How to balance this by oxidation number method $H_{2} O_{2} (a q) + F e^{2} (a q) + ⟶ F e^{3} (a q) + H_{2} O (l)$ [in acidic medium)

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Solution

Step 1: The skeletal ionic equation is

$H_{2} O_{2} (a q) + F e^{2 +} (a q) ⟶ F e^{3 +} (a q) + H_{2} O (l)$

Step 2: Assign oxidation numbers to Fe and O
$H_{2} O_{2}^{- 1} (a q) + F e^{2 +} (a q) ⟶ F e^{3 +} (a q) + H_{2} O^{- 2} (l)$
This indicates that the peroxide ion is the oxidant and the ferrous ion is the reductant.
Step 3: Calculate the increase and decrease of oxidation number, and make them equal:
$2 H_{2} O_{2}^{- 1} (a q) + 3 F e^{2 +} (a q) ⟶ 2 F e^{3 +} (a q) + H_{2} O^{- 2} (l)$
Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both sides, add $2 H^{+}$ on the RIGHT to make ionic charges equal
$2 H_{2} O_{2}^{- 1} (a q) + 3 F e^{2 +} (a q) ⟶ 2 F e^{3 +} (a q) + H_{2} O^{- 2} (l) + 2 H^{+}$
Step 5:Finally, count the hydrogen atoms, and add an appropriate number of water molecules on the right to achieve balanced redox change.
$2 H_{2} O_{2}^{- 1} (a q) + 3 F e^{2 +} (a q) ⟶ 2 F e^{3 +} (a q) + H_{2} O^{- 2} (l)$

In this reaction, the species undergoing oxidation is: Fe3+

(the O.N. of Fe increases from 2 to 3)

The species undergoing reduction is: O2 (the O.N. of O decreases from -1 to -2)

Step2: Write out oxidation and reduction separately and balance the atoms other than H and O.

Oxidation half reaction: Fe2+ → Fe3+

Reduction half reaction: H2O2 → H2O

Step3: Multiply the oxidation reaction with the extent of reduction and reduction reaction by the extent of oxidation and add.

Oxidation half reaction: [Fe2+ → Fe3+]x2

Reduction half reaction: [H2O2 → H2O ]x1

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2Fe2+ +H2O2 → 2Fe3+ + H2O

Step4: Add required number H+ on LHS and H2O on RHS to balance H and O atoms in the acid medium to get a balanced equation.

2 H+ (aq) + H2O2(aq) + 2 Fe2+(aq) → 2 Fe3+(aq) + 2 H2O(l)

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Similar questions

H_{2} O_{2} (a q) + F e^{2 +} (a q) \to F e^{3 +} (a q) + H_{2} O (l)

H_{2} O_{2 (a q)} + C r_{2} O_{7}^{2 -} (a q) \to O_{2 (g)} + C r_{(a q)}^{3 +}

The oxidation half reaction for following reaction is

Fe²⁺(aq) + Cr₂O₇^2-(aq) → Fe³⁺ (aq) + Cr³⁺(aq)

C r_{2} O_{7 (a q)}^{2 -} + I_{(a q)}^{-} \to C r_{(a q)}^{3 +} + I O_{3 (a q)}^{-}

{F e}^{2 +} (a q) + {C r}_{2} O_{7}^{2 -} (a q) (acid medium) - ------- \to {F e}^{3 +} (a q) + {C r}^{3 +} (a q)

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