(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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(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 $+6$. The change in oxidation state for $2$ atoms of $Cr$ is $2 \times (6 - 3) = 6$. 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 $0.2 \times 6 = 1.2 \ mol$.

Class 11 Chemistry Redox Reactions Redox reaction and Method for balancing Redox reaction

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

KCET 2019 All KCET

A
$12$
B
$0.6$
C
$1.2$
D
$6$

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Redox Reactions

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Redox reaction and Method for balancing Redox reaction

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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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