Organic Chemistry : Oxidation-Reduction Reactions

Study concepts, example questions & explanations for Organic Chemistry

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Example Questions

Example Question #32 : Redox Chemistry

The oxidation numbers of calcium and chromium in the compound  are __________ and __________, respectively.

Possible Answers:

Correct answer:

Explanation:

Calcium is in the second group of the periodic table, so its oxidation number will always be . This is true for all alkaline earth metals.

As for chromium, the dichromate anion () has an overall charge of . Each oxygen atom will always have an oxidation number of  as well, so the oxygens would have a total negative charge of . The sum of the oxidation numbers have to equal the overall charge, so chromium must be used to balance the negative charges from the oxygen. To bring the overall charge to , the two chromiums have to have total charge of , giving each an oxidation number of .

Example Question #1 : Finding Oxidation Number

What is the oxidation number of the manganese in potassium permanganate?

Possible Answers:

None of these

Correct answer:

Explanation:

Manganese is a transition metal, so its oxidation number is more variable than the other elements of the compound. Oxygen is almost always . Potassium is almost always . The sum of the oxygens and the potassium is .

The compound is neutral, so the manganese is  to balance the net molecular charge.

Example Question #3 : Oxidation Reduction Reactions

Compared to oxygen in water, the oxygen in hydrogen peroxide has __________ valence electron(s).

Possible Answers:

The same

Two more

One less

One more

Correct answer:

One less

Explanation:

To solve this question we need to calculate the oxidation number of oxygen in both molecules. The formula for water is . The oxidation number of hydrogen is +1. Since there are two of them, the hydrogen atoms contribute to a charge of +2. The water molecule is neutral; therefore, the oxygen must have an oxidation number of  to balance the charge. The formula for hydrogen peroxide is . Using the same logic as water, we can determine that hydrogen contributes +2. We have two oxygen atoms in this case; therefore, each oxygen atom will have an oxidation number of  to give a charge of . This will balance the charges and provide a neutral hydrogen peroxide molecule.

Recall that have a negative charge suggests that an atom has extra valence electrons. A charge of  suggests one extra valence electron and a charge of  suggests two extra valence electrons. An oxygen typically has six valence electrons. The oxygen in water has  oxidation number; therefore, it will have two extra valence electrons (eight total). On the other hand, oxygen in hydrogen peroxide will have one extra valence electron (seven total); therefore, oxygen in hydrogen peroxide has one less valence electron than oxygen in water.

Example Question #2 : Finding Oxidation Number

Which of the following is true regarding the correct oxidation number of potassium in potassium bromide?

Possible Answers:

Correct answer:

Explanation:

Potassium bromide has a formula of . This molecule is made up of an alkali metal (potassium) and a halogen (bromine). Alkali metals have one valence electron that they readily lose to obtain octet whereas halogens have seven valence electrons and they readily gain an electron to obtain octet. Recall that losing an electron will give you a  oxidation number whereas gaining an electron will give you a  oxidation number. This means that alkali metals always have an oxidation number of  whereas halogens always have an oxidation number of ; therefore, potassium has an oxidation number of .

Example Question #3 : Finding Oxidation Number

Which of the following is the correct oxidation number of manganese in Manganese(III) oxide?

Possible Answers:

Correct answer:

Explanation:

Manganese(III) oxide has a formula of . The oxidation number of oxygen is . Since there are three of them, the oxygen atoms contribute to a charge of . To maintain the neutrality of the molecule, manganese must balance the  charge (meaning contribute  charge). There are two manganese atoms; therefore, each manganese atom will have an oxidation number of  and contribute  charge.

Example Question #3 : Finding Oxidation Number

A student reacts sodium chloride and lithium bromide. He collects the products in a jar and performs several tests on them. He concludes that product A has a metal and a nonmetal and that the metal has an oxidation number of  whereas the nonmetal has an oxidation number of . What can you conclude from these results?

Possible Answers:

The other product will have similar oxidation numbers (metal:  and nonmetal: )

The identity of product A is lithium chloride

The results seem invalid

The identity of product A is sodium bromide

Correct answer:

The results seem invalid

Explanation:

The reaction stated in this question is as follows.

This is a double replacement reaction. Both products contain a metal (an alkali metal) and a nonmetal (a halogen). The oxidation number of all alkali metals (first column of periodic table) is  and of all halogens (seventh column of periodic table) is . The results stated in the question seem invalid because a halogen can never have an oxidation number of .

Example Question #3 : Finding Oxidation Number

In the compound , what is the oxidation number of ?

Possible Answers:

Correct answer:

Explanation:

Generally, oxygen has an oxidation number of , and hydrogen has an oxidation number of . If we multiply  by the number of oxygen atoms in the compound (), we get an overall charge of  on the oxygen. If we multiply  by the number of hydrogen atoms in the compound (), we get an overall charge of  on the hydrogen. The sum of  and  is . We can see that  is a neutral molecule, so the charge on  would have to balance out the  charge on the rest of the molecule. To obtain a charge of , the oxidation number on  must then be .

Example Question #1 : Help With Oxidation

Which of the following is NOT an oxidizing agent?

Possible Answers:

Chromic acid 

Potassium permanganate 

Ozone 

Osmium tetroxide

Correct answer:

Explanation:

The answer is .

Lindlar's catalyst enables catalytic hydrogenation, specifically for an alkyne to reduce to a cis-alkene (though the reaction does not continue to produce an alkane). The rest of the answer options are oxidizing agents.

Example Question #4 : Oxidation Reduction Reactions

A water molecule is converted to hydrogen peroxide through a series of reactions. What can you conclude about the oxygen molecule?

Possible Answers:

Oxygen is oxidized because it loses an electron

Oxygen is reduced because it gains an electron

Oxygen is oxidized because it gains an electron

Oxygen is reduced because it loses an electron

Correct answer:

Oxygen is oxidized because it loses an electron

Explanation:

Water, , has an oxygen atom with an oxidation number of  whereas hydrogen peroxide, , has an oxygen atom with an oxidation number of . A less negative oxidation number suggests that the oxygen lost an electron. Recall that oxidation is the loss of electrons whereas reduction is the gain of electrons; therefore, the oxygen atom lost an electron and was oxidized when water was converted to hydrogen peroxide.

Example Question #2 : Help With Oxidation

Which of the following is true regarding oxidation?

I. Oxidation increases the oxidation number

II. Oxidizing agent is always reduced

III. Oxidation of an atom is always spontaneous

Possible Answers:

II and III

I and III

III only

I and II

Correct answer:

I and II

Explanation:

Oxidation is the process of removing electrons from an atom. This increases the oxidation number (makes oxidation number more positive). An atom undergoes oxidation by losing electrons and donating them to another atom. Since it is facilitating reduction (gain of electrons) of another atom, an atom that is oxidized is also called a reducing agent. Oxidizing agents, on the other hand, are reduced (gain electrons) and facilitate the oxidation of other atoms (removal of electrons from other atoms). Oxidation of an atom does not depend on the Gibbs free energy; therefore, oxidation can be spontaneous or nonspontaneous.

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