In acidic medium,the moles of KMnO_4 required | vedclass

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

Question

(D) The oxidation reaction in acidic medium is: $Cu_2S + MnO_4^- \rightarrow Cu^{2+} + SO_2 + Mn^{2+}$.For $Cu_2S$,the oxidation states change as foll

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(D) The oxidation reaction in acidic medium is: $Cu_2S + MnO_4^- ightarrow Cu^{2+} + SO_2 + Mn^{2+}$.For $Cu_2S$,the oxidation states change as follows: $Cu$ from $+1$ to $+2$ (change of $1$ per $Cu$,total $2$ for $2Cu$) and $S$ from $-2$ to $+4$ (change of $6$).Total $n$-factor for $Cu_2S = 2 + 6 = 8$.For $KMnO_4$,$Mn$ changes from $+7$ to $+2$,so the $n$-factor is $5$.According to the law of equivalence: $	ext{moles of } KMnO_4 	imes (n	ext{-factor of } KMnO_4) = 	ext{moles of } Cu_2S 	imes (n	ext{-factor of } Cu_2S)$.$n 	imes 5 = 1.5 	imes 8$.$n = \frac{12}{5} = 2.4 \, mol$.

In acidic medium,the moles of $KMnO_4$ required to oxidize $1.5 \, mol$ of $Cu_2S$ will be ............... $mol$.

Similar Questions

How many moles of $MnO_4^-$ are required to completely oxidize $1 \ mol$ of ferrous oxalate in an acidic medium?

In order to oxidize a mixture of one mole of each of $FeC_2O_4$,$Fe_2(C_2O_4)_3$,$FeSO_4$ and $Fe_2(SO_4)_3$ in acidic medium,the number of moles of $KMnO_4$ required is

$2 Cu_2O_{(s)} + Cu_2S_{(s)} \longrightarrow 6 Cu_{(s)} + SO_{2(g)}$
In the above reaction,the oxidant and reductant respectively are:

In the reaction of sulphur with concentrated sulphuric acid,the oxidised product is $X$ and the reduced product is $Y$. $X$ and $Y$ are respectively:

When hypo solution is added to cupric sulphate solution,the blue colour of the latter is discharged,due to the formation of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module