Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium simultaneously forming a dark brown precipitate of hydrated `MnO_2`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2)SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2)O_(2)`.

Explanation

Step 1. To write balanced equations for the reactions involved.
(i) In acidic medium, `MnO_(4)^(-) ` oxidises `H_(2)O_(2)` to `O_(2)`
`{:(MnO_(4)^(-)+8H^(+)+ 5e^(-) to Mn^(2+) + 4H_(2)O"]"xx2),(" "H_(2)O_(2) to O_(2) + 2H^(+) + 2e^(-) "]"xx5):}/(2MnO_(4)^(-) + 5H_(2)O_(2) + 6H^(+) to 2Mn^(2+) + 5O_(2)+ 8H_(2)O)" "...(i)`
(ii) In neutral medium , `MnO_(4)^(-)` oxidises `Mn^(2+) ` to `MnO_(2)` .
`{:(MnO_(4)^(-)+ 2H_(2)O + 3e^(-) to MnO_(2)+ 4OH^(-)"]"xx2),(" "Mn^(2+)+ 4OH^(-) to MnO_(2) + 2H_(2)O + 2e^(-)"]"xx3):}/(2MnO_(4)^(-)+3Mn^(2+) + 4OH^(-) to 5 MnO_(2) + 2H_(2)O)" "...(ii)`
(iii) In acidic medium. `MnO_(2)` oxidises sodium oxalate to `CO_(2)`
`{:(MnO_(2)+ 4H^(+) + 2e^(-) to Mn^(2+) + 2H_(2)O),(" "C_(2)O_(4)^(2-) to 2CO_(2) + 2e^(-)):}/(MnO_(2) + C_(2)O_(4)^(2-) + 4H^(+) to Mn^(2+) + 2CO_(2) + 2H_(2)O) `
or `5MnO_(2) + 5C_(2)O_(4)^(2-) + 20 H^(+) to 5Mn^(2+) + 10 CO_(2) + 10 H_(2)O" "...(iii)`
From the above three balanced equations, it follows that
`5C_(2)O_(4)^(2-) -=2MnO_(4)^(-) -=5H_(2)O_(2)" "...(iv)`
Step 2. To determine the number of moles of `C_(2)O_(4)^(2-)` present in 10 mL of 0.2 M sodium oxalate.
No. of moles of sodium oxalate =Molarity `xx` volume in litres
`=0.2 xx(10)/(1000)`
`=2xx10^(-3)` mole
Step 3. To calculate the molarity of `H_(2)O_(2)`
From eq. (iv), it follows that `5C_(2)O_(4)^(2-)-= 5H_(2)O_(2)`
or `2xx10^(-3)` mole of `C_(2)O_(4)^(2-) -= 2xx10^(-3)` mole of `H_(2)O_(2)`
Now `2xx10^(-3) ` mole of `H_(2)O_(2)` is present in 10 mL of `H_(2)O_(2)`
`therefore ` Molarity of `H_(2)O_(2)=(2xx10^(-3))/(10)xx1000=0.2` M