Question

Balance the following equations:
 $ I{O_4}^ - + {I^ - } + {H^ + } \to {I_2} + {H_2}O $
(a) $ 2I{O_4}^ - + 14{I^ - } + 16{H^ + } \to 8{I_2} + 8{H_2}O $
(b) $ 2I{O_4}^ - + 14{I^ - } + 4{H^ + } \to 3{I_2} + 8{H_2}O $
(c) $ I{O_4}^ - + 7{I^ - } + 8{H^ + } \to 4{I_2} + 4{H_2}O $
(d) None of the above.

Answer 1

Hint: In these reactions, the oxidation number of the reactants changes. So, the reaction is a Redox reaction and as $ {H^ + } $ is present in the reactant side, the reaction is in an acidic medium. So, we have to balance the redox reaction.

Complete step-by-step answer
We will start with the definition of balanced equation:
Balanced Equation: The equation in which the number of atoms in the reactants side are equal to the number of atoms in the product side is called the balanced chemical equation.
To balance a given redox chemical reaction:
 $ I{O_4}^ - + {I^ - } + {H^ + } \to {I_2} + {H_2}O $
First we have to see that the reaction is in which medium, as $ {H^ + } $ is present in the reaction. It’s a redox reaction in an acidic medium.
This equation is a redox reaction in acidic medium,
Now, we will balance all the atoms except hydrogen and oxygen,
All the atoms other than hydrogen and oxygen are balanced already.
Now, we talk about the change in oxidation number:
The one iodine in the reactant has changed the oxidation number from $ - 1 $ to $ 0 $ . Hence, the change in oxidation number is $ 1 $ .
Similarly, the other iodine in the reactant has changed the oxidation number from $ 7 $ to $ 0 $ . Hence, the change in oxidation number is $ 7 $ .
Now, we balance the charges:
The increase in the oxidation number is balanced by multiplying with the decrease in charge, i.e, $ 7 $ is multiplied with $ {I^ - } $ and $ {I_2} $ with $ 4 $ .
  $ I{O_4}^ - + 7{I^ - } + {H^ + } \to 4{I_2} + {H_2}O $
Now, all the iodine are balanced.
Oxygen atoms are balanced with adding water molecules in the product side:
 $ I{O_4}^ - + 7{I^ - } + {H^ + } \to 4{I_2} + 4{H_2}O $
Now, the oxygen atoms are balanced.
Now, we have to balance the hydrogen atoms,
So, by adding $ {H^ + } $ on the reactants side, the hydrogen atoms are also balanced.
 $ I{O_4}^ - + 7{I^ - } + 8{H^ + } \to 4{I_2} + 4{H_2}O $
Hence, the reaction is now balanced.
The complete balanced equation is:
 $ I{O_4}^ - + 7{I^ - } + 8{H^ + } \to 4{I_2} + 4{H_2}O $
Hence, the correct option is (c) $ I{O_4}^ - + 7{I^ - } + 8{H^ + } \to 4{I_2} + 4{H_2}O $ .

Note
While balancing the redox reactions, we have to first balance the charges and then the atoms other than hydrogen and oxygen and at the end the atoms of the oxygen and hydrogen. There are different methods of balancing redox reactions like the ion electron method.