced equation.

Oxidation and Reduction: from Oxygen to Electrons

The word oxidation first meant "combining with oxygen". Magnesium burning with a dazzling white flame and carbon burning to carbon dioxide are oxidations of magnesium and of carbon:

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

Chemists soon noticed that removing hydrogen has the same effect as adding oxygen — hydrogen sulphide burned in a limited supply of air loses its hydrogen and leaves sulphur — and that other electronegative elements (fluorine, chlorine, sulphur) can play the part of oxygen. The definition was widened step by step, and "reduction" (originally, reducing an ore to its metal) was defined as the opposite process.

📘 Definition 7.1 — Oxidation and reduction (classical)
Oxidation is the addition of oxygen or of an electronegative element to a substance, or the removal of hydrogen or of an electropositive element from it. Reduction is the removal of oxygen or of an electronegative element from a substance, or the addition of hydrogen or of an electropositive element to it. For example:
  • $\mathrm{Mg(s)+F_{2}(g)\rightarrow MgF_{2}(s)}$ — magnesium is oxidised (an electronegative element is added);
  • $\mathrm{2H_{2}S(g)+O_{2}(g)\rightarrow 2S(s)+2H_{2}O(l)}$ — hydrogen sulphide is oxidised (hydrogen is removed) and oxygen is reduced (hydrogen is added);
  • $\mathrm{2HgO(s)\xrightarrow{\Delta}2Hg(l)+O_{2}(g)}$ — mercury(II) oxide is reduced (oxygen is removed);
  • $\mathrm{2FeCl_{3}(aq)+H_{2}(g)\rightarrow}$ $\mathrm{2FeCl_{2}(aq)+2HCl(aq)}$ — iron(III) chloride is reduced (an electronegative element is removed) and hydrogen is oxidised.

Look again at the last two examples: whenever one substance is oxidised, another is reduced. Oxidation and reduction always happen together, which is why such reactions are called redox (reduction–oxidation) reactions. The electronic theory of the atom (Chapter 2) explains why. When sodium burns in chlorine, each sodium atom gives one electron to a chlorine atom:

$$\mathrm{2Na(s)\rightarrow 2Na^{+}+2e^{-}}\qquad\qquad \mathrm{Cl_{2}(g)+2e^{-}\rightarrow 2Cl^{-}}$$

Adding the two lines, the electrons cancel and the ions form the crystal: $\mathrm{2Na(s)+Cl_{2}(g)\rightarrow 2NaCl(s)}$. The same thing happens when magnesium burns in oxygen ($\mathrm{Mg\rightarrow Mg^{2+}+2e^{-}}$, $\mathrm{O_{2}+4e^{-}\rightarrow 2O^{2-}}$). Electrons cannot simply vanish, so a substance can lose electrons only if something else takes them.

📘 Definition 7.2 — Oxidation and reduction as electron transfer
Oxidation is the loss of electrons by an atom, ion or molecule; reduction is the gain of electrons. Each of the two processes, written separately with its electrons shown, is a half-reaction: $$\text{oxidation half-reaction: }\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$$ $$\text{reduction half-reaction: }\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$$ A redox reaction is the sum of an oxidation half-reaction and a reduction half-reaction in which the number of electrons lost equals the number gained, so that no electrons appear in the overall equation.
📘 Definition 7.3 — Oxidising agent and reducing agent
An oxidising agent (oxidant) is a species that oxidises another by accepting electrons from it; in doing so the oxidising agent is itself reduced. A reducing agent (reductant) is a species that reduces another by giving it electrons; the reducing agent is itself oxidised. In $\mathrm{2Na+Cl_{2}\rightarrow 2NaCl}$, sodium is the reducing agent and chlorine the oxidising agent.
⚠️ The agent is the opposite of what happens to it
A reliable memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The common slip is to say "the oxidising agent is oxidised" — it is the other way round. The oxidising agent causes oxidation and is reduced; the reducing agent causes reduction and is oxidised. Notice also that an electron in a half-reaction is written $\mathrm{e^{-}}$ with no state symbol, because free electrons do not exist in solution — they pass directly from one particle to another (or along a wire, Section 7.3).
✏️ Example 1 — Three views of the same reactions
For each reaction, say which substance is oxidised and which is reduced using Definition 7.1, then write the two half-reactions and name the oxidising and reducing agents (Definitions 7.2 and 7.3). (a) $\mathrm{4Na(s)+O_{2}(g)\rightarrow 2Na_{2}O(s)}$   (b) $\mathrm{2Na(s)+H_{2}(g)\rightarrow 2NaH(s)}$   (c) $\mathrm{Zn(s)+Cl_{2}(g)\rightarrow ZnCl_{2}(s)}$
(a) Oxygen is added to sodium: Na is oxidised, $\mathrm{O_{2}}$ is reduced. Half-reactions: $\mathrm{4Na\rightarrow 4Na^{+}+4e^{-}}$ and $\mathrm{O_{2}+4e^{-}\rightarrow 2O^{2-}}$. Reducing agent Na, oxidising agent $\mathrm{O_{2}}$.
(b) Classically, hydrogen is added to sodium, which seems to say sodium is reduced — but sodium hydride is ionic, $\mathrm{Na^{+}H^{-}}$. Electronically: $\mathrm{2Na\rightarrow 2Na^{+}+2e^{-}}$ (oxidation) and $\mathrm{H_{2}+2e^{-}\rightarrow 2H^{-}}$ (reduction). So Na is oxidised and $\mathrm{H_{2}}$, the oxidising agent, is reduced. The electron picture is the one to trust; the "addition of hydrogen" rule is meant for hydrogen bonded to a more electronegative partner.
(c) An electronegative element is added to zinc: Zn is oxidised. $\mathrm{Zn\rightarrow Zn^{2+}+2e^{-}}$, $\mathrm{Cl_{2}+2e^{-}\rightarrow 2Cl^{-}}$. Reducing agent Zn, oxidising agent $\mathrm{Cl_{2}}$.
Check In every case the electrons lost (4, 2, 2) equal the electrons gained, and each overall equation is balanced in atoms and charge ✓.

Competitive Electron Transfer Reactions

Place a strip of zinc in a blue solution of copper(II) sulphate and leave it for an hour. The strip becomes coated with a reddish-brown, spongy layer of copper, and the blue colour, which is due to hydrated $\mathrm{Cu^{2+}}$ ions, slowly fades. If hydrogen sulphide is passed through the solution at the end, a white precipitate of zinc sulphide forms in alkaline medium, showing that zinc ions are now present. The reaction is

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

Each zinc atom has handed two electrons to a copper(II) ion at the surface of the strip: zinc is oxidised and $\mathrm{Cu^{2+}}$ is reduced. The sulphate ions take no part — they are spectator ions and are left out of the ionic equation. Now try the reverse experiment: a copper strip in a colourless zinc sulphate solution. Nothing happens, however long you wait; passing $\mathrm{H_{2}S}$ through the solution gives no black copper sulphide. Electrons flow from zinc to copper(II) ions, but not from copper to zinc ions.

A copper wire in a solution of silver nitrate shows the next step in the competition. Shiny crystals of silver grow on the wire and the colourless solution turns blue as $\mathrm{Cu^{2+}}$ ions form:

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

So copper gives electrons to silver ions, but silver does not give electrons to copper(II) ions. The metals can be ranked by how readily they release electrons — zinc more readily than copper, copper more readily than silver. Every reaction of a metal with the ions of another metal is a contest between two redox couples for the same electrons; the side that holds electrons more weakly loses them. Such reactions are called competitive electron transfer reactions. Even when nothing visible happens (copper in zinc sulphate), a tiny equilibrium amount of reaction occurs; in Section 7.3 the contest is measured as a voltage.

📐 Rule 7.1 — The metal activity series and displacement
Arranging metals in order of their tendency to lose electrons in aqueous solution gives the activity series (hydrogen is included for comparison): $$\mathrm{K\gt Na\gt Ca\gt Mg\gt Al\gt Zn\gt Fe\gt Ni\gt Sn\gt Pb\gt H\gt Cu\gt Hg\gt Ag\gt Au}$$ A metal higher in the series is a stronger reducing agent: it displaces (reduces to the metal) the ions of any metal below it from their solution, and its own ion is a weaker oxidising agent. Metals above hydrogen can displace hydrogen from dilute acids such as $\mathrm{HCl}$ and $\mathrm{H_{2}SO_{4}}$; metals below hydrogen cannot. (Very reactive metals — K, Na, Ca — react with the water itself, so they are not used to displace other metals from aqueous solution.)
✏️ Example 2 — Will it displace?
Use Rule 7.1 to decide whether a reaction occurs. Where it does, write the two half-reactions and the balanced ionic equation, and name the oxidising agent. (a) an iron nail in copper(II) sulphate solution; (b) a silver spoon in copper(II) sulphate solution; (c) magnesium ribbon in zinc sulphate solution; (d) zinc granules in dilute hydrochloric acid; (e) copper turnings in dilute hydrochloric acid.
(a) Fe is above Cu: yes. $\mathrm{Fe(s)\rightarrow Fe^{2+}(aq)+2e^{-}}$; $\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$. Overall $\mathrm{Fe(s)+Cu^{2+}(aq)\rightarrow Fe^{2+}(aq)+Cu(s)}$; the nail turns pink-brown and the blue colour fades. Oxidising agent $\mathrm{Cu^{2+}}$.
(b) Ag is below Cu: no reaction — which is why silver spoons do not dissolve in copper sulphate.
(c) Mg is above Zn: yes. $\mathrm{Mg(s)\rightarrow Mg^{2+}(aq)+2e^{-}}$; $\mathrm{Zn^{2+}(aq)+2e^{-}\rightarrow Zn(s)}$. Overall $\mathrm{Mg(s)+Zn^{2+}(aq)\rightarrow Mg^{2+}(aq)+Zn(s)}$. Oxidising agent $\mathrm{Zn^{2+}}$.
(d) Zn is above H: yes. $\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$; $\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)}$. Overall $\mathrm{Zn(s)+2H^{+}(aq)\rightarrow Zn^{2+}(aq)+H_{2}(g)}$ (the $\mathrm{Cl^{-}}$ ions are spectators). Oxidising agent $\mathrm{H^{+}}$.
(e) Cu is below H: no hydrogen is evolved.
Check In (a), (c) and (d) two electrons are lost and two gained, and the charge is $+2$ on both sides of each overall equation ✓.
Interactive 1 — Metal Strip in a Salt Solution: Who Wins the Electrons?

Choose a metal strip and a solution of metal ions, then press ▶ Play (or drag the time slider). If the strip metal stands higher in the activity series, its atoms give electrons (yellow dots) to the ions in solution: the strip slowly dissolves, metal from the solution plates onto it, and coloured ions such as blue $\mathrm{Cu^{2+}}$ fade. If it stands lower, nothing happens. The readouts give the two half-reactions and the balanced ionic equation.

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Oxidation Number

In the reaction $\mathrm{H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)}$ no electron is completely transferred: hydrogen chloride is a covalent molecule. Yet the reaction looks like the formation of NaCl, and chlorine, the more electronegative element (Section 3.3), pulls the shared pair towards itself, so hydrogen has partly "lost" an electron and chlorine has partly "gained" one. The same happens when hydrogen burns to water, $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)}$. To treat such reactions in the same way as ionic ones, chemists use a bookkeeping charge: they pretend that every shared pair belongs entirely to the more electronegative atom.

📘 Definition 7.4 — Oxidation number
The oxidation number (or oxidation state) of an atom in a substance is the charge the atom would carry if every bond it forms were ionic — that is, if each shared electron pair were given completely to the more electronegative of the two atoms (and a pair shared between identical atoms were split equally). It is written with its sign before the number ($+2$, $-1$), to distinguish it from an ionic charge written as a superscript ($\mathrm{Zn^{2+}}$). In $\mathrm{H_{2}O}$ both O–H pairs go to oxygen, so O is $-2$ and each H is $+1$; in $\mathrm{H_{2}}$ the pair is shared equally and each H is $0$.

Drawing the structure every time would be slow, so in practice oxidation numbers are assigned with a short list of rules, applied in order.

📐 Rule 7.2 — Rules for assigning oxidation numbers
  1. An atom in a free element has oxidation number $0$, whatever its molecular form: $\mathrm{H_{2}}$, $\mathrm{O_{2}}$, $\mathrm{O_{3}}$, $\mathrm{Cl_{2}}$, $\mathrm{P_{4}}$, $\mathrm{S_{8}}$, $\mathrm{Na}$, $\mathrm{Fe}$.
  2. For a monatomic ion the oxidation number equals the charge on the ion: $\mathrm{Na^{+}}$ $+1$, $\mathrm{Fe^{3+}}$ $+3$, $\mathrm{S^{2-}}$ $-2$.
  3. Fluorine, the most electronegative element, is always $-1$ in its compounds. The alkali metals (group 1) are $+1$ and the alkaline earth metals (group 2) are $+2$ in all their compounds; aluminium is $+3$.
  4. Oxygen is $-2$ in almost all its compounds. Exceptions: in peroxides (O–O linkage: $\mathrm{H_{2}O_{2}}$, $\mathrm{Na_{2}O_{2}}$, $\mathrm{BaO_{2}}$) each O is $-1$; in superoxides ($\mathrm{KO_{2}}$, $\mathrm{RbO_{2}}$, ion $\mathrm{O_{2}^{-}}$) each O is $-\tfrac12$; bonded to fluorine, O is positive — $+2$ in $\mathrm{OF_{2}}$ and $+1$ in $\mathrm{O_{2}F_{2}}$.
  5. Hydrogen is $+1$ when bonded to non-metals ($\mathrm{HCl}$, $\mathrm{H_{2}O}$, $\mathrm{NH_{3}}$, $\mathrm{CH_{4}}$) and $-1$ in binary compounds with metals, the metal hydrides ($\mathrm{LiH}$, $\mathrm{NaH}$, $\mathrm{CaH_{2}}$).
  6. The other halogens (Cl, Br, I) are $-1$ in halides. They take positive values when combined with oxygen ($\mathrm{ClO^{-}}$, $\mathrm{ClO_{4}^{-}}$, $\mathrm{IO_{3}^{-}}$) or with a more electronegative halogen (in $\mathrm{ICl}$ iodine is $+1$; in $\mathrm{BrF_{3}}$ bromine is $+3$).
  7. The sum of the oxidation numbers of all the atoms is $0$ for a neutral species and equals the charge for an ion: in $\mathrm{SO_{4}^{2-}}$, $x+4(-2)=-2$, so S is $+6$.
When rules conflict, the one higher in the list wins (rule 3 before rule 4: in $\mathrm{OF_{2}}$, F is $-1$, so O must be $+2$).

For the elements of the s- and p-blocks the oxidation numbers follow the group (Section 3.3). Group 1 and group 2 elements show only $+1$ and $+2$. In groups 13–17 the highest oxidation number is the number of valence electrons, i.e. the group number minus 10 (N and P $+5$, S $+6$, Cl $+7$), and the lowest is that number minus 8 (N $-3$, S $-2$, Cl $-1$). Fluorine (only $-1$) and oxygen (rarely positive) are exceptions. Transition metals show several oxidation states, up to $+7$ for manganese in $\mathrm{MnO_{4}^{-}}$.

✏️ Example 3 — Using the rules
Find the oxidation number of the element named: (a) Mn in $\mathrm{KMnO_{4}}$; (b) Cr in $\mathrm{K_{2}Cr_{2}O_{7}}$ and in $\mathrm{CrO_{4}^{2-}}$; (c) S in $\mathrm{H_{2}SO_{4}}$, $\mathrm{SO_{3}^{2-}}$ and $\mathrm{H_{2}S}$; (d) N in $\mathrm{NH_{4}^{+}}$ and in $\mathrm{NO_{3}^{-}}$, and hence both nitrogen atoms in ammonium nitrate, $\mathrm{NH_{4}NO_{3}}$; (e) P in $\mathrm{H_{3}PO_{4}}$ and in $\mathrm{H_{3}PO_{2}}$.
(a) K $+1$, O $-2$: $1+x+4(-2)=0$, so $x=+7$.
(b) $2(+1)+2x+7(-2)=0$ gives $2x=12$, $x=+6$. In $\mathrm{CrO_{4}^{2-}}$: $x+4(-2)=-2$, $x=+6$ as well — the chromate–dichromate change $\mathrm{2CrO_{4}^{2-}+2H^{+}\rightarrow Cr_{2}O_{7}^{2-}+H_{2}O}$ is not a redox reaction.
(c) $\mathrm{H_{2}SO_{4}}$: $2+x-8=0$, $x=+6$. $\mathrm{SO_{3}^{2-}}$: $x-6=-2$, $x=+4$. $\mathrm{H_{2}S}$: $2+x=0$, $x=-2$ (sulphur's lowest value).
(d) $\mathrm{NH_{4}^{+}}$: $x+4=+1$, $x=-3$. $\mathrm{NO_{3}^{-}}$: $x-6=-1$, $x=+5$. Ammonium nitrate is the salt $\mathrm{NH_{4}^{+}NO_{3}^{-}}$, so its two N atoms are $-3$ and $+5$. (Applying the sum rule to $\mathrm{N_{2}H_{4}O_{3}}$ as a whole would give the meaningless average $+1$.)
(e) $\mathrm{H_{3}PO_{4}}$: $3+x-8=0$, $x=+5$. $\mathrm{H_{3}PO_{2}}$: $3+x-4=0$, $x=+1$.
Check Every value lies between the lowest and highest values allowed by the group: Mn up to $+7$, Cr up to $+6$, S $-2$ to $+6$, N $-3$ to $+5$, P $-3$ to $+5$ ✓.
✏️ Example 4 — The special cases, and fractional oxidation numbers
Find the oxidation number of oxygen or hydrogen in (a) $\mathrm{BaO_{2}}$, (b) $\mathrm{KO_{2}}$, (c) $\mathrm{OF_{2}}$, (d) $\mathrm{CaH_{2}}$. Then find the oxidation number of (e) Fe in magnetite, $\mathrm{Fe_{3}O_{4}}$; (f) S in the tetrathionate ion, $\mathrm{S_{4}O_{6}^{2-}}$; (g) C in carbon suboxide, $\mathrm{C_{3}O_{2}}$. What does a fractional answer mean?
(a) Ba is $+2$ (rule 3, which outranks rule 4): $2+2x=0$, $x=-1$ — barium peroxide contains the $\mathrm{O_{2}^{2-}}$ ion.
(b) K is $+1$: $1+2x=0$, $x=-\tfrac12$ — potassium superoxide contains $\mathrm{O_{2}^{-}}$.
(c) F is always $-1$: $x+2(-1)=0$, $x=+2$.
(d) Ca is $+2$ and this is a metal hydride: $2+2x=0$, $x=-1$ (the $\mathrm{H^{-}}$ ion).
(e) $3x+4(-2)=0$, $x=+\tfrac83$.
(f) $4x+6(-2)=-2$, $4x=10$, $x=+2.5$.
(g) $3x+2(-2)=0$, $x=+\tfrac43$.
Meaning No atom can carry a fraction of an electron's charge. A fractional value is the average over atoms of the same element that are in different environments. Magnetite is $\mathrm{FeO\cdot Fe_{2}O_{3}}$: one Fe is $+2$ and two are $+3$, and $\tfrac{2+3+3}{3}=\tfrac83$ ✓.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — oxidation numbers from structures

Not required by the CBSE syllabus. JEE Main and NEET list oxidation numbers under redox reactions, and JEE Advanced lists redox calculations; assigning them from a structure is a common exam step.

When the sum rule gives a fraction or an impossibly high value, the structure tells you the individual oxidation numbers. In $\mathrm{S_{4}O_{6}^{2-}}$ ($\mathrm{{}^{-}O_{3}S{-}S{-}S{-}SO_{3}^{-}}$) the two central S atoms are bonded only to sulphur and are $0$, while each terminal S is $+5$: average $\tfrac{5+0+0+5}{4}=2.5$ ✓. In peroxomonosulphuric acid $\mathrm{H_{2}SO_{5}}$ the sum rule gives $+8$, above sulphur's maximum of $+6$; the molecule contains one O–O linkage (two O at $-1$, three at $-2$), so S is $+6$. In the blue "butterfly" oxide $\mathrm{CrO_{5}}$ there are two peroxo groups and one oxo oxygen: $x+4(-1)+(-2)=0$, so Cr is $+6$, not $+10$. Likewise $\mathrm{Br_{3}O_{8}}$ has an average of $+\tfrac{16}{3}$ for Br (two Br at $+6$ and one at $+4$).

Oxidation numbers also give a systematic way to name compounds of elements that show more than one oxidation state. The older names (ferrous / ferric, cuprous / cupric, stannous / stannic) only distinguish two states and say nothing about their values.

📘 Definition 7.5 — Stock notation
In Stock notation (after the German chemist Alfred Stock) the oxidation number of a metal is written as a Roman numeral in brackets after its name or symbol: $\mathrm{FeCl_{2}}$ iron(II) chloride and $\mathrm{FeCl_{3}}$ iron(III) chloride; $\mathrm{Cu_{2}O}$ copper(I) oxide and $\mathrm{CuO}$ copper(II) oxide; $\mathrm{SnCl_{4}}$ tin(IV) chloride; $\mathrm{Hg_{2}Cl_{2}}$ mercury(I) chloride; $\mathrm{MnO_{2}}$ manganese(IV) oxide; $\mathrm{Cr_{2}O_{3}}$ chromium(III) oxide; $\mathrm{Au^{III}}$ for gold in the $+3$ state.
Interactive 2 — Oxidation-Number Calculator

Pick a species or type a formula — write charges with ^, e.g. Cr2O7^2-, NH4^+, K4[Fe(CN)6]. The calculator applies Rule 7.2 one step at a time (drag the step slider or press ▶ Play): each atom turns red for a positive, blue for a negative and grey for a zero oxidation number, and the sum check at the bottom must equal the charge. Try the special cases — $\mathrm{H_{2}O_{2}}$, $\mathrm{KO_{2}}$, $\mathrm{OF_{2}}$, $\mathrm{CaH_{2}}$ — and the fractional averages $\mathrm{Fe_{3}O_{4}}$ and $\mathrm{S_{4}O_{6}^{2-}}$.

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Oxidation and Reduction in Terms of Oxidation Number

When zinc becomes $\mathrm{Zn^{2+}}$ its oxidation number rises from $0$ to $+2$; when $\mathrm{Cu^{2+}}$ becomes copper its oxidation number falls from $+2$ to $0$. Losing electrons always raises the oxidation number, and gaining them lowers it. For covalent substances the oxidation number tracks the electron shift in the same way, so the most general definitions are written in terms of oxidation numbers.

📐 Rule 7.3 — Oxidation and reduction in terms of oxidation number
  • Oxidation is an increase in the oxidation number of an element; reduction is a decrease.
  • An oxidising agent is a reagent that increases the oxidation number of an element in another substance; the oxidising agent contains an element whose oxidation number decreases. A reducing agent lowers the oxidation number of an element in another substance and contains an element whose oxidation number increases.
  • A reaction is a redox reaction if, and only if, the oxidation number of at least one element changes. The total increase in oxidation number equals the total decrease, because both count the same electrons.

In $\mathrm{\overset{0}{H}_{2}(g)+\overset{0}{Cl}_{2}(g)\rightarrow 2\overset{+1}{H}\overset{-1}{Cl}(g)}$ hydrogen rises from $0$ to $+1$ (oxidised) and chlorine falls from $0$ to $-1$ (reduced): a redox reaction, although no ions are formed. By contrast, in neutralisation ($\mathrm{H^{+}+OH^{-}\rightarrow H_{2}O}$), precipitation ($\mathrm{Ag^{+}+Cl^{-}\rightarrow AgCl}$) and the decomposition of limestone ($\mathrm{CaCO_{3}\rightarrow CaO+CO_{2}}$) no oxidation number changes, so these are not redox reactions.

✏️ Example 5 — Reading a redox reaction off its oxidation numbers
(a) In the final step of copper extraction, $\mathrm{2Cu_{2}O(s)+Cu_{2}S(s)\rightarrow 6Cu(s)+SO_{2}(g)}$, which element is oxidised, which is reduced, and what are the oxidising and reducing agents? (b) Which of these are redox reactions? (i) $\mathrm{NaOH(aq)+HCl(aq)\rightarrow NaCl(aq)+H_{2}O(l)}$ (ii) $\mathrm{2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(aq)}$ (iii) $\mathrm{Cr_{2}O_{7}^{2-}(aq)+2OH^{-}(aq)\rightarrow}$ $\mathrm{2CrO_{4}^{2-}(aq)+H_{2}O(l)}$ (iv) $\mathrm{PCl_{3}(l)+Cl_{2}(g)\rightarrow PCl_{5}(s)}$
(a) ONs Cu is $+1$ in both $\mathrm{Cu_{2}O}$ and $\mathrm{Cu_{2}S}$ and $0$ in Cu; S is $-2$ in $\mathrm{Cu_{2}S}$ and $+4$ in $\mathrm{SO_{2}}$; O stays $-2$.
(a) result Copper is reduced ($+1\rightarrow0$), sulphur is oxidised ($-2\rightarrow+4$). $\mathrm{Cu_{2}O}$ is an oxidising agent; $\mathrm{Cu_{2}S}$ is the reducing agent (through its sulphur) and an oxidising agent (through its copper). Check: six Cu each fall by 1 (total $-6$), one S rises by 6 (total $+6$) ✓.
(b) (i) No: Na $+1$, O $-2$, H $+1$, Cl $-1$ throughout (acid–base). (ii) Yes: Fe $+3\rightarrow+2$, I $-1\rightarrow0$. (iii) No: Cr is $+6$ on both sides. (iv) Yes: P $+3\rightarrow+5$, Cl $0\rightarrow-1$.
✏️ Example 6 — Can it oxidise, reduce, or both?
An element in its highest oxidation number can only be reduced, in its lowest only oxidised, and in between it can do either. Use this to say whether each species can act as an oxidising agent only, a reducing agent only, or both: (a) $\mathrm{KMnO_{4}}$; (b) $\mathrm{H_{2}S}$; (c) $\mathrm{SO_{2}}$; (d) $\mathrm{HNO_{3}}$; (e) $\mathrm{NaNO_{2}}$; (f) $\mathrm{H_{2}O_{2}}$.
(a) Mn $+7$, its maximum: oxidising agent only.
(b) S $-2$, its minimum: reducing agent only.
(c) S $+4$, between $-2$ and $+6$: both. It reduces acidified dichromate (S → $+6$) and oxidises $\mathrm{H_{2}S}$ to sulphur (S → $0$).
(d) N $+5$, its maximum: oxidising agent only (as the nitrate part; the $\mathrm{H^{+}}$ also acts as an acid).
(e) N $+3$ in $\mathrm{NO_{2}^{-}}$, between $-3$ and $+5$: both.
(f) O $-1$, between $-2$ and $0$: both. Hydrogen peroxide oxidises $\mathrm{Fe^{2+}}$ (O → $-2$, giving water) but reduces acidified $\mathrm{KMnO_{4}}$ (O → $0$, giving $\mathrm{O_{2}}$). Section 7.2 shows that it can even oxidise and reduce itself.

Exercises

I
Oxidation and reduction as electron transfer.
  • 1 Define oxidation and reduction (a) in terms of oxygen and hydrogen, (b) in terms of electrons.
  • 2 In $\mathrm{2Mg(s)+O_{2}(g)\rightarrow 2MgO(s)}$, write the two half-reactions and name the reducing agent.
  • 3 Classify each change as oxidation or reduction: (a) $\mathrm{Fe^{2+}\rightarrow Fe^{3+}}$ (b) $\mathrm{Cl_{2}\rightarrow 2Cl^{-}}$ (c) $\mathrm{2I^{-}\rightarrow I_{2}}$ (d) $\mathrm{Sn^{4+}\rightarrow Sn^{2+}}$. Add the electrons needed to complete each half-reaction.
  • 4 In $\mathrm{2K_{4}[Fe(CN)_{6}](aq)+H_{2}O_{2}(aq)}$ $\mathrm{\rightarrow 2K_{3}[Fe(CN)_{6}](aq)+2KOH(aq)}$, use Definition 7.1 to say which substance is oxidised.
  • 5 Why must oxidation and reduction always occur together?
  • 6 In $\mathrm{2Na(s)+H_{2}(g)\rightarrow 2NaH(s)}$, which species is the oxidising agent? Explain using electrons.
  • 7 "An oxidising agent is a substance that is easily oxidised." Correct the statement.
1 (a) Oxidation: addition of oxygen or an electronegative element, or removal of hydrogen or an electropositive element; reduction: the reverse. (b) Oxidation is loss of electrons, reduction is gain of electrons.
2 $\mathrm{2Mg\rightarrow 2Mg^{2+}+4e^{-}}$; $\mathrm{O_{2}+4e^{-}\rightarrow 2O^{2-}}$. Reducing agent: Mg.
3 (a) Oxidation: $\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^{-}}$. (b) Reduction: $\mathrm{Cl_{2}+2e^{-}\rightarrow 2Cl^{-}}$. (c) Oxidation: $\mathrm{2I^{-}\rightarrow I_{2}+2e^{-}}$. (d) Reduction: $\mathrm{Sn^{4+}+2e^{-}\rightarrow Sn^{2+}}$.
4 Potassium (an electropositive element) is removed from $\mathrm{K_{4}[Fe(CN)_{6}]}$, so it is oxidised (and $\mathrm{H_{2}O_{2}}$ is reduced).
5 Electrons lost by one species cannot exist free in solution; they must be taken up by another species at the same moment.
6 $\mathrm{H_{2}}$: it accepts the electrons from sodium, $\mathrm{H_{2}+2e^{-}\rightarrow 2H^{-}}$, and is reduced.
7 An oxidising agent is a substance that is easily reduced — it accepts electrons readily.
II
Competitive electron transfer and the activity series (Rule 7.1).
  • 1 Write the balanced ionic equation for the reaction of zinc with copper(II) sulphate solution and describe two observations.
  • 2 Why does a copper strip not react with zinc sulphate solution?
  • 3 A copper wire is left in silver nitrate solution. Write the half-reactions and the overall equation, and explain why the solution turns blue.
  • 4 Predict whether a reaction occurs: (a) Pb in $\mathrm{AgNO_{3}(aq)}$; (b) Ni in $\mathrm{ZnSO_{4}(aq)}$; (c) Al in $\mathrm{CuSO_{4}(aq)}$; (d) Fe in dilute $\mathrm{H_{2}SO_{4}}$; (e) Ag in dilute HCl.
  • 5 Write the balanced ionic equation for part 4(c).
  • 6 Metals X, Y and Z: X displaces Y from a solution of $\mathrm{Y^{2+}}$; Z displaces neither X nor Y but reacts with no dilute acid; Y reacts with dilute acid. Arrange X, Y, Z and H in decreasing order of reducing power.
  • 7 Why should copper(II) sulphate solution not be stored in an iron container?
1 $\mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}$. A reddish-brown deposit of copper forms on the zinc, and the blue colour of the solution fades (the solution may also warm slightly).
2 Copper lies below zinc in the activity series: it holds its electrons more tightly than zinc does, so it cannot reduce $\mathrm{Zn^{2+}}$.
3 $\mathrm{Cu\rightarrow Cu^{2+}+2e^{-}}$; $\mathrm{2Ag^{+}+2e^{-}\rightarrow 2Ag}$; overall $\mathrm{Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)}$. The $\mathrm{Cu^{2+}(aq)}$ ions formed are blue.
4 (a) Yes (Pb above Ag). (b) No (Ni below Zn). (c) Yes (Al above Cu). (d) Yes, $\mathrm{H_{2}}$ is evolved (Fe above H). (e) No (Ag below H).
5 $\mathrm{2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)}$ — six electrons lost and six gained; charge $+6$ on both sides.
6 X $\gt$ Y $\gt$ H $\gt$ Z.
7 Iron is above copper: it would reduce $\mathrm{Cu^{2+}}$ ($\mathrm{Fe+Cu^{2+}\rightarrow Fe^{2+}+Cu}$), so the container would dissolve and the solution would be contaminated.
III
Assigning oxidation numbers (Rule 7.2).
  • 1 Find the oxidation number of (a) S in $\mathrm{Na_{2}S_{2}O_{3}}$ (b) Cl in $\mathrm{HClO_{4}}$ (c) Cl in $\mathrm{NaClO}$ (d) I in $\mathrm{IO_{3}^{-}}$.
  • 2 Find the oxidation number of C in (a) $\mathrm{CH_{4}}$, (b) $\mathrm{CH_{3}OH}$, (c) $\mathrm{HCHO}$, (d) $\mathrm{HCOOH}$, (e) $\mathrm{CO_{2}}$. What pattern do you see?
  • 3 Find the oxidation number of N in $\mathrm{N_{2}O}$, $\mathrm{NO}$, $\mathrm{NO_{2}}$, $\mathrm{N_{2}O_{5}}$, $\mathrm{HNO_{2}}$, $\mathrm{N_{2}H_{4}}$ and $\mathrm{NH_{2}OH}$.
  • 4 Find the oxidation number of oxygen in $\mathrm{Na_{2}O_{2}}$, $\mathrm{RbO_{2}}$, $\mathrm{O_{2}F_{2}}$ and $\mathrm{O_{3}}$.
  • 5 Find the oxidation number of hydrogen in $\mathrm{LiH}$, $\mathrm{NH_{3}}$ and $\mathrm{LiAlH_{4}}$.
  • 6 Find the oxidation number of Fe in $\mathrm{K_{4}[Fe(CN)_{6}]}$ and $\mathrm{K_{3}[Fe(CN)_{6}]}$ (the cyanide ion $\mathrm{CN^{-}}$ carries charge $-1$).
  • 7 Find the average oxidation number of Pb in red lead, $\mathrm{Pb_{3}O_{4}}$, and explain it in terms of $\mathrm{2PbO\cdot PbO_{2}}$.
  • 8 Give the Stock names of $\mathrm{Fe_{2}O_{3}}$, $\mathrm{CuCl}$, $\mathrm{PbO_{2}}$, $\mathrm{SnCl_{2}}$ and $\mathrm{Mn_{2}O_{7}}$.
  • 9 Find the oxidation number of Mn in $\mathrm{MnO_{4}^{2-}}$, $\mathrm{MnO_{2}}$ and $\mathrm{Mn_{3}O_{4}}$.
  • 10 What are the highest and lowest oxidation numbers of phosphorus and of chlorine? Give one example of a compound or ion for each.
1 (a) $2+2x-6=0$, $x=+2$ (an average). (b) $1+x-8=0$, $+7$. (c) $1+x-2=0$, $+1$. (d) $x-6=-1$, $+5$.
2 (a) $-4$ (b) $-2$ (c) $0$ (d) $+2$ (e) $+4$. Each step from methane to carbon dioxide (removing H or adding O) raises the oxidation number of carbon by 2: these are successive oxidations.
3 $\mathrm{N_{2}O}$ $+1$; NO $+2$; $\mathrm{NO_{2}}$ $+4$; $\mathrm{N_{2}O_{5}}$ $+5$; $\mathrm{HNO_{2}}$ $+3$; $\mathrm{N_{2}H_{4}}$ $-2$; $\mathrm{NH_{2}OH}$ $-1$.
4 $\mathrm{Na_{2}O_{2}}$ $-1$ (peroxide); $\mathrm{RbO_{2}}$ $-\tfrac12$ (superoxide); $\mathrm{O_{2}F_{2}}$ $+1$; $\mathrm{O_{3}}$ $0$ (element).
5 $\mathrm{LiH}$ $-1$; $\mathrm{NH_{3}}$ $+1$; $\mathrm{LiAlH_{4}}$: $1+3+4x=0$, $x=-1$.
6 $4(+1)+x+6(-1)=0$, $x=+2$; $3+x-6=0$, $x=+3$.
7 $3x-8=0$, $x=+\tfrac83$. In $\mathrm{2PbO\cdot PbO_{2}}$ two Pb are $+2$ and one is $+4$: $\tfrac{2+2+4}{3}=\tfrac83$.
8 Iron(III) oxide; copper(I) chloride; lead(IV) oxide; tin(II) chloride; manganese(VII) oxide.
9 $\mathrm{MnO_{4}^{2-}}$: $x-8=-2$, $+6$. $\mathrm{MnO_{2}}$: $+4$. $\mathrm{Mn_{3}O_{4}}$: $3x=8$, $+\tfrac83$ (one Mn(II), two Mn(III)).
10 Phosphorus (group 15): highest $+5$ ($\mathrm{H_{3}PO_{4}}$, $\mathrm{PCl_{5}}$), lowest $-3$ ($\mathrm{PH_{3}}$). Chlorine (group 17): highest $+7$ ($\mathrm{ClO_{4}^{-}}$, $\mathrm{Cl_{2}O_{7}}$), lowest $-1$ ($\mathrm{NaCl}$, $\mathrm{HCl}$).
IV
Identifying redox reactions (Rule 7.3).
  • 1 In $\mathrm{CuO(s)+H_{2}(g)\rightarrow Cu(s)+H_{2}O(g)}$ give the oxidation-number changes and name the oxidising agent.
  • 2 In $\mathrm{MnO_{2}(s)+4HCl(aq)\rightarrow}$ $\mathrm{MnCl_{2}(aq)+Cl_{2}(g)+2H_{2}O(l)}$, which element is oxidised and which reduced? What fraction of the chlorine atoms change oxidation number?
  • 3 Is $\mathrm{AgNO_{3}(aq)+NaCl(aq)\rightarrow}$ $\mathrm{AgCl(s)+NaNO_{3}(aq)}$ a redox reaction? Justify.
  • 4 Is $\mathrm{2O_{3}(g)\rightarrow 3O_{2}(g)}$ a redox reaction? Explain.
  • 5 In $\mathrm{Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)}$ find the total increase and the total decrease in oxidation number, and show that they are equal.
  • 6 In $\mathrm{2H_{2}S(g)+SO_{2}(g)\rightarrow 3S(s)+2H_{2}O(l)}$, identify the oxidising and reducing agents.
  • 7 Which of $\mathrm{Cl^{-}}$, $\mathrm{ClO^{-}}$, $\mathrm{ClO_{4}^{-}}$ can act only as a reducing agent, only as an oxidising agent, or as both? Explain.
  • 8 Applied. Breath-alcohol testers once used orange acidified dichromate, which turns green as $\mathrm{Cr_{2}O_{7}^{2-}}$ becomes $\mathrm{Cr^{3+}}$ while ethanol, $\mathrm{C_{2}H_{5}OH}$, becomes ethanoic acid, $\mathrm{CH_{3}COOH}$. Find the change in oxidation number of Cr and the change in average oxidation number of C, and say which is the reducing agent.
1 Cu $+2\rightarrow0$ (reduced); H $0\rightarrow+1$ (oxidised). Oxidising agent: CuO.
2 Mn $+4\rightarrow+2$ (reduced); Cl $-1\rightarrow0$ in $\mathrm{Cl_{2}}$ (oxidised). Of the four Cl atoms, two end up in $\mathrm{Cl_{2}}$ and two stay at $-1$ in $\mathrm{MnCl_{2}}$: half of them. Check: decrease 2 (one Mn), increase $2\times1$ ✓.
3 No. Ag $+1$, N $+5$, O $-2$, Na $+1$, Cl $-1$ on both sides; it is a precipitation (ion-combination) reaction.
4 No. Oxygen is $0$ in both $\mathrm{O_{3}}$ and $\mathrm{O_{2}}$ (both are forms of the element); it is a change of allotrope.
5 Fe: $2\times(+3\rightarrow0)$, decrease $6$. C: $3\times(+2\rightarrow+4)$, increase $6$. Equal ✓.
6 S falls from $+4$ (in $\mathrm{SO_{2}}$) to $0$ and rises from $-2$ (in $\mathrm{H_{2}S}$) to $0$. Oxidising agent $\mathrm{SO_{2}}$; reducing agent $\mathrm{H_{2}S}$.
7 $\mathrm{Cl^{-}}$ ($-1$, lowest): reducing agent only. $\mathrm{ClO_{4}^{-}}$ ($+7$, highest): oxidising agent only. $\mathrm{ClO^{-}}$ ($+1$): both.
8 Cr: $+6\rightarrow+3$, a decrease of 3 per Cr (6 per $\mathrm{Cr_{2}O_{7}^{2-}}$). C: ethanol $2x+6-2=0$, average $-2$; ethanoic acid $2x+4-4=0$, average $0$; an increase of 2 per C (4 per molecule). Ethanol is the reducing agent.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The oxidation number of oxygen is $+2$ in (A) $\mathrm{H_{2}O_{2}}$ (B) $\mathrm{KO_{2}}$ (C) $\mathrm{OF_{2}}$ (D) $\mathrm{O_{2}F_{2}}$

Idea F is always $-1$, so oxygen is positive only in its fluorides. $\mathrm{OF_{2}}$: $x-2=0$, $x=+2$; $\mathrm{O_{2}F_{2}}$: $2x-2=0$, $x=+1$. $\mathrm{H_{2}O_{2}}$ gives $-1$ and $\mathrm{KO_{2}}$ gives $-\tfrac12$.
Answer (C).

E2. JEE Main Numerical answer. The sum of the oxidation numbers of Cr in $\mathrm{K_{2}Cr_{2}O_{7}}$ and $\mathrm{CrO_{4}^{2-}}$, of Mn in $\mathrm{KMnO_{4}}$, and of S in $\mathrm{S_{2}O_{3}^{2-}}$ (average) is ____.

Each Cr in $\mathrm{K_{2}Cr_{2}O_{7}}$: $+6$. Cr in $\mathrm{CrO_{4}^{2-}}$: $+6$. Mn in $\mathrm{KMnO_{4}}$: $+7$. S in $\mathrm{S_{2}O_{3}^{2-}}$: $2x-6=-2$, $x=+2$.
Answer $6+6+7+2=$ 21.

E3. Board · 3 marks (a) Define oxidation and reduction in terms of oxidation number. (b) In $\mathrm{Zn(s)+2Ag^{+}(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}$, write the oxidation and reduction half-reactions. (c) Name the oxidising agent and explain why a silver strip placed in zinc sulphate solution shows no change.

1 mark Oxidation is an increase in the oxidation number of an element; reduction is a decrease.
1 mark Oxidation: $\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$. Reduction: $\mathrm{2Ag^{+}(aq)+2e^{-}\rightarrow 2Ag(s)}$.
1 mark The oxidising agent is $\mathrm{Ag^{+}}$. Silver is far below zinc in the activity series, so it cannot give electrons to $\mathrm{Zn^{2+}}$: the reverse reaction does not occur.

E4. JEE Main NEET In which compound does the element shown have the lowest oxidation number? (A) N in $\mathrm{N_{2}H_{4}}$ (B) S in $\mathrm{S_{8}}$ (C) C in $\mathrm{CH_{3}OH}$ (D) P in $\mathrm{PH_{4}^{+}}$

Values (A) $2x+4=0$, $-2$. (B) $0$ (element). (C) $x+3+1-2=0$, $-2$. (D) $x+4=+1$, $-3$.
Answer (D), $-3$.

E5. JEE Advanced One or more options correct. Which statements are correct? (A) In $\mathrm{Fe_{3}O_{4}}$ the iron atoms have oxidation numbers $+2$ and $+3$ in the ratio $1:2$. (B) The two kinds of sulphur in $\mathrm{S_{4}O_{6}^{2-}}$ have oxidation numbers $+5$ and $0$. (C) The oxidation number of Cr in $\mathrm{CrO_{5}}$ is $+10$. (D) In $\mathrm{NH_{4}NO_{3}}$ the two nitrogen atoms have oxidation numbers $-3$ and $+5$.

(A) True: $\mathrm{FeO\cdot Fe_{2}O_{3}}$, one $+2$ and two $+3$ (average $+\tfrac83$).
(B) True: the terminal S atoms (bonded to three O) are $+5$, the two bridging S atoms are $0$.
(C) False: $\mathrm{CrO_{5}}$ has two peroxo groups (four O at $-1$) and one oxo O ($-2$), so Cr is $+6$; no element of group 6 can exceed $+6$.
(D) True: $\mathrm{NH_{4}^{+}}$ $-3$ and $\mathrm{NO_{3}^{-}}$ $+5$.
Answer (A), (B), (D).

E6. Board · Case study In a blast furnace iron ore is reduced by carbon monoxide: $\mathrm{Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(l)+3CO_{2}(g)}$. Some of the ore is magnetite, which reacts as $\mathrm{Fe_{3}O_{4}(s)+4CO(g)\rightarrow 3Fe(l)+4CO_{2}(g)}$. (i) Give the oxidation number of iron in each ore. (2 marks) (ii) Identify the reducing agent and the element it reduces. (1 mark) (iii) Show that in the magnetite reaction the total decrease in oxidation number equals the total increase. (1 mark)

(i) $\mathrm{Fe_{2}O_{3}}$: $2x-6=0$, $+3$. $\mathrm{Fe_{3}O_{4}}$: $3x-8=0$, average $+\tfrac83$ (one Fe(II), two Fe(III)).
(ii) CO (carbon $+2\rightarrow+4$) is the reducing agent; it reduces iron to the metal ($0$).
(iii) Decrease: $3\times\tfrac83=8$ (or $2+3+3$). Increase: $4\times(4-2)=8$ ✓.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Nitrogen moves between the air, the soil, rivers and living things in a cycle driven largely by bacteria. Which steps of the nitrogen cycle are oxidations and which are reductions — and which of them release the greenhouse gas nitrous oxide, $\mathrm{N_{2}O}$, from over-fertilised fields?

Why it matters. Fertiliser use roughly feeds half the world's population, but nitrate run-off causes algal blooms and "dead zones" in the Gulf of Mexico, the Baltic Sea and the Bay of Bengal, and $\mathrm{N_{2}O}$ is about 270 times as powerful a greenhouse gas as $\mathrm{CO_{2}}$ over a century. Agronomists in India, China and Europe now design fertiliser schedules around the redox chemistry of soil bacteria.

What to do.

  1. Assign the oxidation number of nitrogen in $\mathrm{N_{2}}$, $\mathrm{NH_{3}}$, $\mathrm{NH_{4}^{+}}$, $\mathrm{NO_{2}^{-}}$, $\mathrm{NO_{3}^{-}}$, $\mathrm{NO}$ and $\mathrm{N_{2}O}$, and place them on a vertical "oxidation-number ladder" from $-3$ to $+5$.
  2. Find out what happens in nitrogen fixation, nitrification (two steps), denitrification and ammonification, and draw each as an arrow on your ladder. Label each arrow "oxidation" or "reduction".
  3. For nitrification, $\mathrm{NH_{4}^{+}\rightarrow NO_{2}^{-}\rightarrow NO_{3}^{-}}$, find the number of electrons lost per nitrogen atom in each step.
  4. Which steps can produce $\mathrm{N_{2}O}$ as a by-product? Where does $\mathrm{N_{2}O}$ sit on your ladder?

Tools: Rule 7.2 and Rule 7.3, an environmental-science or soil-science review of the nitrogen cycle, a drawing program or squared paper.

Hand in: The labelled ladder diagram with electron counts, and a 150-word note explaining why waterlogged (oxygen-poor) soils lose nitrate to the air.

Going further. Nitrifying bacteria get their energy from oxidising ammonium with oxygen. Which substance is their "fuel" (reducing agent) and which their "oxidant"? Compare with the glucose and oxygen of your own respiration.

Instructor note: N is $0$ in $\mathrm{N_{2}}$, $-3$ in $\mathrm{NH_{3}}$ and $\mathrm{NH_{4}^{+}}$, $+3$ in $\mathrm{NO_{2}^{-}}$, $+5$ in $\mathrm{NO_{3}^{-}}$, $+2$ in NO, $+1$ in $\mathrm{N_{2}O}$. Fixation ($\mathrm{N_{2}\rightarrow NH_{3}}$, $0\rightarrow-3$) and denitrification ($\mathrm{NO_{3}^{-}\rightarrow NO_{2}^{-}\rightarrow NO\rightarrow N_{2}O\rightarrow N_{2}}$) are reductions; nitrification is oxidation, losing 6 electrons per N ($-3\rightarrow+3$) and then 2 ($+3\rightarrow+5$); ammonification (organic N, $-3$, to $\mathrm{NH_{4}^{+}}$) involves no change. $\mathrm{N_{2}O}$ ($+1$) leaks from incomplete denitrification and as a side product of nitrification. In waterlogged soil bacteria use nitrate instead of $\mathrm{O_{2}}$ as their oxidising agent. Going further: $\mathrm{NH_{4}^{+}}$ is the reducing agent and $\mathrm{O_{2}}$ the oxidising agent, as glucose and $\mathrm{O_{2}}$ are in respiration.

🧪 Self-check quiz

0/5 correct

Q1. In $\mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}$, the oxidising agent is

Q2. The oxidation number of S in $\mathrm{Na_{2}S_{4}O_{6}}$ (average) is

Q3. Hydrogen has oxidation number $-1$ in

Q4. Which reaction is not a redox reaction?

Q5. A metal M displaces copper from $\mathrm{CuSO_{4}}$ solution but does not displace zinc from $\mathrm{ZnSO_{4}}$ solution. M could be

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