The correct balanced reaction, assuming that the complex is Mn F 4 − , is | TestHub

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The correct balanced reaction, assuming that the complex is $Mn F_{4}^{-}$ , is

Options:

Answer:

B

Solution:

$M n^{2 +}$ $\to$ Mn(III) loses 1 $e^{-}$ each; $Mn O_{4}^{-}$ $\to$ Mn(III) gains 4 $e^{-}$ , so 4 $M n^{2 +}$ per $Mn O_{4}^{-}$ . Balancing charge and atoms with $8 H^{+} /4 H_{2} O$ and adding $20 F^{-}$ to complex all 5 Mn(III) as $Mn F_{4}^{-}$ gives $4M n^{2 +}$ + $Mn O_{4}^{-}$ + $8 H^{+}$ + $20 F^{-}$ $\to$ $5Mn F_{4}^{-}$ + $4 H_{2} O$ , matching (b).

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Balancing of Redox Reactions

⏱ 2mℹ️ Source: QB

Chemistry - Redox Reactions Question with Solution | TestHub

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