To balance the oxygen and hydrogen for a redox reaction that takes place in acidic solution, it is necessary to use which of the following?A. H2O and H+
B. H2O only
C. H2O and OH-
D. OH- only
I can tell you from many years of experience that MOST of the time the correct answer is A (H^+ and H2O) BUT that isn't always the correct answer.
For example, in the half reaction of
O2 ==> H2O we do the following:
first multiply H2O x 2.
O2 -->2H2O
O2 is zero, O in H2O is -2 for -4 total, add 4e to the left.
O2 + 4e ==>2H2O
Charge on left is -4, on right is zero, add 4H^+ to the left.
4H^+ + 4e + O2 --> 2H2O
Balanced by adding just H^+ and no water.
To balance the oxygen and hydrogen in a redox reaction that occurs in an acidic solution, it is necessary to use option A, H2O and H+.
Here's how you balance the redox reaction:
1. Start by writing down the unbalanced equation.
2. Identify the oxidized and reduced species.
3. Balance the atoms other than hydrogen and oxygen.
4. Balance the oxygen atoms by adding H2O molecules to the side that needs more oxygen.
5. Balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen.
6. Balance the charge by adding electrons to the appropriate side of the equation.
7. Finally, balance the electrons by multiplying one or both half-reactions to equalize the electron transfer.
For example, let's say we have the reaction:
MnO4- + H2O2 → Mn2+ + O2
First, we balance the non-hydrogen and non-oxygen atoms:
MnO4- + H2O2 → Mn2+ + 2O2
Next, we balance the oxygen atoms by adding H2O molecules:
MnO4- + H2O2 → Mn2+ + 2H2O + O2
Then, we balance the hydrogen atoms by adding H+ ions:
MnO4- + H2O2 + 4H+ → Mn2+ + 2H2O + O2
Finally, we balance the charge by adding electrons (e-) to the appropriate side of the equation. In this case, the MnO4- ion has a total charge of -1, so we need 5 electrons on the left side:
MnO4- + 5e- + H2O2 + 4H+ → Mn2+ + 2H2O + O2
Now the equation is balanced in acidic solution, using H2O and H+ ions.