Balancing redox reaction by half-reaction method

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Homework Statement

Balance the following equation by the half-reaction method. The reaction takes place in basic solution.

Fe(OH)₂(s) + O₂(g) --> Fe(OH)₃(s)

The Attempt at a Solution

All the other problems like this had 2 reactants and 2 products. What's confusing me about this problem is that there is only 1 product. So I'm not really sure how to set up the two half-reactions. What I tried was:

Fe(OH)₂(s) --> Fe(OH)₃(s)
O₂(g) --> Fe(OH)₃(s)

But the problem with that is that I cannot balance the second one since there is Fe on only one side. Any help would be really appreciated.

Thanks

 

[nate808]

for posting this problem! Balancing equations using the half-reaction method can be tricky, but I'll do my best to explain the process.

First, let's take a look at the overall reaction:

Fe(OH)2(s) + O2(g) --> Fe(OH)3(s)

We can see that there is only one reactant (Fe(OH)2) and one product (Fe(OH)3). This means that we will only need to balance one half-reaction.

To begin, we need to identify which elements are being oxidized and reduced. In this case, iron (Fe) is being oxidized from a +2 oxidation state in Fe(OH)2 to a +3 oxidation state in Fe(OH)3. Oxygen (O) is being reduced from a 0 oxidation state in O2 to a -2 oxidation state in Fe(OH)3.

Now, we can write the two half-reactions:

Oxidation half-reaction: Fe(OH)2(s) --> Fe(OH)3(s) + 2e-
Reduction half-reaction: O2(g) + 4e- --> 2OH-

Notice that in the oxidation half-reaction, we have balanced the iron atoms by adding a coefficient of 1 to Fe(OH)3. In the reduction half-reaction, we have balanced the oxygen atoms by adding a coefficient of 2 to OH-.

Next, we need to balance the number of electrons in each half-reaction by multiplying one or both of the reactions by a coefficient. In this case, we can multiply the oxidation half-reaction by 4 and the reduction half-reaction by 2, so that the number of electrons is equal in both reactions:

Oxidation half-reaction: 4Fe(OH)2(s) --> 4Fe(OH)3(s) + 8e-
Reduction half-reaction: 2O2(g) + 8e- --> 4OH-

Now, we can add the two half-reactions together to get the balanced overall reaction:

4Fe(OH)2(s) + 2O2(g) --> 4Fe(OH)3(s) + 4OH-

Finally, we need to balance the number of hydroxide ions (OH-) on both sides of the equation. We can do this by adding 4OH- to the reactants side