In the standardization of Na_2S_2O_3 using K_ | vedclass

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(C) In the reaction of $K_2Cr_2O_7$ with $KI$ in an acidic medium,the oxidation state of $Cr$ changes from $+6$ to $+3$.$Cr_2O_7^{2-} + 14H^+ + 6e^- \

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$MW/6$

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(C) In the reaction of $K_2Cr_2O_7$ with $KI$ in an acidic medium,the oxidation state of $Cr$ changes from $+6$ to $+3$.$Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$Since $2$ atoms of $Cr$ are involved,the total change in oxidation number is $2 	imes (6 - 3) = 6$.Therefore,the $n$-factor for $K_2Cr_2O_7$ is $6$.Equivalent weight = $\frac{	ext{Molecular Weight (MW)}}{n	ext{-factor}} = \frac{MW}{6}$.

In the standardization of $Na_2S_2O_3$ using $K_2Cr_2O_7$ by iodometry,the equivalent weight of $K_2Cr_2O_7$ is:

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