Balance the following chemical equation by the oxidation number method.
$KI + H_{2} {SO}_{4} \to K_{2} {SO}_{4} + I_{2} + {SO}_{2} + H_{2} O$

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Solution

Step 1: Identify atoms which undergo oxidation and reduction:
The reaction of potassium iodide with sulphuric acid gives potassium sulphate iodine and sulphur dioxide gas.
$KI (aq) + H_{2} {SO}_{4} (aq) \to K_{2} {SO}_{4} (aq) + I_{2} (g) + {SO}_{2} (g) + H_{2} O (l)$
Step 2: Write the oxidation state of each atom:
$+ 1 - 1 + 2 + 6 - 8 + 2 + 6 - 8 0 + 6 - 8 + 2 - 2 K I (aq) + H_{2} S O_{4} (aq) \to K_{2} S O_{4} (aq) + I_{2} (g) + S O_{2} (g) + H_{2} O (l)$
Step 3: Check for the oxidation and reduction reaction:
The increase in the oxidation number indicates an oxidation reaction and the decrease in the oxidation state indicates a reducing reaction.
In the reaction, the potassium oxidation number increases on the product side. So, it undergoes an oxidation reaction.
There is a decrease in the oxidation number in the case of iodine, so it undergoes a reduction reaction.
Step 4: Balancing charge by multiplication:
The potassium needs to be balanced when it is cross multiplied, the whole equation gets balanced i.e

$+ 1 - 1 + 2 + 6 - 8 + 2 + 6 - 8 0 + 6 - 8 + 2 - 2 K I (aq) + H_{2} S O_{4} (aq) \to K_{2} S O_{4} (aq) + I_{2} (g) + S O_{2} (g) + H_{2} O (l) ⋮ ⋮ + 1 + 2 + 2 - 2 + 2 + 6 - 8 + 2 + 6 - 8 0 + 6 - 8 + 2 - 2 2 K I (aq) + H_{2} S O_{4} (aq) \to K_{2} S O_{4} (aq) + I_{2} (g) + S O_{2} (g) + H_{2} O (l)$
Step 5: Balancing other atoms:
On balancing the other atoms, the whole equation is balanced,
$2 K I (aq) + H_{2} S O_{4} (aq) \to K_{2} S O_{4} (aq) + I_{2} (g) + S O_{2} (g) + H_{2} O (l) ⋮ ⋮ ⋮ 4 Oxygen 7 oxygen 2 K I (aq) + 2 H_{2} S O_{4} (aq) \to K_{2} S O_{4} (aq) + I_{2} (g) + S O_{2} (g) + 2 H_{2} O (l) ⋮ ⋮ 8 oxygen 8 oxygen$
Hence, the balanced equation is
$2 K I (aq) + 2 H_{2} S O_{4} (aq) \to K_{2} S O_{4} (aq) + I_{2} (g) + S O_{2} (g) + 2 H_{2} O (l)$

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