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Determine the oxidation numbers of all of the elements in the unbalanced reactions. Then, balance each redox reaction in basic solution. #1 Please!

Determine the oxidation numbers of all of the elements in the unbalanced reactions-example-1
User Kris Hunt
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1. According to the explanation given in our previous session, in this question we have the following reaction:

MnO4- + C2O42- -> MnO2 + CO2

O = 2-

MnO4- = we have an overall charge of -1, 4 oxygens, therefore we will have -8, and Mn must be = 7+

Mn (reactant) = 7+

C2O42- = overall charge -2, 4 oxygens = -8, therefore carbon must give a final value of +6, since we have 2 carbons, the oxidation number will be +3

C (reactant) = 3+

Mn (products) = 4+

C (products) = 4+

Now for the balanced redox equation:

2 MnO4- + 3 C2O42- + 4 H2O -> 2 MnO2 + 6 CO2 + 8 OH-

User Luchux
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