The equivalent weight of K_{2}Cr_{2}O_{7} in | vedclass

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(C) In acidic medium, the reduction reaction is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_{2}O$ Here, the oxidation state of

Vedclass · Accepted Answer

$M / 6$

Vedclass · Answer

(C) In acidic medium, the reduction reaction is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_{2}O$ Here, the oxidation state of $Cr$ changes from $+6$ to $+3$. Since there are two $Cr$ atoms in $K_{2}Cr_{2}O_{7}$, the total change in oxidation state is $2 	imes (6 - 3) = 6$. Therefore, the n-factor is $6$. Equivalent weight = $\frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{M}{6}$.

Compound	Oxidation number
$1$. $C_6H_{12}O_6$ (Carbon)	$a$. $+1$
$2$. $CCl_4$ (Carbon)	$b$. $0$
$3$. $NH_4Cl$ (Nitrogen)	$c$. $+4$
$4$. $Ba(H_2PO_2)_2$ (Phosphorus)	$d$. $-3$

The equivalent weight of $K_{2}Cr_{2}O_{7}$ in acidic medium is expressed in terms of its molecular weight $(M)$ as

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