2KMnO_4 + 3H_2SO_4 + 5H_2O_2 rightarrow K_2SO | vedclass

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(D) The reaction is a redox titration between $KMnO_4$ and $H_2O_2$ in acidic medium.For $KMnO_4$,the change in oxidation state of $Mn$ is from $+7$ t

Vedclass · Accepted Answer

$8 	imes 10^{-2} \ N$

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(D) The reaction is a redox titration between $KMnO_4$ and $H_2O_2$ in acidic medium.For $KMnO_4$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so the n-factor is $5$.For $H_2O_2$,the change in oxidation state of $O$ is from $-1$ to $0$,so the n-factor is $2$.At the equivalence point,the number of milliequivalents of $H_2O_2$ equals the number of milliequivalents of $KMnO_4$.$N_1 V_1 = N_2 V_2$Here,$N_2 = M_2 	imes 	ext{n-factor} = 0.02 	imes 5 = 0.1 \ N$.$V_2 = 16 \ mL$,$V_1 = 20 \ mL$.$N_1 = \frac{N_2 V_2}{V_1} = \frac{0.1 	imes 16}{20} = \frac{1.6}{20} = 0.08 \ N = 8 	imes 10^{-2} \ N$.

$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$. Find the normality of $H_2O_2$ solution,if $20 \ mL$ of it is required to react completely with $16 \ mL$ of $0.02 \ M \ KMnO_4$ solution. (Molar mass of $KMnO_4 = 158 \ g \ mol^{-1}$)

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