Oxidation, reduction and redox equations

Oxidation and reduction

Metal oxides

Reactive metals will react with oxygen to produce metal oxides.

metal(s) + oxygen(g) metal oxide(s)
No doubt at some time in your science lessons you have held a piece of magnesium ribbon in a Bunsen flame and cautiously observed the bright flash from the burning magnesium.

magnesium burning in air

The product of this reaction is magnesium oxide. No doubt from your gcse chemistry you will recall that we defined oxidation as the addition of oxygen to a substance and reduction as the removal of oxygen from a compound, while these definitions of oxidation and reduction are useful they are by no means the only way in which we can talk about substances being oxidised and reduced.
In the example above the magnesium metal has had oxygen added to it to form the compound magnesium oxide, we would say that the magnesium has been oxidised. We can write a word equation to show this simple reaction:

magnesium(s) + oxygen(g) magnesium oxide(s)
and a symbolic equation for this reaction is:
2Mg + O2(g) 2MgO(s)
Now magnesium oxide is an ionic compound which contains Mg2+ and O2- ions. If we consider how these ions are formed:
Mg -2eMg2+
or the preferred way to show oxidation is with the electrons on the right hand side of the equation:
MgMg2+ + 2e
The magnesium atoms are losing 2 electrons to form magnesium ions (Mg2+). The addition of oxygen or the loss of electrons is also called oxidation. So in this example magnesium atoms have been oxidised to form magnesium ions (Mg2+). So if losing electrons is another way in which to define oxidation then if a substance gains electrons it will be reduced.
The oxygen atoms in this reaction have gained 2 electrons from the magnesium atoms to form an oxide ion (O2-). We can show this as:
O2e2O2-
or since oxygen is a diatomic gas we should really multiply this equation by x2 to give:
O24eO2-
The gain of electrons is called reduction and when a substance gains electrons we say it has been reduced.

Burning copper metal

copper metal burning If a square of copper metal is held with a pair of tongs in a hot Bunsen flame for about 30 seconds the shiny bronze coloured copper turns black. No flames or bright flashes are produced. The copper metal is reacting with the oxygen in the air, it is being oxidised. An equation for this reaction is:

copper(s) + oxygen(g) → copper oxide(s)

Copper oxide is an ionic compound which contains Cu2+ ions and O2- ions. The copper ions are formed when the copper atoms are oxidised, that is they lose 2 electrons to form copper ions:

CuCu2++ 2e
and the oxide ions are formed when the oxygen atoms gain 4 electrons from the copper atoms:
O24e2O2-
The oxygen atoms have been reduced (gain electrons) by the copper atoms to form oxide ions (O2-). A substance which donates electrons is called a reducing agent. In this example the copper atoms are the reducing agent, they supply or donate 2 electrons to the oxygen atoms. Confusing the reducing agent is oxidised when it reacts, that is when it loses or gives away electrons.
An oxidising agent is an electron acceptor, it will accept electrons from another substance and it in turn will be reduced. It is easy to see why some people can easily get these terms mixed up but with practice you will not make that mistake!

The equation for the oxidation of copper atoms:

CuCu2++ 2e
and the reduction of oxygen atom into oxide ions:
O24e 2O2-
These equations are called ion-electron half equations or simply half-equations for obvious reasons, they represent only half of the reaction taking place. To get the overall equation for the reaction we simply add together the two half equations. However the 2 equations need to balance in terms of atoms and also electrons lost and gained. The copper atoms lose 2 electrons but the oxygen atoms need to gain 4 electrons, so we need to x2 the half-equation for the oxidation of copper to balance off the electrons:
2Cu2Cu2++ 4e   oxidation
O2 +4e2O2-   reduction
2Cu + O22Cu2+O2-   overall equation

Example 2- The displacement reaction between copper sulfate solution and iron

displacement reaction between copper sulfate solution and 
an iron nail When an iron nail is placed in a beaker containing a solution of copper(II) sulfate a displacement reaction occurs, an equation for the reaction is shown below:
iron(s) + copper sulfate(aq)iron sulfate(aq) + copper(s)
Fe(s) + CuSO4(aq)Fe2+SO4(aq)2- + Cu
If we now write an ionic equation it may allow us to better understand what has been oxidised and what has been reduced in this redox reaction:
Fe + Cu2+ SO42-Fe2+SO42- + Cu
the sulfate ion, SO42- is unchanged in the reaction, it appears on both the reactant and product side of the equation and it is unchanged. It takes no part in the reaction and is only present to balance off the charges present. Ions like this which take no part in the reaction are called spectator ions. To simplify the equation above we can remove the spectator ions, this will give a clearer view of what is actually happening in the reaction:
Fe + Cu2+Fe2+ + Cu

We can break this down into 2 separate half equations. The iron atoms lose 2 electrons and are being oxidised to form iron ions (Fe2+). The half equation for this is:

FeFe2+ + 2e

While the copper ions (Cu2+) in the copper sulfate solution gain 2 electrons and are reduced to form metallic copper. This is the brown substance that coats the iron nail and collects in the bottom of the beaker in the image above. The half equation for this reduction reaction is:

Cu2+ + 2eCu

Example 3 - The reaction of sodium and chlorine

Sodium is an alkali metal with an electronic configuration of 1s22s22p63s1, so when it reacts it will lose its outer 3s1 electron and in the process be oxidised. Chlorine is a reactive halogen in group 7 with an electronic configuration of 1s22s22p63s23p5. To gain a np6 or noble gas electronic configuration chlorine only has to gain 1 electron, this means it will be looking to gain 1 electron in its reactions, this will make it an excellent oxidising agent and by contrast sodium will be looking to lose its outer 3s1 electron to gain a noble gas electronic configuration. This means sodium, like most metals, is looking to lose electrons. This means that metals are good reducing agents.
If a small piece of sodium metal is placed on a pile of sand in a flask filled with chlorine gas you might a violent reaction to occur, however the reaction is slow. To start the reaction a few drops of water from a pipette, as shown in the image below, are added to the sodium metal, a violent reaction between the sodium and chlorine gas then starts. The flask quickly fills with white fumes of sodium chloride.

sodium and chlorine react violently

A half equation for the oxidation of sodium is:

NaNa+ + e
and a half equation for the reduction of chlorine gas (Cl2) is:
Cl2 + 2e 2Cl-
Since chlorine is a diatomic gas it will require 2 moles of electrons to reduce it to form 2 moles of chloride ions (2Cl-) however the half equation for the oxidation of sodium only releases 1 mole of electrons. This means that in order to balance these two equations the half equation for the oxidation of sodium will have to be multiplied by x2.
oxidation half equation: 2Na 2Na+ + 2e
reduction half equation: Cl2 + 2e2Cl-
We can simply cancel out the electrons, as they appear on both sides of the equation to get the overall equation for the reaction:
overall equation: 2Na + Cl22NaCl

Key Points

Practice questions

Check your understanding - Redox reactions.

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