In the half-reaction Cr_2O_7^{2-} + 14H^+ + 6 | vedclass

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(C) The equivalent weight of an oxidizing or reducing agent is calculated by dividing its molecular weight $(M)$ by the number of electrons gained or

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The equivalent weight of an oxidizing or reducing agent is calculated by dividing its molecular weight $(M)$ by the number of electrons gained or lost per molecule ($n$-factor).In the given half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$.The total number of electrons gained by one molecule of $Cr_2O_7^{2-}$ is $6$.Therefore,the $n$-factor for $Cr_2O_7^{2-}$ is $6$.Equivalent weight = $\frac{	ext{Molecular weight}}{6}$.

In the half-reaction $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,the equivalent weight of $Cr_2O_7^{2-}$ is obtained by dividing its molecular weight by:

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