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........

#"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.