how do you balance this redox equation?
Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + NaOH(aq)
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
To balance a redox equation, you need to follow a few steps. First, assign oxidation numbers to all the elements in the equation. Then, identify the elements being oxidized and reduced. Finally, balance the equation by adding appropriate coefficients.
Let's go through the process for balancing the given redox equation:
Step 1: Assign oxidation numbers
Co(OH)2(aq): The oxidation number of Co is +2. The oxidation number of H is +1, and since there are two of them, the total oxidation number is +2. The oxidation number of O in OH is -2, and since there are two OH groups, the total oxidation number is -4. So, the overall oxidation number of Co(OH)2(aq) is 2+(-4) = -2.
Na2O2(aq): The oxidation number of Na is +1, and since there are two of them, the total oxidation number is +2. The oxidation number of O in O2 is -1, so the overall oxidation number of Na2O2(aq) is 2+(-1) = +1.
Co(OH)3(s): The oxidation number of Co is +3. The oxidation number of O in OH is -2, and since there are three OH groups, the total oxidation number is -6. So, the overall oxidation number of Co(OH)3(s) is 3+(-6) = -3.
NaOH(aq): The oxidation number of Na is +1. The oxidation number of O in OH is -2, so the overall oxidation number of NaOH(aq) is 1+(-2) = -1.
Step 2: Identify the elements being oxidized and reduced
In this equation, the element Co changes its oxidation number from +2 to +3, so it is being oxidized. The element O changes its oxidation number from -1 to -2, so it is being reduced.
Step 3: Balance the equation
Start by balancing the atoms other than hydrogen and oxygen:
Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + 2NaOH(aq)
To balance oxygen atoms, add water (H2O):
Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + 2NaOH(aq) + H2O
To balance hydrogen atoms, add H⁺ ions:
Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + 2NaOH(aq) + H2O + 2H⁺
To balance the charge, add electrons:
Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + 2NaOH(aq) + H2O + 2H⁺ + 2e⁻
Finally, multiply the half-reactions by appropriate coefficients to balance the electrons. In the given equation, the number of electrons transferred is the same on both sides, so we don't need to multiply any half-reactions.
The balanced redox equation is:
Co(OH)2(aq) + Na2O2(aq) → Co(OH)3(s) + 2NaOH(aq) + H2O + 2H⁺ + 2e⁻