In acidic medium,the equivalent weight of K_2 | vedclass

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(C) The reduction half-reaction for $K_2Cr_2O_7$ in an acidic medium is:$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} 	o 2Cr^{3+} + 7H_2O$In this reaction,the cha

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$M/6$

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(C) The reduction half-reaction for $K_2Cr_2O_7$ in an acidic medium is:$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} 	o 2Cr^{3+} + 7H_2O$In this reaction,the change in oxidation state of $Cr$ is from $+6$ to $+3$. Since there are two $Cr$ atoms,the total change in oxidation number is $2 	imes (6 - 3) = 6$.Thus,the $n$-factor for $K_2Cr_2O_7$ is $6$.The equivalent weight is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{n	ext{-factor}} = \frac{M}{6}$.

In acidic medium,the equivalent weight of $K_2Cr_2O_7$ $(mol. wt. = M)$ is

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