How many Faraday are required to reduce one mol of ${M n O}_{4}^{-}$ to ${M n}^{2 +}$ ?

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Solution

The correct option is D 5
$M n O^{- 4} = M n^{+ 2} + 5 e^{-}$
$M n$ in $M n O_{4}^{-}$ has an oxidation state of +7.
Going from +7 to +2 requires an additional 5 electrons
1 mol of $e = 1 F$
5 mol of $e = 5 F$

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How many Faraday are required to reduce one mol of MnO−4 to Mn2+?

The correct option is D 5MnO−4=Mn+2+5e−Mn in MnO4− has an oxidation state of +7.Going from +7 to +2 requires an additional 5 electrons1 mol of e=1 F5 mol of e=5 F

How many Faraday are required to reduce one mol of ${M n O}_{4}^{-}$ to ${M n}^{2 +}$ ?

The correct option is D 5
$M n O^{- 4} = M n^{+ 2} + 5 e^{-}$
$M n$ in $M n O_{4}^{-}$ has an oxidation state of +7.
Going from +7 to +2 requires an additional 5 electrons
1 mol of $e = 1 F$
5 mol of $e = 5 F$