For the reaction Cr_2O_7^{2-} xrightarrow{H^+ | vedclass

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(A) The balanced half-reaction for the reduction of dichromate is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ In this reaction,the cha

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Molecular weight / $6$

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(A) The balanced half-reaction for the reduction of dichromate is: $Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$ 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 $2 	imes (6 - 3) = 6$. Therefore,the number of electrons gained per mole of $Cr_2O_7^{2-}$ is $6$. The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{M}{6}$

For the reaction $Cr_2O_7^{2-} \xrightarrow{H^+} Cr^{3+}$,what is the equivalent weight of $Cr_2O_7^{2-}$?

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