redox equations

Redox equations

Aluminium (Al) and iodine (I2) react together in a spectacular reaction to produce the ionic compound aluminium iodide (AlI3). When powdered aluminium and iodine are mixed and placed on a watch glass little or no apparent reaction happens - a little disappointing! However if a few drops of water (which catalyses the reaction) is added to the mixture, then within about 20-30 seconds later a violent exothermic reaction happens. The mixture of aluminium and iodine react to produce an intense bright glow, lots of heat is released. However since iodine is a small covalent diatomic molecule lots of purple vapour is released as the heat of the reaction causes some of the iodine to sublime. The reaction is best carried out in a fume cupboard as lots of dense purple iodine vapour is released during the reaction.

reaction between aluminium powder and iodine

Ionic half-equations

Aluminium reacts with iodine to form aluminium iodide AlI3.
This is a redox reaction in which the aluminium atoms are oxidised to form aluminium ions:

Al(s)Al3+(s) + 3e

The normal convention when writing oxidation equations is that the electrons are written on the right hand side of the equation and NOT as shown below:

Al(s) -3eAl3+(s)

When writing half-equations it is important to ensure that not only do the numbers of atoms balance but that the charges on both sides of the equation balance. In the example above this is obvious, the charge on the aluminium atoms is 0 (since its an element) and on the product side the positive charges on the Al3+ ions are cancelled out by the negative charge on the 3 electrons.

The iodine atoms in this reaction accept electrons from the aluminium atoms and are reduced to form iodide ions. The reduction half-equation is usually written with the electrons on the left hand side of the equation:

I2(s) + 2e2I-(s)


To get the overall equation for the reaction we simply add the 2 half-equations together:

Al(s) Al3+(s) + 3e
I2(s) + 2e2I-(s)

However before we get to the overall equation we need to balance off the electrons being lost and gained. The aluminium atoms act as the reducing agent in this reaction and lose 3 electrons , the iodine molecules act as the oxidising agent and gain 2 electrons. However we need to ensure that the number of electrons lost by the aluminium equals the number of electrons gained by the iodine. The simplest way to balance these equations is the multiply equation 1 by 2 and equation 2 by 3. This will give 6 electrons in each equation:

2Al(s)2Al3+(s) + 6e
3I2(s) + 6e6I-(s)

The overall equation is gained by simply adding together the 2 half-equations, the electrons on each side of the equations will simply cancel out:

2Al + 3I22Al3+ + 6l-

Or simply:
2Al + 3I2 → 2AlI3

Metal acid reactions

Recall that all acids are solutions of H+(aq). Reactive metals such as magnesium will react with acids to produce a salt and water, for example magnesium reacts with hydrochloric acid to form:

Magnesium + hydrochloric acid → magnesium chloride + hydrogen
Mg + 2HCl → MgCl2 + H2
metals and acids react to produce a salt and hydrogen

This reaction is a redox reaction, the magnesium is the reducing agent and it will supply electrons to reduce the hydrogen ions present in the acid to hydrogen gas. The hydrogen ions are the oxidising agent in this reaction. The chloride ions are spectator ions, they take no part in the reaction and remain unchanged. They are simply present to balance off the charges present.

Oxidation half-equation: Mg → Mg2+ + 2e
reduction half-equation: 2H+ + 2e H2
Overall equation omitting spectator ions:
Mg + 2H+ → Mg2+ + H2(g)

Halogen displacement reactions

When bromine is added to a solution of potassium iodide a displacement reaction occurs. Here the more reactive bromine will displace or remove the less reactive halide ion, the iodide ion from the potassium iodide solution. The bromide ion will take the place of the iodide in the solution. If a little organic solvent is added then the displaced iodine will dissolve in this rather than staying in the aqueous solution. In an organic solvent such as cyclohexane the halogens appear as bright clear colours, for example iodine dissolves in cyclohexane to produce a bright clear purple coloured solution. This is shown in the image below:

2KI(aq) + Br2(aq) → 2KBr(aq) + I2(aq)

halogen displacement reaction betweem potassium iodide and bromine water

This reaction as well as being a displacement reaction is also a redox reaction. We can write two half-equations to show the oxidation and reduction reactions taking place. The potassium ions are spectator ions and can be ignored as they take no part in the reaction; they are only present to balance off the charges on the ions present in the equation:

Oxidation reaction: 2I- I2 + 2e
reduction reaction: Br2 + 2e → 2Br-
The overall equation for the reaction can be found by simply cancelling out the electrons which appear on opposite sides of the 2 half-equations
overall equation: 2I- + Br2 I2 + 2Br-

Common oxidising agents

As already mentioned an oxidising agent is an electron acceptor, it will oxidise a substance when it reacts and it will be reduced by gaining electrons. Common oxidising agents you are likely to meet are listed below with half equations to show the reduction reactions of oxidising agents:

Oxidising agent Half-equation
chlorine Cl2(g) + 2e → 2Cl-(aq)
bromine Br2(g) + 2e → 2Br-(aq)
iodine I2(g) + 2e → 2I-(aq)
oxygen O2(g) + 4e → 2O2-(aq)
There are other common oxidising agents such as acidified potassium permanganate, acidified potassium dichromate, hydrogen peroxide and concentrated sulfuric acid. The half equations for these reactions are a little more complicated but easy to work out if you follow some simple rules. However before we look at these equations we need to discuss oxidation numbers.

Key Points

Practice questions

Check your understanding - Questions on redox reactions