The quantity of electricity in Faraday needed to reduce 1mol of Cr...

The quantity of electricity in Faraday needed to reduce

1mol

{Cr}_{2}{O}_{7}^{2 -}

{Cr}^{3 +}

Correct answer

Step-by-step solution

1 faraday of charge is the magnitude of charge of one mole of electrons.

According to the given reaction:
${Cr}_{2}{O}_{7}^{2 -} + {6e}^{-} ⟶ {2Cr}^{3 +}$

Reduction of $1mol$ of ${Cr}_{2}{O}_{7}^{2 -}$ to ${Cr}^{3 +}$ requires $6$ moles of electrons. Hence, charge required $= 6Faraday$

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