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Question

(A) The balanced half-reaction for the reduction of dichromate in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$From

Vedclass · Accepted Answer

$6.369 	imes 10^5 \ C$

Vedclass · Answer

(A) The balanced half-reaction for the reduction of dichromate in acidic medium is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$From the stoichiometry,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons for reduction.Therefore,for $1.1 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is $1.1 	imes 6 = 6.6 \ mol$.The total charge $Q$ is given by $Q = n 	imes F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C/mol)$.$Q = 6.6 \ mol 	imes 96500 \ C/mol = 636900 \ C = 6.369 	imes 10^5 \ C$.

Calculate the amount of electricity required to convert $1.1 \ mol$ of $Cr_2O_7^{2-}$ to $Cr^{3+}$ in acidic medium.

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