You visited us 1 times! Enjoying our articles? Unlock Full Access!

Oxidation Number Method

VO2+ is oxidi...

Question

$V O^{2 +}$ is oxidised to $V O_{2}^{+}$ by $M n O_{4}^{-}$ in acidic medium, which in turn is reduced to $M n^{2 +}$ . What is balanced equation for this reaction ?

V O^{2 +} + M n O_{4}^{-} + H_{2} O \to V O_{2}^{+} + M n^{2 +} + H^{+}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 V O^{2 +} + M n O_{4}^{-} + H_{2} O \to 5 V O_{2}^{+} + M n^{2 +}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 V O^{2 +} + M n O_{4}^{-} + H_{2} O \to 5 V O_{2}^{+} + M n^{2 +} + 2 H^{+}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $5 V O^{2 +} + M n O_{4}^{-} + H_{2} O \to 5 V O_{2}^{+} + M n^{2 +} + 2 H^{+}$
$V O_{2 +} + M n O_{4}^{-} \to M n^{2 +} + V O_{2}^{+}$
First consider two half reactions of oxidation and reduction,
$V^{4 +} \to V^{5 +} + e^{-}$
$M n^{7 +} + 5 e^{-} \to M n^{2 +}$
Balance the electrons at both half cell reactions,
$5 V^{4 +} \to 5 V^{5 +} + 5 e^{-}$
$M n^{7 +} + 5 e^{-} \to M n^{2 +}$
Add both the reactions and balance the oxygen using water and acid ions,
$5 V O^{2 +} + M n O_{4}^{-} + H_{2} O \to 5 V O_{2}^{+} + M n^{2 +} + 2 H^{+}$

Suggest Corrections

Similar questions

{M n O}_{4}^{-} \to M n^{2 +} + H_{2} O

M n O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +} + H_{2} O

M n O_{4}^{-} + a F e^{2 +} + b H^{+} \to M n^{2 +} + c F e^{3 +} + d H_{2} O

M n^{2 +} + S_{2} O_{8}^{2 -} \to S O_{4}^{2 -} + M n O_{4}^{-}

S_{2} O_{8}^{2 -}

2

M n^{2 +}

100

H_{2} O_{2}

100

0.01 M

K M n O_{4}

M n O_{4}^{⊝}

M n^{2 +}

100

H_{2} O_{2}

V

0.01 M

K M n O_{4}

M n O_{4}^{⊝}

M n O_{2}

V

F e^{2 +} \to F e^{3 +} + e^{-}; M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}

F e^{2 +}

M n O_{4}^{-}

Join BYJU'S Learning Program