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Half equations are used to identify and represent redox
The atoms and charges
must be balanced
are used to balance
the charges on the two sides of the equation.
They can be combined
to give an overall picture of a reaction or left as individual equations to illustrate the oxidation or the reduction steps separately.
Half equations are empirical
equations. When they are written as a pair, the minimum stoichiometric
values should be used. These are then scaled accordingly when combining them.
Consider for example the reaction between magnesium and hydrochloric acid.
The molecular equation is:
Mg + 2HCl à MgCl2 + H2
The ionic equation is:
Mg + 2H+ + 2Cl- à Mg2+ + 2Cl- + H2
So you see straight away that Cl- is not actually involved since it is present on both sides. The actual reaction would therefore be:
Mg + 2H+ à Mg2+ + H2
Now to redox. It is evident that the magnesium has lost electrons since neutral Mg atoms have become Mg2+ cations.
The half equation for this is:
Mg à Mg2+ + 2e-
Electrons have been released (Mg oxidised).
The other species that has changed is the H+. The + charge has been removed therefore an electron has been added i.e. reduction. The half equation is:
2H+ + 2e- à H2
When writing the half equations, you deal only with the atoms and ions that are reacting.
The process can also be reversed, you can work back from the half equations and construct the ionic equation.
The link is the number of electrons.
See how it works using this example.
(i) Al à Al3+ + 3e-
(ii) 2H+ + 2e- à H2
To combine these equations, the link is the electrons. You would need 3 lots of equation (ii) to 2 lots of equation (i). 6e- would be involved in each case. Thus:
6H+ + 2Al à 2Al3+ + 3H2
Note that atoms and charges balance. As a check, work out the ionic equation (minus spectator ions) from the (unbalanced) molecular equation:
Al + HCl à AlCl3 + H2
Write half equations for the following:
The formation of silver halides from KF, KCl, KBr and KI.
The displacement reactions for the following pairs of chemicals Zn/CuSO4; Li/Pb(NO3)2; Fe/AgNO3; Al/FeSO4; Mg/ZnSO4
For each of the above examples (displacements), explain, with the aid of suitable equations
, why you have ignored certain ions in the half equations.
Potassium dichromate (K2Cr2O7) acts as an oxidising agent in acidic conditions. In a similar way to the manganate(VII) ion, Cr3+ is produced. Write a balanced half equation to show what is happening.
Sn2+ can be oxidised to Sn4+ in acidic conditions. Write a half equation and use it, and your previous answer, to deduce the number of moles of potassium dichromate that would oxidise 0.25 moles of a tin(II) salt.