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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 $KMnO_4 = 158 \ g \ mol^{-1}$)

Similar Questions

Which of the following reactions are correct? (Ignore balancing of the reactions)
$(I)$ $Al + NaOH \xrightarrow{\Delta} NaAlO_2 + H_2$
$(II)$ $P_4 + NaOH + H_2O \to NaH_2PO_2 + PH_3$
$(III)$ $S + NaOH \to Na_2S_2O_3 + Na_2S + H_2O$

$Zn_{(s)} + CuSO_{4(aq)} \rightarrow ZnSO_{4(aq)} + Cu_{(s)}$ is an example of which type of process?

The gaseous product of the reaction between $Sn$ and conc. $H_2SO_4$ is

$6 \times 10^{-3} \text{ mole}$ of $K_2Cr_2O_7$ reacts completely with $9 \times 10^{-3} \text{ mole}$ of $X^{n+}$ to give $XO_3^-$ and $Cr^{3+}$. The value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module