Balance the following redox reaction in acidic and basic solutions:
Fe(OH)2(s) + O2(g) --> Fe(OH)3(s)
These are the steps my instructor gave us to balance redox reactions:
1. Divide into 1/2 rxns (reduction & oxidation)
2. Balance atoms other than H & O
3. Balance O with H2O
4. Balance H with H+
5. Balance charge by adding e-
6. Equalize # of e- and add reactions
-- in basic soln --
7. Balance H+ by adding OH-
I only included the steps she gave us because everyone seems to have a different way of balancing these reactions. My problem with this reaction though is just plain writing the half reactions! I have a feeling that the oxidation reaction is Fe2+ --> Fe3+, but I am not sure how to write the reduction reaction. Any help is greatly appreciated!
To balance the given redox reaction in acidic solution, we need to follow the steps provided by your instructor. Let's go through each step and solve the reaction:
Step 1: Divide into 1/2 reactions
The given reaction is already a half-reaction, as it shows the oxidation and reduction happening simultaneously.
Step 2: Balance atoms other than H and O
The atoms other than hydrogen (H) and oxygen (O) are iron (Fe) and hydroxide (OH).
Fe(OH)2(s) + O2(g) --> Fe(OH)3(s)
The Fe and OH are already balanced on both sides of the equation.
Step 3: Balance oxygen (O) with water (H2O)
The reaction currently has no oxygen on the left side, and three oxygens on the right side. To balance the oxygen atoms, we need to add three water molecules (H2O) to the left side:
Fe(OH)2(s) + O2(g) --> Fe(OH)3(s) + 3H2O
Step 4: Balance hydrogen (H) with hydrogen ions (H+)
The reaction currently has four hydrogen atoms on the left side (from the two hydroxide ions, OH), and six hydrogen atoms on the right side (from the three water molecules, H2O). To balance the hydrogen atoms, we need to add two hydrogen ions (H+) to the left side:
Fe(OH)2(s) + O2(g) + 2H+ --> Fe(OH)3(s) + 3H2O
Step 5: Balance charge by adding electrons (e-)
Now we need to balance the charges on both sides of the equation. Iron (Fe) changes from a +2 oxidation state to a +3 oxidation state, which means it loses one electron. So, we add one electron (e-) to the left side:
Fe(OH)2(s) + O2(g) + 2H+ + e- --> Fe(OH)3(s) + 3H2O
Step 6: Equalize the number of electrons and add the reactions
The number of electrons on both sides is now equal. We can combine the half-reactions by multiplying the oxidation half-reaction by three to equalize the number of electrons:
3Fe(OH)2(s) + O2(g) + 6H+ + 3e- --> 3Fe(OH)3(s) + 9H2O
Now, the electrons can cancel out:
3Fe(OH)2(s) + O2(g) + 6H+ --> 3Fe(OH)3(s) + 9H2O
This is the balanced redox reaction in acidic solution.
If you want to balance it in basic solution (as specified in Step 7), you need to proceed with another step:
Step 7: Balance H+ with OH-
To balance the H+ ions in basic solution, you need to add an equal number of OH- ions to both sides of the equation. In this case, you would add six OH- ions to the right side:
3Fe(OH)2(s) + O2(g) + 6H+ + 6OH- --> 3Fe(OH)3(s) + 9H2O + 6OH-
Lastly, you can simplify the equation by canceling out water molecules:
3Fe(OH)2(s) + O2(g) + 6H2O --> 3Fe(OH)3(s) + 6H2O
The resulting balanced redox reaction in basic solution is:
3Fe(OH)2(s) + O2(g) + 6H2O --> 3Fe(OH)3(s) + 6H2O + 6OH-