Question

Balance the following redox reaction in the basic medium.
$C{l_2} + IO_3^ - + O{H^ - } \to IO_4^ - + C{l^ - }$
A. $C{l_2} + IO_3^ - + 3O{H^ - } \to IO_4^ - + 5C{l^ - } + {H_2}O$
B. $C{l_2} + IO_3^ - + 2O{H^ - } \to IO_4^ - + 2C{l^ - } + {H_2}O$
C. $C{l_2} + 2IO_3^ - + 2O{H^ - } \to IO_4^ - + 5C{l^ - } + {H_2}O$
D. None of these

Answer 1

Hint: When an ion loses electron, then there is an increase in its oxidation state and it is termed as oxidation and when an ion gains electron, then there is decrease in its oxidation state and it is known as reduction. A chemical reaction in which the reduction and oxidation of ions occur in the same reaction simultaneously is known as a redox reaction.

Complete answer: As per question, the given reaction is as follows:
$C{l_2} + IO_3^ - + O{H^ - } \to IO_4^ - + C{l^ - }$
Steps to balance the given reaction in basic medium are as follows:
Step-1: Write oxidation half reaction and reduction half reaction separately for the given redox reaction.
Oxidation half reaction: $IO_3^ - \to IO_4^ - $
Reduction half reaction: $C{l_2} \to C{l^ - }$
Step-2: Except oxygen and hydrogen, balance all other elements in both oxidation and reduction half reaction i.e., the number of atoms of element must be equal in product as well as reactant.
Oxidation half reaction: $IO_3^ - \to IO_4^ - $
Reduction half reaction: $C{l_2} \to 2C{l^ - }$
Step-3: Add a water molecule to balance the oxygen atoms in the required half reaction.
Oxidation half reaction: $IO_3^ - + {H_2}O \to IO_4^ - $
Reduction half reaction: $C{l_2} \to 2C{l^ - }$
Step-4: Balance the number of hydrogen atoms by adding protons in the required half reaction.
Oxidation half reaction: $IO_3^ - + {H_2}O \to IO_4^ - + 2{H^ + }$
Reduction half reaction: $C{l_2} \to 2C{l^ - }$
Step-5: Balance the charge by adding the required number of electrons in both half reactions.
Oxidation half reaction: $IO_3^ - + {H_2}O \to IO_4^ - + 2{H^ + } + 2{e^ - }$
Reduction half reaction: $C{l_2} + 2{e^ - } \to 2C{l^ - }$
Step-6: Combine both half reactions and cancel the electrons in the reaction
$IO_3^ - + {H_2}O + C{l_2} + 2{e^ - } \to 2C{l^ - } + IO_4^ - + 2{H^ + } + 2{e^ - }$
On cancelling electrons, reaction will be as follows:
$IO_3^ - + {H_2}O + C{l_2} \to 2C{l^ - } + IO_4^ - + 2{H^ + }$
Step-7: Add hydroxide ions to balance the hydrogen ions or protons in the reaction. As in the reaction, two protons are present, so two hydroxide ions will be added to reactants as well as product. The reaction is as follows:
$IO_3^ - + {H_2}O + C{l_2} + 2O{H^ - } \to 2C{l^ - } + IO_4^ - + 2{H^ + } + 2O{H^ - }$
Step-8: ${H^ + }$ and $O{H^ - }$ ions combine to form water molecules. The reaction is as follows:
$IO_3^ - + {H_2}O + C{l_2} + 2O{H^ - } \to 2C{l^ - } + IO_4^ - + 2{H_2}O$
Step-9: As there are two moles of water in the product and one mole of water in reactant. So, one mole of water in the reactant is cancelled and only one mole of water remains in the product. The reaction is as follows:
$IO_3^ - + C{l_2} + 2O{H^ - } \to 2C{l^ - } + IO_4^ - + {H_2}O$
Hence, the balanced chemical reaction for the given ionic equation is $C{l_2} + IO_3^ - + 2O{H^ - } \to IO_4^ - + 2C{l^ - } + {H_2}O$
So, option (B) is the correct answer.

Note:
It is important to note that bases when dissolved in aqueous solution give hydroxide ions i.e., $O{H^ - }$ ions, hence balancing redox reaction in basic medium requires adding hydroxide ions to balance the reaction whereas in acidic medium, ${H^ + }$ are added to balance the ionic redox reaction.