Balancing More Complex Redox Equations - The redox reaction between bromine and sulfite ions and separately the reduction of dichromate ions.
When brown bromine solution (bromine water) is added to colourless sodium sulfite solution, the bromine rapidly loses its colour and sulfite ions change into colourless sulfate ions. Sulfite ions are sometimes called sulfate (IV) ions and sulfate ions are sometimes called sulfate (VI) ions
These equations can be found in the data book and reversed as required.
The standard reduction potential series gives the following reaction involving bromine and bromide ions.
Br2 + 2e- ---> 2Br-
This half reaction describes the reaction of bromine into bromide ions and is an example of a reduction reaction as electrons are gained by bromine molecules. 2 moles of electrons are gained per mole of bromine that is reduced
However, the conversion of sulfite ions into sulfate ions can be used to illustrate how to write certain ion-electron equations involving 'complex' ions such as those ending in -ATE or -ITE.
SO32-(aq) ---> SO42-(aq)
This equation could not be balanced conventionally by putting appropriate numbers in front of formulae.
Three rules are used to balance this type of equation:
Check that elements other than oxygen are balanced. In this example, sulfur is balanced.
(If it was not balanced the equation would need to be balanced before applying the first rule below.)
1. Balance oxygen by adding molecules of water to the equation:
SO32-(aq) + H2O(l) ---> SO42-(aq)
2. Balance hydrogen by adding hydrogen ions to the equation:
SO32-(aq) + H2O(l) ---> SO42-(aq) + 2H+(aq)
3. Balance charge by adding electrons to the equation.
To determine how many electrons are needed, first add up the charges on each side of the equation separately.
On the left-hand side one sulfite ion carries 2 negative charges, while the water molecule is neutral. There are two negative charges on the left-hand side.
On the right-hand side one sulfate ion carries 2 negative charges and two hydrogen ions each have one positive charge giving two positive charges. There is no charge on the right-hand side.
There are two negative charges more on the left-hand side than on the tight-hand side, and these are balanced by adding two electrons to the right-hand side:
SO32-(aq) + H2O(l) ---> SO42-(aq) + 2H+(aq) + 2e-
This is the oxidation half reaction and it produces 2 moles of electrons per mole of dichromate ions.
As the same number of electrons are made in the oxidation half reaction that are used in the reduction half reaction, the two half reaction equation can be added to give the overall redox equation.
Br2 + SO32-(aq) + H2O(l) ---> 2Br- + SO42-(aq) + 2H+(aq)
A half reaction that causes problems sometimes is the conversion of dichromate ions into chromium(III) ions.
Cr2O72- ---> Cr3+
Before you start, you need to first balance the element chromium.
Cr2O72- ---> 2Cr3+
Chromium is now balanced. Let's balance oxygen as described above.
Cr2O72- ---> 2Cr3+ + 7H2O
Oxygen is now balanced. Let's balance hydrogen as described above.
Cr2O72-+ 14H+ ---> 2Cr3+ + 7H2O
Oxygen and hydrogen are now balanced. Let's balance charge as described above.
Cr2O72-+ 14H+ + 6e- ---> 2Cr3+ + 7H2O