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Determine the oxidation numbers of the elements in the unbalanced reactions. Then, balance each redox reaction in acid solution.

Determine the oxidation numbers of the elements in the unbalanced reactions. Then-example-1

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4. In this question we need to find the oxidation number of each individual element, first, let's write the reaction:

P + Cu2+ -> Cu + H2PO4

Important things to have in mind

Oxidation number of single atoms is = 0

Oxidation number of Oxygen = -2

Oxidation number of Hydrogen = +1

These are informations that help us determine other oxidation numbers, therefore:

P (reactant) = 0

Cu (reactant) = +2 (it has a charge already)

Cu (product) = 0

H2 = 1*2 = 2

O = -2*4 = -8

P (product) = here we have to match the value given from H and O, therefore +2 -8 = -6, but H2PO4 is an anion, having as a final charge -1, therefore P must have a charge that when calculated with -6 it must give -1 as a result, the only value possible is +5, -6 + 5 = -1, and +5 is one of the possible values of oxidation number for P

Now balancing this redox reaction in acidic solution, we will have to add water in this case, and the reaction will be the following:

2 P + 5 Cu2+ + 8 H2O -> 2 H2PO4- + 5 Cu + 12 H+

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