Reduction and Oxidation Reactions

Lessons

Senior 4 · Chemistry

Reduction and Oxidation Reactions

5 lessons

Lesson 1 of 5

Oxidation, Reduction and Electrochemistry

30 min

A silver tree on a copper wire

Hang a coil of bare copper wire in a colourless solution of silver nitrate. Within minutes the wire grows a fur of glittering silver crystals, and the solution slowly turns blue. Nothing has been added but a metal and a salt, yet two new substances have appeared: silver metal and copper(II) ions. Each copper atom has given two electrons, and two silver ions have taken them. Reactions in which electrons pass from one substance to another are called redox reactions. They rust iron, burn fuel, release energy from food in our cells, and, when the electrons are made to travel through a wire, run every battery. This lesson defines oxidation and reduction and shows how redox reactions and electricity are connected.

Definition

Oxidation: loss of electrons (also: gain of oxygen, or loss of hydrogen). Example: $\mathrm{Cu \rightarrow Cu^{2+} + 2e^-}$ .

Reduction: gain of electrons (also: loss of oxygen, or gain of hydrogen). Example: $\mathrm{Ag^+ + e^- \rightarrow Ag}$ .

Redox reaction: a reaction in which oxidation and reduction happen together; the electrons lost by one species are gained by another.

Half-equation: an equation showing either the oxidation or the reduction part of a redox reaction, with electrons written in.

Electrochemistry: the study of how redox reactions produce electricity (in galvanic, or voltaic, cells) and how electricity drives redox reactions (in electrolytic cells).

Three definitions, one idea

Chemists first described oxidation as combining with oxygen. The definition was widened twice, and the electron definition includes the other two.

Definition	Oxidation is	Reduction is	Example
Oxygen	gain of oxygen	loss of oxygen	$\mathrm{CuO + H_2 \rightarrow Cu + H_2O}$ : CuO is reduced, $\mathrm{H_2}$ is oxidised
Hydrogen	loss of hydrogen	gain of hydrogen	$\mathrm{H_2S + Cl_2 \rightarrow 2HCl + S}$ : $\mathrm{H_2S}$ is oxidised, $\mathrm{Cl_2}$ is reduced
Electrons	loss of electrons	gain of electrons	$\mathrm{2Na + Cl_2 \rightarrow 2NaCl}$ : Na loses $e^{-}$ (oxidised), Cl gains $e^{-}$ (reduced)

The sodium–chlorine reaction involves no oxygen or hydrogen at all, yet it is plainly the same kind of change: only the electron definition covers it.

Writing half-equations

For copper in silver nitrate, the nitrate ions take no part (they are spectator ions). Split the ionic equation into its two halves and balance the charge with electrons:

\text{oxidation: } \mathrm{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-}

\text{reduction: } \mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)} \quad (\times 2)

The electrons lost must equal the electrons gained, so the reduction is doubled before adding: $\mathrm{Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}$ Electrons never appear in the overall equation.

Making the electrons go through a wire

If zinc is dipped straight into copper(II) sulfate solution, electrons jump from zinc atoms to copper(II) ions on the metal surface and the energy is released as heat. In an electrochemical (galvanic) cell the two half-reactions are separated into two half-cells, so the electrons must travel through an external wire, where they can do useful work (Figure 16.1).

Figure 16.1: A zinc–copper (Daniell) cell. Oxidation happens at the anode, reduction at the cathode; the salt bridge lets ions move between the solutions so that each stays electrically neutral.

The electrode where oxidation happens is the anode; where reduction happens is the cathode. This is true in every kind of cell.
In a galvanic cell the anode is the negative electrode, because electrons are released there.
Under standard conditions this cell gives 1.10 V. It can be predicted from the standard electrode potentials of the two half-cells (Lesson 5): $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (+0.34) - (-0.76) = +1.10\ \mathrm{V}$ .

In electrolysis, which you met in Senior 3, the reverse happens: an external supply forces electrons into one electrode and out of the other, driving a redox reaction that would not happen by itself. For molten sodium chloride, the cathode reduces sodium ions and the anode oxidises chloride ions: $\text{cathode: } \mathrm{Na^+ + e^- \rightarrow Na} \qquad \text{anode: } \mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$

Worked example 1: zinc in copper(II) sulfate

Zinc powder is stirred into blue copper(II) sulfate solution. The blue colour fades, a pink-brown solid forms and the mixture warms up. Write the ionic equation and the two half-equations, and identify what is oxidised and what is reduced.

Solution. Sulfate ions are spectators. $\text{oxidation: } \mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-} \qquad \text{reduction: } \mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$ Each half transfers 2 electrons, so they add directly: $\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$ Zinc is oxidised (loses electrons); copper(II) ions are reduced (gain electrons). The blue $\mathrm{Cu^{2+}}$ ions are used up, the pink-brown solid is copper, and the energy is released as heat because the electrons transfer directly.

Worked example 2: counting the electrons

In the silver-tree experiment, 0.635 g of copper dissolves. Calculate (a) the mass of silver deposited and (b) the number of electrons transferred. (Cu = 63.55, Ag = 107.87, $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$ )

Solution. $n(\mathrm{Cu}) = \frac{0.635}{63.55} = 9.99 \times 10^{-3}\ \mathrm{mol}$ (a) From $\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}$ : $n(\mathrm{Ag}) = 2 \times 9.99 \times 10^{-3} = 2.00 \times 10^{-2}\ \mathrm{mol}$ ; mass $= 2.00 \times 10^{-2} \times 107.87 = 2.16\ \mathrm{g}$ .

(b) Each Cu atom loses 2 electrons: $n(e^-) = 2.00 \times 10^{-2}\ \mathrm{mol}$ , so the number of electrons $= 2.00 \times 10^{-2} \times 6.022 \times 10^{23} = 1.20 \times 10^{22}$ .

Warning

Common mistakes. (1) Thinking oxidation can happen alone: every oxidation is matched by a reduction. (2) Writing electrons on the wrong side: in an oxidation half-equation they are products; in a reduction they are reactants. (3) Adding half-equations without first making the electrons equal. (4) Saying "the anode is always positive": in a galvanic cell it is negative; in electrolysis it is positive. What never changes is that oxidation happens at the anode.

Safety. Silver nitrate solution is corrosive, stains skin black and is toxic to aquatic life: wear gloves and goggles and collect silver residues for recovery, not down the sink. Copper(II) sulfate is harmful if swallowed and irritates eyes. Wash your hands after the practical.

Tip

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). AN OX, RED CAT: oxidation at the ANode, reduction at the CAThode, in every cell.

Raquel Yoong: Displacement Reaction of Metals, Zinc in Copper(II) Sulfate (6 min). Watch the copper appear on the zinc and the explanation of the electron transfer at the level of particles.

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Practice

Define oxidation and reduction in terms of electrons.
In $\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}$ , state, using oxygen transfer, which substance is oxidised and which is reduced.
Write half-equations for the reaction $\mathrm{Mg(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2(g)}$ .
Combine these half-equations into an ionic equation: $\mathrm{Al \rightarrow Al^{3+} + 3e^-}$ and $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$ .
In the Daniell cell (Figure 16.1), state the direction of electron flow in the wire, and name the anode and the cathode.
Explain why the salt bridge is needed.
In the electrolysis of molten lead(II) bromide, write the half-equation at each electrode and say which is oxidation.
0.327 g of zinc reacts completely with excess copper(II) sulfate solution. Calculate the mass of copper formed and the number of moles of electrons transferred (Zn = 65.38, Cu = 63.55).

Show answersHide answers

Oxidation is loss of electrons; reduction is gain of electrons.
CO gains oxygen, so it is oxidised; $\mathrm{Fe_2O_3}$ loses oxygen, so it is reduced.
$\mathrm{Mg \rightarrow Mg^{2+} + 2e^-}$ (oxidation); $\mathrm{2H^+ + 2e^- \rightarrow H_2}$ (reduction).
Multiply to 6 electrons: $\mathrm{2Al \rightarrow 2Al^{3+} + 6e^-}$ ; $\mathrm{3Cu^{2+} + 6e^- \rightarrow 3Cu}$ ; overall $\mathrm{2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu}$ .
From zinc to copper. Zinc is the anode (oxidation, negative); copper is the cathode (reduction, positive).
It completes the circuit by letting ions move between the half-cells, so that the solutions stay electrically neutral as $\mathrm{Zn^{2+}}$ forms in one and $\mathrm{Cu^{2+}}$ is removed from the other; without it the current stops.
Cathode: $\mathrm{Pb^{2+} + 2e^- \rightarrow Pb}$ (reduction); anode: $\mathrm{2Br^- \rightarrow Br_2 + 2e^-}$ (oxidation).
$n(\mathrm{Zn}) = \frac{0.327}{65.38} = 5.00 \times 10^{-3}\ \mathrm{mol}$ = n(Cu); mass of Cu $= 5.00 \times 10^{-3} \times 63.55 = 0.318\ \mathrm{g}$ ; electrons $= 2 \times 5.00 \times 10^{-3} = 1.00 \times 10^{-2}\ \mathrm{mol}$ .

Quick check

In Cu + 2Ag⁺ → Cu²⁺ + 2Ag, which species is oxidised?

Key points

Oxidation is loss of electrons (or gain of O, loss of H); reduction is gain of electrons (or loss of O, gain of H). They always occur together in a redox reaction.
A redox reaction can be split into half-equations; electrons lost must equal electrons gained before the halves are added.
In a galvanic cell a spontaneous redox reaction drives electrons through a wire (Zn–Cu: 1.10 V); in electrolysis electricity drives a non-spontaneous one.
Oxidation always happens at the anode and reduction at the cathode.

Sources and credits

Image: Arbre de Diane (début de réaction) 03.jpg by ΛΦΠ, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Image: Daniell cell demonstration.jpg by Taweetham, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Electrode potentials from Standard electrode potential (data page), Wikipedia contributors, licensed under CC BY-SA 4.0.
Video: [4K] Displacement Reaction of Metals - Zinc in Copper (II) Sulfate - with explanation at micro level, Raquel Yoong, embedded from YouTube.
The text, table, diagram, worked examples and Practice items were written and drawn by Ganzaa.
This lesson uses CC BY-SA material and is shared under CC BY-SA 4.0.

1A silver tree on a copper wire

2Three definitions, one idea

3Writing half-equations

4Making the electrons go through a wire

A silver tree on a copper wire

Three definitions, one idea

Writing half-equations

Making the electrons go through a wire