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 $K_2Cr_2O_7$: $C

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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 $K_2Cr_2O_7$: $Cr$ changes from $+6$ to $+3$. The $n$-factor $= 2 	imes (6 - 3) = 6$.For $Sn^{2+}$: $Sn$ changes from $+2$ to $+4$. The $n$-factor $= 4 - 2 = 2$.According to the law of equivalence,the equivalents of $K_2Cr_2O_7$ must equal the equivalents of $Sn^{2+}$.Let $x$ be the moles of $K_2Cr_2O_7$.$x 	imes (n	ext{-factor of } K_2Cr_2O_7) = 	ext{moles of } Sn^{2+} 	imes (n	ext{-factor of } Sn^{2+})$.$x 	imes 6 = 1 	imes 2$.$x = \frac{2}{6} = \frac{1}{3} \, mol$.

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

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