40 text{ mL} of x text{ M } KMnO_4 solution i | vedclass

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

Question

(A) In acidic medium,the balanced redox reaction is:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$Using the equivalence princ

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

(A) In acidic medium,the balanced redox reaction is:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ ightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$Using the equivalence principle,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid:$n_1 	imes M_1 	imes V_1 = n_2 	imes M_2 	imes V_2$For $KMnO_4$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so $n_1 = 5$.For oxalic acid $(H_2C_2O_4)$,the change in oxidation state of $C$ is from $+3$ to $+4$ for two carbon atoms,so $n_2 = 2$.Substituting the given values:$5 	imes x 	imes 40 = 2 	imes 0.02 	imes 200$$200x = 8$$x = \frac{8}{200} = 0.04 	ext{ M}$

$40 \text{ mL}$ of $x \text{ M } KMnO_4$ solution is required to react completely with $200 \text{ mL}$ of $0.02 \text{ M}$ oxalic acid solution in acidic medium. The value of $x$ is:

Similar Questions

Balance the following redox reactions using the oxidation number and ion-electron method:
$(1) \ Zn + NO_3^- + H^+ \to Zn^{2+} + N_2O + H_2O$
$(2) \ I^- + O_2 + H_2O \to I_2 + OH^-$
$(3) \ MnO_4^- + C_2H_5OH \to Mn^{2+} + CH_3COOH$

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$BaCO_3 \downarrow + 2CH_3COOH \longrightarrow Ba(CH_3COO)_2 + CO_2 \uparrow + H_2O$

Assign $A, B, C, D$ from the given type of reaction.
$ZnSO_4 \, (aq) + MgCl_2 \, (aq) \longrightarrow$ No reaction

Assign $A, B, C, D$ from the given type of reaction.
$NO + NO_2 \xrightarrow{-11 \ ^{\circ}C} N_2O_3$

For the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$,which option represents twice the equivalent weight of the oxidizing agent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module