Balance the following redox equations by the oxidation number method. The reactions occur in acidic medium.
$H_{2} O_{2 (a q)} + C r_{2} O_{7}^{2 -} (a q) \to O_{2 (g)} + C r_{(a q)}^{3 +}$

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Solution

For the equation:
$C r_{2} O_{7}^{2 -} + H_{2} O_{2} ⟶ C r^{3 +} + O_{2}$

Step $1$ : Assign oxidation number for each atom in the equation

$_{2} - 2 O_{7}^{2 -} +_{2}_{2} ⟶^{3 +} +_{2}$

Step $2$ : Divide equation in two half $- (1)$ reduction half where reactant being reduced (oxidation no decreases ) and $(2)$ oxidation half (oxidation no increases)
And also write the transfer of electrons to make numbers of oxidized & reduced atoms equal.

Oxidation half:

$_{2}_{2} ⟶_{2} + {2 e}^{-}$ (here oxygen is oxidized & oxidation number increases)

Reduction half:

$_{2}_{7} + {6 e}^{-} ⟶^{3 +}$ (here $C r$ reduced by $3$ electrons each)

Step $3 - (a)$ : Balance the charge is acidic medium add $H^{+}$ ion to the side deficient in positive charge

$_{2}_{2} ⟶_{2} + {2 e}^{-} + {2 H}^{+}$

$_{2} - 2 O_{7}^{2 -} + {6 e}^{-} + {14 H}^{+} ⟶ {2 C r}^{3 +}$

$(b)$ : Balance oxygen atom if oxygen atoms are not balanced add water molecules for it

$H_{2} O_{2} ⟶ O_{2} + 2 e^{-} + 2 H^{+}$

$_{2} - 2 O_{7}^{2 -} + {6 e}^{-} + {14 H}^{+} ⟶^{3 +} + {7 H}_{2} O$

Step $4$ : The electron lost in oxidation half must be equal to electron gained in reduced early so we multiply oxidation half by $3$ and reduction early by $1$ .

$O : 3 {+ 1 H}_{2} {- 1 O}_{2} ⟶ 3 {0 O}_{2} + {6 e}^{-} + {6 H}^{+}$

$R : {C r}_{2} O_{7}^{2 -} + {6 e}^{-} + 14 H^{+} ⟶ 2 {C r}^{3 +} + {7 H}_{2} O$

Step $5$ : Add half-reaction by adding all the reactants on one side and product on the other side.

$3 {0 H}_{2} O_{2} + {C r}_{2} O_{7}^{2 -} + {6 e}^{-} + 14 H^{+} ⟶ 3 O_{2} + {6 e}^{-} + {6 H}^{+} + 2 {C r}^{3 +} + 7 H_{2} O$

$\Rightarrow 3 {+ 1 H}_{2} {- 1 O}_{2} +_{2} O_{7}^{2 -} + {8 H}^{+} ⟶ {0 3 O}_{2} + 2 {C r}^{3 +} + {7 H}_{2} O$

Finally, always check the equation is balanced and contain substance type & the number of atoms both sides and sum of charges on both sides are equal.

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