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Balancing Chemical Equations by Partial Equation Method

Balance the f...

Question

Balance the following equation taking place in basic solution.

$F e {(O H)}_{2} (s) + O_{2} (g) ⟶ F e {(O H)}_{3} (s)$

4 F e {(O H)}_{2} (s) + 2 H_{2} O (l) + O_{2} (g) ⟶ 4 F e {(O H)}_{3} (s)

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4 F e {(O H)}_{2} (s) + H_{2} O_{2} (l) + O_{2} (g) ⟶ 4 F e {(O H)}_{3} (s)

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4 F e {(O H)}_{3} (s) + 2 H_{2} O (l) + O_{2} (g) ⟶ 4 F e {(O H)}_{3} (s)

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Solution

The correct option is A $4 F e {(O H)}_{2} (s) + 2 H_{2} O (l) + O_{2} (g) ⟶ 4 F e {(O H)}_{3} (s)$
The unbalanced redox equation is as follows:
$F e {(O H)}_{2} (s) + O_{2} (g) ⟶ F e {(O H)}_{3} (s)$
All atoms other than H and O are balanced.
The oxidation number of Fe changes from 2 to 3. The change in the oxidation number is 1.
The oxidation number of changes from 0 to -2. The change in the oxidation number per O atom is 2. For 2 O atoms, the change in the oxidation number is 4.
To balance the increase in the oxidation number with decrease in the oxidation number multiply $F e {(O H)}_{2} (s)$ and $F e {(O H)}_{3} (s)$ with 4.
$4 F e {(O H)}_{2} (s) + O_{2} (g) ⟶ 4 F e {(O H)}_{3} (s)$
To balance O atoms, add 2 water molecules on LHS.
$4 F e {(O H)}_{2} (s) + O_{2} (g) + 2 H_{2} O ⟶ 4 F e {(O H)}_{3} (s)$
This is the balanced chemical equation.

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