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

1 Answer
Jul 31, 2017

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)