# college chemistry

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how do you balance this redox equation?

Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + NaOH(aq)

• college chemistry -

Instead of giving you the answer, let me show you how to do this type. When they get a little complicated, I like to break them up into half reactions.
Co(OH)2 ==> Co(OH)3 .
It's obvious that Co changes from +2 to +3 so we need a one electron on the right and we need to balance the OH^- which we can do directly.

Co(OH)2 + OH^- ==> Co(OH)3 + e
Note that the equation balances
a. elements.
b. electron change.
c. charge.
Next half cell.
O2^-2 ==> OH^-
First we place a 2 coefficient for OH^- and compute changes.
O2^-2 ==> 2OH^-
Now O2^-2 has changed from -2 on the left (for both oxygens) to -4 on the right (for both oxygens) (which is why I stuck that two before starting any of this--we must compare the same number of oxygen atoms). So the change in electrons is -2 to -4 or +2; i.e.,

O2^-2 + 2e ==> 2OH^-
The charge on the left is -2, on the right is -4 so we must add 2OH^- to the right.
O2^-2 + 2e ==> 4OH^- and add water to the left.
2H2O + O2^-2 + 2e ==> 4OH^-
a. by atoms. yes.
b. by electron change. yes.
c. by charge. yes.

Now note the first half reaction changes by 1 e, the second half reaction by 2e; therefore, we multiply the first one by 2 and second one by 1 and add. You should do this but you should get this.

2Co(OH)2 + 2OH^- + O2^-2 + 2H2O ==>2Co(OH)3 + 4OH^-

We can cancel 2OH&- to make it
2Co(OH)2 + O2^-2 + 2H2O ==>2Co(OH)3 + 2OH^-

Now we can add Na^+ to the left for the Na2O2 and the right for the NaOH in the problem.
2Co(OH)2 + Na2O2 + 2H2O ==> 2Co(OH)3 + 2NaOH

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