In acidic medium
2KMn+7O4 + 5H2 \(O^{-1}_2\) + 3H2SO4 \(\rightarrow\)5\(O^0_2\) + 2 Mn+2SO4 + K2SO4 + 8H2O
We have given,
Molarity of KMnO4 = 0.01M.
For reaction in acidic medium the normality of
KMnO4 = Molarity x Change in Oxidation state of Mn
= 0.01m x 5
=0.05 N
Using Normality equation
N1V1 = N2V2
\(\Rightarrow\) 0.05 N x 100ml = \(N_{{H_2}{O_2}}\,\times\,100 ml\)
\(\Rightarrow\) \(N_{{H_2}{O_2}}\) = 0.05 N
Therefore the Normality of H2O2 = 0.05 N
In basic medium
2KMn+7O4 + 3H\(O_2^-1\) \(\rightarrow\) 3\(O^0_2\) +2Mn+4O2 + 2KOH
There is no change in the Normality of H2O2. because there is same change in the oxidation state of oxygen ( In acidic medium as well as basic medium )
Therefore Normality of H2O2 = 0.05 N
but, There is change in the Normality of KMnO4 .
because , there is different change in oxidation state of Mn atom ( In acidic = 5 , In basic = 3 )
Therefore Normality of KMnO4 = molarity x n
= 0.01 \(\times\) 3
= 0.03 N
Using Normality equation
\((N_1V_1)_{KMnO_4} = (N_2V_2)_{H_2O_2}\)
\(\Rightarrow\) 0.03 N \(\times\,\)V1 = 0.05 \(\times
\) 100 ml
\(\Rightarrow\) v1 = \(\frac{500}{3} ml\)
Hence, \(\frac{500}{3}\) ,of same KMnO4 required in alkaline medium to oxidize 0.1 L of same H2O2.