ISSR CLASSES
Checkpoint Science
iGCSE Chemistry
iGCSE Physics
iGCSE Coordinated Science
A-Level Chemistry
PRACTICALS
Practicals Home
Practicals A-Z
Study Plan for Practical Work
|
Cl2(aq) + 2KI(aq) ⇒ I2(aq) + 2KCl(aq) |
First split the ionic solutions into their constituent ions. Don't be tempted to try and put a negative charge on the halogens. The halogens are neutral molecules. The potassium salts, however, are ionic salts made up of positive and negative ions. No charge is seen in the above equation as the charge on the ions cancels out. When we split the compound up into its ions the charges on the ions are then shown:
|
Cl2(aq) + 2K+(aq) + 2I-(aq) ⇒ I2(aq) + 2K+(aq) + 2Cl-(aq) |
Note that 2KI contains 2K+ ions and 2I- ions. You should see that 2K+(aq) appears on both sides of the equation. The K+ is the spectator ion. It is done nothing. It is simply cancelled out to leave the balanced ionic equation:
|
Cl2(aq) + 2I-(aq) ⇒ I2(aq) + 2Cl-(aq) |
Using the same method:
|
Cl2(aq) + 2KBr(aq) ⇒ Br2(aq) + 2KCl(aq) |
becomes:
|
Cl2(aq) + 2Br-(aq) ⇒ Br2(aq) + 2Cl-(aq) |
and
|
Br2(aq) + 2KI(aq) ⇒ I2(aq) + 2KBr(aq) |
becomes:
|
Br2(aq) + 2I-(aq) ⇒ I2(aq) + 2Br-(aq) |
We can split these ionic equations into two half equations. The half equations show that one of the reactants has gained electrons and one of them has lost electrons. Using the last example:
|
Br2(aq) + 2I-(aq) ⇒ I2(aq) + 2Br-(aq) |
|
Br2(aq) + 2e- ⇒ 2Br-(aq) |
|
2I-(aq) ⇒ I2(aq) + 2e- |
You should remember that:
|
Oxidation is loss of electrons and reduction is gain of electrons |
This shows that the halogen (Br2 in this case) gains electrons to become the negative halide ion - it is reduced. The halide ions (I- in this case) loses electrons to become the halogen molecule - they are oxidized.