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(C) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $

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$1.2$

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(C) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} ightarrow 2Cr^{3+} + 7H_2O$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. The change in oxidation state for $2$ atoms of $Cr$ is $2 	imes (6 - 3) = 6$. Thus,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons. For $0.2 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is $0.2 	imes 6 = 1.2 \ mol$.

The number of moles of electrons required to reduce $0.2 \ mol$ of $Cr_2O_7^{2-}$ to $Cr^{3+}$ is:

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