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

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

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Answer 1 · 2017-07-31T20:29:19

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

Explanation:

$"Reduction half equation...."$

$Ce^(4+) + e^(-) rarr Ce^(3+)$ $(i)$

$"Oxidation half equation...."$

$H_2O_2 rarr O_2 + 2H^(+)+ 2e^(-)$ $(ii)$

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

$2Ce^(4+) + H_2O_2 rarr 2Ce^(3+)+O_2 + 2H^(+)$

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

For another $"redox equation"$ see [here.](https://socratic.org/questions/how-do-you-balance-this-equation-kmno4-na2s2o3-h2so4-k2so4-mnso4-na2so4-h2o)

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

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

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