Balance the following in basic medium- Fe3O4 - MnO-4 - Fe2O3 - MnO2

The correct option is A 6Fe_3O_4 + 2MnO^ _4 + H_2O → 9Fe_2O_3 + 2MnO_2 + 2OH^ The unbalanced redox equation is as follows: Fe_3O_4 + MnO^ _4 → Fe_2O_ ...

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Standard XII

Chemistry

S Block Elements

Balance the f...

Question

Balance the following in basic medium:

$F e_{3} O_{4} + {M n O}_{4}^{-} \to F e_{2} O_{3} + M n O_{2}$

6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} + H_{2} O \to 9 F e_{2} O_{3} + 2 M n O_{2} + 2 O H^{-}

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6 F e_{3} O_{4} + 4 {M n O}_{4}^{-} + H_{2} O \to 9 F e_{2} O_{3} + 5 M n O_{2} + O H^{-}

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6 F e_{3} O_{4} + {M n O}_{4}^{-} + H_{2} O \to 9 F e_{2} O_{3} + 3 M n O_{2} + 2 O H^{-}

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Solution

The correct option is A $6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} + H_{2} O \to 9 F e_{2} O_{3} + 2 M n O_{2} + 2 O H^{-}$
The unbalanced redox equation is as follows:
$F e_{3} O_{4} + {M n O}_{4}^{-} \to F e_{2} O_{3} + M n O_{2}$
Balance all atoms other than H and O.
$2 F e_{3} O_{4} + {M n O}_{4}^{-} \to 3 F e_{2} O_{3} + M n O_{2}$
The oxidation number of Fe changes from 2.67 to 3. The change in the oxidation number of 1 Fe atom is 0.33. Total change in the oxidation number for 6 C atoms is 2.
The oxidation number of Mn changes from +7 to 4. The change in the oxidation number of Mn is 3.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $F e_{3} O_{4}$ and $F e_{2} O_{3}$ with 3 and by multiplying ${M n O}_{4}^{-}$ and $M n O_{2}$ with 2.
$6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} \to 9 F e_{2} O_{3} + 2 M n O_{2}$
O atoms are balanced by adding 1 water molecule on RHS.
$6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} \to 9 F e_{2} O_{3} + 2 M n O_{2} + H_{2} O$
Hydrogen atoms are balanced by adding 2 $H^{+}$ atoms on the LHS.
$6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} + 2 H^{+} \to 9 F e_{2} O_{3} + 2 M n O_{2} + H_{2} O$
Since the reaction occurs in basic medium, add 2 hydroxide ions on both sides of the equation.
$6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} + 2 H^{+} + 2 O H^{-} \to 9 F e_{2} O_{3} + 2 M n O_{2} + H_{2} O + 2 O H^{-}$
On LHS, 2 $H^{+}$ ions combine with 2 $O H^{-}$ ions to form 2 water molecules out of which 1 water molecule cancels with 1 water molecule on RHS.
$6 F e_{3} O_{4} + 2 {M n O}_{4}^{-} + H_{2} O \to 9 F e_{2} O_{3} + 2 M n O_{2} + 2 O H^{-}$
This is the balanced chemical equation.

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Similar questions

Q. Balance the following chemical equation in basic medium:

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

9 F e_{2} O_{3} + 2 N H_{3} \to 6 F e_{3} O_{4} + N_{2} + 3 H_{2} O

According to the balanced equation above , when 1 mole of

N H_{3}

reacts completely, which of the following is true?

Balance the following reaction in basic medium.

C_{2} H_{5} O H + {M n O}_{4}^{-} ⟶ C_{2} H_{3} O^{-} + M n O_{2} (s) + H_{2} O

Q. The equation

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

when balanced in a basic medium by oxidation number method will be :

The complete balanced equation is:

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

(Basic medium)

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Balance the following in basic medium:Fe3O4+MnO−4→Fe2O3+MnO2

Balance the following in basic medium:

$F e_{3} O_{4} + {M n O}_{4}^{-} \to F e_{2} O_{3} + M n O_{2}$