The number of moles of K_2Cr_2O_7 reduced by | vedclass

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

Question

(B) The balanced redox reaction involves the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$ and the oxidation of $Sn^{2+}$ to $Sn^{4+}$.For $K_2Cr_2O_7$: Th

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) The balanced redox reaction involves the reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$ and the oxidation of $Sn^{2+}$ to $Sn^{4+}$.For $K_2Cr_2O_7$: The change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change is $2 	imes (6 - 3) = 6$. Thus,the $n$-factor is $6$.For $Sn^{2+}$: The change in oxidation state of $Sn$ is from $+2$ to $+4$. The change is $4 - 2 = 2$. Thus,the $n$-factor is $2$.Using the law of equivalence: $	ext{Equivalents of } K_2Cr_2O_7 = 	ext{Equivalents of } Sn^{2+}$$n_{K_2Cr_2O_7} 	imes 6 = n_{Sn^{2+}} 	imes 2$Given $n_{Sn^{2+}} = 1 \ mol$,we have:$n_{K_2Cr_2O_7} 	imes 6 = 1 	imes 2$$n_{K_2Cr_2O_7} = \frac{2}{6} = \frac{1}{3} \ mol$.

The number of moles of $K_2Cr_2O_7$ reduced by $1 \ mol$ of $Sn^{+2}$ is:

Similar Questions

In the balanced reaction $x\,MnO_4^- + y\,H_2O \to 2Mn^{2+} + 5H_2O + 9O_2 + z\,e^-$,what are the values of $x$,$y$,and $z$?

Which product is obtained during the reaction of $MnO_4^-$ with $I^-$ in a faintly alkaline condition?

$2 MnO_4^{-} + bI^{-} + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^{-}$
If the above equation is balanced with integer coefficients,the value of $z$ is . . . . . .

Moles of $K_2Cr_2O_7$ used to oxidise $1$ $mole$ of $Fe_{0.92}O$ to $Fe^{3+}$ are -

If an iron rod is dipped in $CuSO_4$ solution,what happens?

Vedclass Test Series

Exam Paper Generator

Online Exam Module