How many faradays are required to reduce 1 m | vedclass

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(D) The reduction half-reaction is: $Cr_{2}O_{7}^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$. In this reaction, the oxidation state of $Cr$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The reduction half-reaction is: $Cr_{2}O_{7}^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$. In this reaction, the oxidation state of $Cr$ changes from $+6$ to $+3$. Since there are $2$ atoms of $Cr$ in $Cr_{2}O_{7}^{2-}$, the total change in oxidation state is $2 	imes (6 - 3) = 6$. Therefore, $6 \ mol$ of electrons are required to reduce $1 \ mol$ of $Cr_{2}O_{7}^{2-}$, which corresponds to $6 \ F$ of electricity.

How many faradays are required to reduce $1 \ mol$ of $Cr_{2}O_{7}^{2-}$ to $Cr^{3+}$ in acidic medium?

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