Question

Complete and balance the following equations:
(a)$KMn{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+MnS{{O}_{4}}+{{H}_{2}}O+...$
(b)$C{{u}^{2+}}+{{I}^{-}}\to C{{u}^{+}}+{{I}_{2}}$

Answer 1

Hint: A balanced chemical equation consists of the same amount of the atoms on the products as well as the reactants side. When there is increase and decrease in the oxidation states of the reactants, then the reaction is called a redox reaction. Redox reactions can be balanced by half reaction methods where the oxidation and the reduction halves are separately balanced.

Complete answer:
A reaction where there is increase as well as decrease in the oxidation states is known to be a redox reaction. These reactions can be balanced by a half reaction method. For this, the reduction and the oxidation half are separately balanced. The increase in oxidation number occurs in oxidation half, and decrease in oxidation number occurs in reduction half. The complete and balanced equations are:
(a) $KMn{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+MnS{{O}_{4}}+{{H}_{2}}O+...$
In this equation, $Mn{{O}_{4}}^{-}$ undergoes reduction and ${{O}^{-}}$ undergoes oxidation as:
Reduction half - $Mn{{O}_{4}}^{-}+5{{e}^{-}}\to M{{n}^{2-}}$
Oxidation half - $2{{O}^{-}}\to {{O}_{2}}+2{{e}^{-}}$
Now, in both the equations, balancing the charges by adding ${{H}^{+}}$ ions to balance positive charges and${{H}_{2}}O$for balancing the hydrogen ions we have,
$Mn{{O}_{4}}^{-}+8{{H}^{+}}+5{{e}^{-}}\to 4{{H}_{2}}O+M{{n}^{2-}}......(1)$and
$2{{O}^{-}}\to {{O}_{2}}+2{{e}^{-}}....(2)$
Now, to balance the electrons, multiplying equation 1 by 2 and equation 2 by 5 we have
$2Mn{{O}_{4}}^{-}+16{{H}^{+}}+10{{e}^{-}}\to 8{{H}_{2}}O+2M{{n}^{2-}}......(1)$
$10{{O}^{-}}\to 5{{O}_{2}}+10{{e}^{-}}....(2)$
Cancelling out the similar quantities and converting to simplest whole numbers, thereby adding equation 1 and 2 we have a balanced and complete chemical equation as:
$2Mn{{O}_{4}}^{-}+16{{H}^{+}}+10{{O}^{-}}\to 5{{O}_{2}}+8{{H}_{2}}O+2M{{n}^{2-}}$
So, the balanced equation is $2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+5{{H}_{2}}{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+5{{O}_{2}}$
(b) $C{{u}^{2+}}+{{I}^{-}}\to C{{u}^{+}}+{{I}_{2}}$
Considering the reduction and the oxidation halves and balancing their respective charges we have:
The oxidation half is $2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}....(1)$
Reduction half is $2C{{u}^{2+}}+2{{e}^{-}}\to 2C{{u}^{+}}.....(2)$
Adding up both the equations 1 and 2 due to balanced charges we have a balanced chemical equation as:
 $2C{{u}^{2+}}+2{{I}^{-}}\to 2C{{u}^{+}}+{{I}_{2}}$
Hence, the equations are balanced and completed.

Note:
The equation in (a) part consists of hydrogen peroxide and sulphuric acid that makes the medium acidic, so there will be addition of hydrogen ions to balance the positive charges as well as water molecules to maintain neutrality of the acid. This reaction is also called the ion electron method.