How would one balance the following redox reaction using the half reaction method?
Fe(OH)2 + O2 + H2O ---> Fe(OH)3
This is in my chemistry teacher's homework, and it comes up with an answer, but I don't see how to use the half reaction method when there is only one substance on the right side. Please help. Thank you very much.
4Fe(OH)2 + O2 + 2H2O ==> 4Fe(OH)3
Stepwise: Balanced half cell for Fe is
equn 1 is Fe^2+ ==> Fe^3+ + e
O2 ==> 2OH^-
O2 changes from zero to -4 on the right. Add 4e to the left side.
O2 + 4e ==> 2OH^-
Count up the charge. -4 on the left and -2 on the right. Add 2OH^- on the right to balance which makes it
O2 + 4e ==> 4OH^-. Now add water to balance the H which makes it
equn 2 balanced half cell is O2 + 4e + 2H2O ==> 4OH^-
Multiply equn 1 by 4 and equan 2 by 1 and add the two.
4Fe^2+ O2 + 2H2O ==> 4OH^- + 4Fe^3+
Add 8OH^- on the left and right.
4Fe(OH)2 + O2 + 2H2O --> 12OH^- + 4Fe^3+
4Fe(OH)2 + O2 + 2H2O ==> Fe(OH)3
Voila.
When balancing redox reactions using the half-reaction method, we first split the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
To balance the given equation:
Fe(OH)2 + O2 + H2O ---> Fe(OH)3
First, let's assign oxidation numbers to each element:
Fe(OH)2 (Fe: +2)
O2 (O: 0)
H2O (H: +1, O: -2)
Fe(OH)3 (Fe: +3)
Next, we identify the atoms undergoing oxidation and reduction. In this case, iron (Fe) is being oxidized from +2 to +3, while oxygen (O) is being reduced from 0 to -2.
For the oxidation half-reaction, we write the Fe atoms on both sides of the equation and balance the oxygen atoms by adding water molecules (H2O):
Fe(OH)2 ---> Fe(OH)3 + H2O
Next, we balance the hydrogen atoms by adding H+ ions:
Fe(OH)2 + 2H+ ---> Fe(OH)3 + H2O
For the reduction half-reaction, we balance the oxygen atoms by adding water molecules (H2O):
O2 + 2H2O ---> 4OH-
Next, we balance the hydrogen atoms by adding H+ ions:
O2 + 4H2O ---> 4OH- + 4H+
Now, we need to balance the number of electrons lost and gained in each half-reaction. Since the iron (Fe) is being oxidized from +2 to +3, it is losing one electron. And since oxygen (O) is being reduced from 0 to -2, it is gaining four electrons.
To balance the electrons, we multiply the oxidation half-reaction by 4 and the reduction half-reaction by 1:
4(Fe(OH)2 + 2H+ ---> Fe(OH)3 + H2O)
O2 + 4H2O ---> 4OH- + 4H+ + 4e-
Now, we can combine the two half-reactions by canceling out the electrons:
4Fe(OH)2 + 8H+ + O2 + 4H2O ---> 4Fe(OH)3 + 4OH-
Finally, we balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen:
4Fe(OH)2 + 8H+ + O2 + 4H2O ---> 4Fe(OH)3 + 4OH- + 8H+
And that's the balanced equation using the half-reaction method.
To balance a redox reaction using the half reaction method, you need to split the overall reaction into separate oxidation and reduction half reactions.
In this case, the iron (Fe) is being oxidized from a +2 oxidation state in Fe(OH)2 to a +3 oxidation state in Fe(OH)3. To balance the oxidation half reaction, you need to determine the number of electrons transferred.
The reduction half reaction involves the oxygen (O2) being reduced from a 0 oxidation state to a -2 oxidation state in Fe(OH)3. Again, you need to determine the number of electrons transferred.
Now, let's determine and balance the individual half reactions:
Oxidation half reaction:
Fe(OH)2 --> Fe(OH)3
The oxidation state of Fe changes from +2 to +3, which means it lost one electron. So, the half reaction for the oxidation is:
Fe(OH)2 --> Fe(OH)3 + e-
Reduction half reaction:
O2 + 2H2O + 4e- --> 4OH-
To balance the reduction half reaction, we added 4 electrons to the left side to balance the charges.
Now, we can balance the number of electrons in both half reactions by multiplying them as necessary:
4(Fe(OH)2 --> Fe(OH)3 + e-)
O2 + 2H2O + 4e- --> 4OH-
Now, if we combine these two half reactions, we can see that the electrons on both sides cancel out:
4Fe(OH)2 + O2 + 2H2O --> 4Fe(OH)3
Finally, check that all other atoms are balanced. In this case, we have 4 iron atoms, 8 hydrogen atoms, and 4 oxygen atoms on both sides of the equation.
Therefore, the balanced equation using the half reaction method is:
4Fe(OH)2 + O2 + 2H2O --> 4Fe(OH)3
4Fe(OH)2 + O2 + 2H2O ==> 4Fe(OH)3
Stepwise: Balanced half cell for Fe is
equn 1 is Fe^2+ ==> Fe^3+ + e
O2 ==> 2OH^-
O2 changes from zero to -4 on the right. Add 4e to the left side.
O2 + 4e ==> 2OH^-
Count up the charge. -4 on the left and -2 on the right. Add 2OH^- on the right to balance which makes it
O2 + 4e ==> 4OH^-. Now add water to balance the H which makes it
equn 2 balanced half cell is O2 + 4e + 2H2O ==> 4OH^-
Multiply equn 1 by 4 and equan 2 by 1 and add the two.
4Fe^2+ O2 + 2H2O ==> 4OH^- + 4Fe^3+
Add 8OH^- on the left and right.
4Fe(OH)2 + O2 + 2H2O --> 12OH^- + 4Fe^3+
4Fe(OH)2 + O2 + 2H2O ==> 4 Fe(OH)3