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Oxidation is losing electrons and reduction gaining them; they always happen together, and the electrons lost equal the electrons gained.
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By the end of this lesson you will be able to identify oxidation and reduction from changes in oxidation number, name the oxidising and reducing agents, balance a redox equation by matching electrons, balance a half equation in acid with hydrogen ions and water, count the electrons an oxidant takes, and use a redox ratio in a titration calculation.
Last lesson you assigned oxidation numbers, and in Chemistry 1 you balanced half equations with electrons. The acids unit followed protons from donor to acceptor. Redox follows electrons the same way: one species gives them, another takes them, and the numbers must match.
Oxidation is loss of electrons; the oxidation number of the atom rises. Reduction is gain of electrons; the oxidation number falls. A redox reaction is one in which both happen. An oxidising agent (oxidant) takes electrons from another species and is itself reduced; a reducing agent (reductant) gives electrons to another species and is itself oxidised. A half equation shows one half of the transfer with the electrons written in.
Drop a strip of zinc into blue copper(II) sulfate solution. The blue fades, and the zinc grows a brown coat of copper:
$$\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$$
No oxygen is gained or lost by anything, yet this is a textbook redox reaction. Follow the oxidation numbers: zinc goes from $0$ to $+2$, so each atom loses two electrons — it is oxidised. Copper goes from $+2$ to $0$, so each ion gains two electrons — it is reduced. The two half equations show it:
$$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-} \qquad \mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$$
Electrons do not appear from nowhere or vanish, so oxidation and reduction always happen together, and the electrons lost equal the electrons gained. That is the rule for combining half equations: multiply each until the electrons match, add, and cancel the electrons. For aluminium and copper(II) ions, aluminium loses three and copper gains two, so six electrons must move: $\mathrm{2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu}$.
The agents are named for what they do to the other species. Zinc gives copper(II) ions electrons, reducing them, so zinc is the reducing agent — and in doing so zinc is oxidised. Copper(II) ions take electrons from zinc, oxidising it, so they are the oxidising agent — and they are reduced. Every agent undergoes the opposite of its name.
Another way: picture
A mnemonic many chemists use is OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons. For the agents, think of a travel agent, who arranges your journey but does not travel: the oxidising agent arranges the oxidation of something else and is itself reduced.
Another way: steps
To analyse a redox reaction:
Many oxidising agents contain oxygen that becomes water when they are reduced, and they work in acid. Their half equations are balanced in a fixed order. For manganate(VII) becoming manganese(II):
$$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$$
The five electrons match the fall in manganese's oxidation number from $+7$ to $+2$ — a useful check. Combined with $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$ multiplied by five, it gives the equation behind the manganate(VII) titration: one manganate(VII) ion oxidises five iron(II) ions. The purple manganate(VII) turns colourless as it is reduced, so it is its own indicator: the end point is the first drop that stays purple.
Oxidation was first named for reactions with oxygen: metals rusting, fuels burning. When magnesium burns in oxygen it becomes magnesium ions, losing two electrons each, and the oxygen atoms gain them to become oxide ions. The electron definition includes all of those reactions and many more: magnesium burning in chlorine is the same change for magnesium, with chlorine taking the electrons. Gaining oxygen or losing hydrogen are useful signs of oxidation in organic chemistry, but the definition that always works is electrons, tracked by oxidation numbers.
Not every reaction is redox. In neutralisation, precipitation and the decomposition of carbonates, every element keeps its oxidation number. Protons and ions move, but no electrons change hands.
One element can even be oxidised and reduced in the same reaction. When chlorine dissolves in cold sodium hydroxide solution to make household bleach, $\mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O}$, one chlorine atom falls from $0$ to $-1$ and the other rises from $0$ to $+1$. This is called disproportionation, and it is still one electron lost for one gained — the same bookkeeping, with both halves happening to the same element.
Oxidation means gaining oxygen. That is one case. Oxidation is losing electrons, with or without oxygen.
The oxidising agent is oxidised. It is reduced; it oxidises the other species.
Oxidation can happen on its own. The electrons have to go somewhere, so something else is always reduced at the same time.
Balance atoms and forget the electrons. In an ionic redox equation the charge must balance too, which only happens when electrons lost equal electrons gained.
Every reaction with ions is redox. Neutralisation and precipitation move ions without changing any oxidation number.
Chlorine gas is bubbled into iron(II) chloride solution: $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$ and $\mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-}$.
One half loses one electron; the other gains two.
Double the iron half: $\mathrm{2Fe^{2+} \rightarrow 2Fe^{3+} + 2e^-}$.
Now two electrons are lost and two gained.
Add and cancel: $\mathrm{2Fe^{2+} + Cl_2 \rightarrow 2Fe^{3+} + 2Cl^-}$. Charge: $+4$ each side.
Chlorine is the oxidising agent; iron(II) is the reducing agent.
How many moles of electrons does $0.020$ mol of dichromate(VI), $\mathrm{Cr_2O_7^{2-}}$, gain on becoming $\mathrm{Cr^{3+}}$?
Two chromium atoms per ion.
Each chromium falls from $+6$ to $+3$, three electrons; two chromiums, six electrons per ion.
Fall in oxidation number times the number of atoms.
$0.020 \times 6 = 0.12$ mol of electrons.
Scale by the amount.
Copper goes from $0$ to $+2$, so it loses electrons and is oxidised.
Oxidation number rises.
Silver ions go from $+1$ to $0$, so they are reduced, which makes them the $\ldots$
…oxidising agent: the silver ions take the electrons that copper loses, two for each copper atom.
For the reaction $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$, name the species for each role.
| species | |
|---|---|
| species oxidised | |
| species reduced | |
| oxidising agent | |
| reducing agent |
Balance the ionic equation for aluminium displacing copper from copper(II) ions. The atoms and the total charge must both balance.
This task has no paper form; do it on a device.
Balance the half equation for nitrate ions being reduced to nitrogen monoxide in acid solution. Hydrogen ions and water are supplied, and electrons appear as a species.
This task has no paper form; do it on a device.
Magnesium ribbon burns in chlorine to form a white ionic solid. No oxygen is involved. Is the magnesium oxidised?
How many moles of electrons are gained when $0.04$ mol of $\mathrm{NO_3^-}$ is reduced to $\mathrm{NO}$?
Answer: mol
Sort each reaction: is it a redox reaction or not?
| redox | not redox | |
|---|---|---|
| zinc powder added to copper(II) sulfate solution | ||
| hydrochloric acid neutralised by sodium hydroxide | ||
| hydrogen burning in oxygen to make water | ||
| silver nitrate and sodium chloride solutions giving a white precipitate | ||
| calcium carbonate heated to calcium oxide and carbon dioxide | ||
| magnesium fizzing in hydrochloric acid |
An iron supplement tablet is dissolved in dilute sulfuric acid and titrated with $0.0100$ mol/L potassium manganate(VII). The purple colour stops disappearing after $15.00$ mL. In the reaction each $\mathrm{MnO_4^-}$ gains five electrons and each $\mathrm{Fe^{2+}}$ loses one. What mass of iron was in the tablet? (Iron: $56$ g/mol.)
Answer: unit: g / kg / mg
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In the reaction $\mathrm{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$, which species is the oxidising agent?
You can follow electrons through a redox reaction and name both agents. Say out loud why the oxidising agent is the species that is reduced. Next: separating the two halves in space, so the electrons travel through a wire — electrochemical cells.
10. Your turn: in $\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}$, which species is oxidised and which is the oxidising agent?, step 3