The number of moles of K_2Cr_2O_7 reduced by | vedclass

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(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

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$\frac{1}{3}$

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(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:

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