Chapter 7 Source Answers and Examples
Chapter problems → · Original reading →
This chapter has no answer key. Its worked examples follow.
Worked examples and reviewed revisions
Example 7.1.1
Original example and steps
Example 7.1.1
Let’s assign some oxidation states for the following compounds
i). SO2
ii.) PCl5
iii.) CO32-
Alright, let’s use the rules that we just learned and apply them to these compounds.
i). The rule that is the most important here is the one about oxygen oxidation states. Remember that means that the oxidation state for oxygen is -2. In the case of this compound, there are two oxygen atoms and it is not a peroxide compound so the oxygen atoms have an overall
oxidation state of -4. Since the SO2 compound is a neutral compound, then we must use rule number 5. That means that the sum of the oxidation states for all the atoms must be zero. So since oxygen has a -4 oxidation state, that must mean that the sulfur atom has an oxidation state of +4, therefore giving an overall sum of zero.
ii.) The rule to apply here is rule three. Chlorine has a greater affinity for electrons therefore its oxidation state will be equal to it’s negative charge in its ionic state, so in this case each chlorine atom will have an oxidation state of -1; meaning that the sum of chlorine oxidation states is -5. Since PCl5 is neutral as well, P must have a +5 oxidation state to achieve a sum of zero.
iii.) The thing that’s different here is the overall charge of the compound is 2- meaning that the compound is not neutral so there is an imbalance in the oxidation states. We know that oxygen must remain at -2 oxidation states so since there are three of them that means the sum of the states is -6. In order to meet the 2- charge of the compound, the carbon atom must have an oxidation state that meets those requirements. In this case, it is +4, as it allows for two electrons to remain thus giving us the overall charge that we’re looking for.
Let’s recap these solutions in a more visual form where you can actually see the compounds. View the gallery on the next page.
Example 7.2.1
Worked example
Example 7.2.1
Let’s consider the combustion of methanol, CH₃OH, using the unbalanced skeleton CH₃OH+O₂→CO₂+H₂O. This deliberately corrects the original ethanol label so that the following one-carbon formula and electron count agree.
Balance this equation using the oxidation states method.
For H2O, O has a -2 oxidation state and H has +1, so therefore it is balanced.
Notice how carbon goes from an oxidation state of -2 to +4, meaning that 6 electrons were transferred, an increase. On the other hand, oxygen goes of 0 to -2, a decrease of 2. In order to balance this, that means that 3 oxygen atoms are needed to balance the single increase of oxidation for the carbon atom. This gives us this general equation:
The half is there to “cancel” the 2 subscript on the oxygen to give us a solid number of 3 oxygen atoms. The main thing you have to understand from this is that the oxidation number decrease (2) has to equal to oxidation number increase (6), which is where the balancing comes in (having to have 3 oxygen atoms)
Now the rest of the balancing is just book keeping and making sure that the numbers all match up:
Multiplying by 2 to get rid of the fraction gives us the final balanced equation:
For H2O, O has a -2 oxidation state and H has +1, so therefore it is balanced.
Methanol’s carbon changes from −2 to +4, a loss of six electrons per carbon atom. An oxygen atom from O₂ changes from 0 to −2, a gain of two electrons, so three oxygen atoms receive six electrons. Accounting for all atoms gives 2CH₃OH+3O₂→2CO₂+4H₂O. Check carbon, hydrogen and oxygen counts after using the electron ratio; redox accounting alone does not replace atom balance.
The half is there to “cancel” the 2 subscript on the oxygen to give us a solid number of 3 oxygen atoms. The main thing you have to understand from this is that the oxidation number decrease (2) has to equal to oxidation number increase (6), which is where the balancing comes in (having to have 3 oxygen atoms)
Now the rest of the balancing is just book keeping and making sure that the numbers all match up:
Multiplying by 2 to get rid of the fraction gives us the final balanced equation:
Example Problem 7.2.2
Original example and steps
Example Problem 7.2.2
Use the oxidation state method to balance the following reaction:
Alright, so at first glance this is a little intimidating because the compounds seem larger, but again, its not that bad once we break it down.
Step 1:
First see if you can easily balance this reaction through inspection. This should always be your first step, why makes thing harder than it should be? Problem is that you can’t balance this that easily. Sorry, if you could do that, then there would be no point in teaching you this method.
Step 2:
Find the oxidation states for each atom in the equation. Breaking it down atom by atom let’s look at the oxidation states for just the reactants:
For HNO3:
- O: has a -2 oxidation state. There are 3 of those so it totals -6
- H: There is only 1, so it has a +1 oxidation state
- N: Since HNO3 is neutral, N has to have a oxidation state of +5
For H3AsO3:
- H: +1 and since there are 3, it totals at +3
- O: -2 and since there are 3, it totals at -6
- As: In order to achieve neutrality, it must have a +3 oxidation state
Now, let’s look at the oxidation states for the products:
For NO:
- O: it has a -2 oxy. state
- N: To be neutral, N must have an oxidation state of +2
For H3AsO4:
- H: Totaling +3 (+1 for each H atom)
- O: Totaling -8 (-2 for each O atom)
- As: in order to be neutral, it must have an oxidation state of +5
For H2O:
- O: -2 oxidation state
- H: +1 for each atom, meaning +2 total
Step 3:
Now that we know the oxidation states, lets have a more detailed look at the changes in oxidation state for each atom:
Now since oxygen and hydrogen keep the same oxidation state in the reaction, they are not considered in contributing to the balancing of the reaction.
Step 4
Now we have to remember that the decrease of oxidation state has to equal the increase in oxidation state meaning that there must be 3 As atoms for every 2 N atoms to achieve an overall balance.
Step 5
Since we know this, we can write the reaction for the reactant side:
Step 6
Use inspection method to balance the rest of the equation, which should be easy enough to yield your final, balanced equation:
Example 7.3.1
Worked example
AP edition reviewed replacement.
Balance Cu+HNO₃+H⁺→Cu²⁺+NO+H₂O in acidic solution by half-reactions, using HNO₃ as the explicitly chosen neutral-form bookkeeping reagent. Then state the equivalent fully molecular nitric-acid equation.
New independently computed worked answer.
Oxidation:Cu→Cu²⁺+2e⁻. Reduction:HNO₃+3H⁺+3e⁻→NO+2H₂O. Multiply oxidation by3and reduction by2; cancel6e⁻. The sum is3Cu+2HNO₃+6H⁺→3Cu²⁺+2NO+4H₂O. Equivalent molecular bookkeeping is3Cu+8HNO₃→3Cu(NO₃)₂+2NO+4H₂O. The original3H⁺ is valid for the neutralHNO₃ half-reaction; a nitrate-ion half-reaction would require4H⁺ because its reagent already has a different hydrogen count and charge. Nitric acid is a strong aqueous electrolyte; these are equivalent balanced writing conventions, not different mechanisms.
Example Problem 7.3.2
Original example and steps
Example Problem 7.3.2
Balance the equation in a basic solution using the half reaction method,
Step 1
We’re going to have the same approach as before. Write down the two half reactions that are oxidized and reduced.
Reduced:
Manganese is reduced going from +7 oxidation state to +4
Oxidized:
Carbon is oxidized because it goes from a +2 oxidation state to +4
Step 2
We will still balance the O’s by adding H2O to the equations. Let’s look at the reduced half reaction first.
And we will still add H+ ions to balance the hydrogens,
Step 3
What we’re going to do differently here is add OH- ions to balance the H+ ions, thus it’ll look like,
Step 5
Now to simplify this equation, we can combine the OH- and H+ ions to make H2O atoms, which in our reduced half reaction, we take the 4 OH- and 4H+ions to make 4H2O atoms,
So we can replace this reduced half reaction further to
Step 6
We can simplify things even further by eliminating the excess H2O atoms, thus giving us,
In addition, the balancing will continue by ensuring that the charges will remain balanced, and we will do this by adding electrons to equalize the charges. We will need to add 3 electrons to counter the -4 charge on the right side
Step 7
Use the same approach for the oxidized reaction:
After balancing with H2O, H+, OH - and electrons, we get the final oxidized half reaction of:
Step 8
Next we must balance the amount of electrons like we did in the previous example so that the electrons will cancel, so we multiply the reduced half reaction by 2 and the oxidized half reaction by 3,
giving us,
(reduced)
(oxidized)
Step 9
Combine the equations, while making sure to cancel the extra OH- ions, the extra H2O molecules and eliminate the electrons to get the final balanced reaction in a basic solution which is our answer,
Notice that the only difference between this solution and the previous examples solution is that this has OH- ions in its final balanced reaction rather than H+ ions.