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

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Solution

${C r}_{2} O_{7}^{2 -} + H_{2} O_{2} ⇌ {C r}^{3 +} + O_{2}$
1) Separating into half reaction
${+ 1 H}_{2} {- 1 O}_{2} ⟶ {0 O}_{2} + 2 e^{-}$ (Oxidation)
${+ 6 C r}_{2} O_{7}^{2 -} + {6 e}^{-} ⟶ 2^{3 +}$ (Reduction)
2) Balancing the atoms & charge
$H_{2} O_{2} ⟶ O_{2} + {2 e}^{-} + 2 H^{+}$
${C r}_{2} O_{7}^{2 -} + 6 e^{-} + 14 H^{+} ⟶ {2 C r}^{3 +} + 7 H_{2} O$
4) Making electron gain equivalent to electron lost,
$(H_{2} O_{2} ⟶ O_{2} + 2 e^{+} + 2 H^{+}) \times 3$
$({C r}_{2} O_{7}^{2 -} + 6 e^{-} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 7 H_{2} O) \times 1 - ---------------------------------------------------- -$
Adding,
$3 H_{2} O_{2} + {C r}_{2} O_{7}^{2 -} + 8 H^{+} ⟶ 3 O_{2} + 2 {C r}^{3 +} + 7 H_{2} O$

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