Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 6- B:

The method of half-reaction in acid aqueous solution

We consider the following redox-reaction that we wish to balance in an acid solution:

MnO 4 - (aq) + Fe2+(aq)  Fe3+(aq) + Mn2+(aq)

This reaction is often used to analyse the contents of iron in iron ore. The first step is to identify and write

the reaction equations for the half-reaction. First we write the half-reactions for the oxidation reaction. It is

clear from the overall reaction that it is the iron ion that is oxidised which is why the oxidation equation

becomes:

Fe2+(aq)  Fe3+(aq) (oxidation)

Similarly the reductions half reaction is associated with the manganese-ion and is written as follows:

MnO 4 - (aq)  Mn2+(aq) (reduction)

The next step is to balance each of the half-reactions in order to match the charge on each side. The

equations are balanced in terms of electrons:

Fe2+ Fe3+ + e-(oxidation)

MnO 4 - + 5 e- Mn2+ (reduction)

Now the charges match on each side of the half-reactions. In order to make the equations ready for

addition the oxidation reaction should by multiplied with ”5” as the reduction reaction involves 5 electrons

while the oxidation only involves 1 electron. We thereby gets:

5 Fe2+ 5 Fe3+ + 5 e-(oxidation)

MnO 4 - + 5 e- Mn2+ (reduction)

Now the equations are added:

5 Fe2+ + MnO 4 -  Mn2+ + 5 Fe3+

The next step is to balance the reaction in order to match the charges on an overall basis. When we have an

acid solution we balance with H+-ions while we balance with OH- in basic solutions. The charges on both

sides of the reaction arrow are calculated as follows:

5×(+2) + (-1)  (+2) + 5×(+3) =

+9  +17

We thereby have to place 8 H+-ions on the left side in order to make sure that there is the same number of

charges of on the left and the right side:

Electrochemistry