# How do you balance H_2O_2 + Ce^(4+) -> O_2 + Ce^(3+)?

Jul 31, 2017

Separate the reduction and oxidation reactions........

#### Explanation:

$\text{Reduction half equation....}$

$C {e}^{4 +} + {e}^{-} \rightarrow C {e}^{3 +}$ $\left(i\right)$

$\text{Oxidation half equation....}$

${H}_{2} {O}_{2} \rightarrow {O}_{2} + 2 {H}^{+} + 2 {e}^{-}$ $\left(i i\right)$

And we add these equations together so that electrons do not appear in the final equation: $2 \times \left(i\right) + \left(i i\right)$............

$2 C {e}^{4 +} + {H}_{2} {O}_{2} \rightarrow 2 C {e}^{3 +} + {O}_{2} + 2 {H}^{+}$

Charge and mass are balanced as is required..............

For another $\text{redox equation}$ see here.
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