Question

Balance by oxidation number method and ion electron method?
$KMn{{O}_{4}}+{{C}_{2}}{{O}_{4}}{{H}_{2}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+MnS{{O}_{4}}+C{{O}_{2}}+{{H}_{2}}O$

Answer 1

Hint Oxidation number or oxidation state of an atom in a molecule or compound is defined as the actual charge on the atom if it exists in a monatomic ion.
Permanganate ion oxidises the oxalate ion in acidic medium to carbon dioxide and reduces itself to $M{{n}^{2+}}$.

Complete Step by step solution:
There are two types of method for the balancing of redox reaction and they are, (i) Oxidation number change method (ii) ion electron method
(i) Balancing equation by oxidation number method-
$KMn{{O}_{4}}+{{C}_{2}}{{O}_{4}}{{H}_{2}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+MnS{{O}_{4}}+C{{O}_{2}}+{{H}_{2}}O$
In this method firstly we will know the oxidation number of all the atoms.
Manganese in potassium permanganate is in +7 oxidation state and carbon in ${{C}_{2}}{{O}_{4}}{{H}_{2}}$ has +3 oxidation state whereas sulphur in sulphuric acid has +6 oxidation state on the reactant side.
On the product side, sulphur of ${{K}_{2}}S{{O}_{4}}$ has +6 state, $Mn$ is +2 and $C$ in +4state.
Thus we can observe that that there is change in the oxidation state of $Mn$ and $C$
$KMn{{O}_{4}}\to MnS{{O}_{4}}$ …….(i) (oxidation number of $Mn$ decreases by 5)
${{C}_{2}}{{O}_{4}}{{H}_{2}}\to C{{O}_{2}}$ …………(ii) (oxidation number of $C$ decreases by 1)
To make increase and decrease equal, equation (i) multiplied by 2 and equation (ii) multiplied by 5.
$2KMn{{O}_{4}}\to 2MnS{{O}_{4}}$ …….(I)
$5{{C}_{2}}{{O}_{4}}{{H}_{2}}\to 10C{{O}_{2}}$ …….(II)
So overall balance reaction is –
$2KMn{{O}_{4}}+5{{C}_{2}}{{O}_{4}}{{H}_{2}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+10C{{O}_{2}}+{{H}_{2}}O$
After balancing oxalate ion, hydrogen and oxygen of the following reaction, the final balance reaction is-
$2KMn{{O}_{4}}+5{{C}_{2}}{{O}_{4}}{{H}_{2}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+10C{{O}_{2}}+8{{H}_{2}}O$

(ii)Balancing the equation by ion electron method-
In this method firstly we will know the oxidation number of all the atoms.
Manganese in potassium permanganate is in +7 oxidation state and carbon in ${{C}_{2}}{{O}_{4}}{{H}_{2}}$ has +3 oxidation state whereas sulphur in sulphuric acid has +6 oxidation state on the reactant side.
On the product side, sulphur of ${{K}_{2}}S{{O}_{4}}$ has +6 state, $Mn$ is +2 and $C$ in +4state.
In the second step we will write the oxidation and reduction half reaction.
We will add electron to make the difference in oxidation number
In next step we will add ${{H}_{2}}O$ molecule to balance the$O$atom
Balance the $H$ atom in both the equations by adding ${{H}^{+}}$ ions.

& {{C}_{2}}{{O}_{4}}\to 2C{{O}_{2}}+2{{e}^{-}} \\\ & Mn{{O}_{4}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O \\\ \end{aligned}$$ We will multiply the oxidation half reaction by 2 and reduce half reaction by 5 to equalize the electron loss and gain and add the two half reaction. $$\begin{aligned} & [{{C}_{2}}{{O}_{4}}\to 2C{{O}_{2}}+2{{e}^{-}}]\times 5 \\\ & [Mn{{O}_{4}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O]\times 2 \\\ \end{aligned}$$ To write the final equation we will balance the ${{H}_{2}}S{{O}_{4}}$ hydrogen by adding water molecule and final equation will be – $2KMn{{O}_{4}}+5{{C}_{2}}{{O}_{4}}{{H}_{2}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+10C{{O}_{2}}+8{{H}_{2}}O$ **Note:** oxidation number is not the same as valency. Valency represents the charge in whole numbers with no plus and minus sign, while oxidation number may or may not be equal to valency, it may be positive, negative, zero and a whole number or fractional.