Oxidation State - AP Chemistry

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Question

What is the oxidation number of chlorine in perchlorate?

Answer

Perchlorate is a complex ion with the formula .

The most important rule for oxidation number is that the sum of the oxidation states of the atoms must equal the overall molecular charge.

This compound has four oxygens, which always have a oxidation state.

Chlorine usually has an oxidation state of , but in this case it must balance out the oxygens.

Chlorine must be .

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Question

When is the oxidation number of H (-1)?

Answer

We typically think of Hydrogen as having an oxidation number of +1. However when it is bonded to a less electronegative element such as Na it is actually assigned an oxidation number of -1.

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Question

What is the oxidation number of Cr, S, and Fe in the following substances: (a) K2Cr2O7 (b) H2SO4 (c) Fe2O3.

Answer

(a) Since O has a –2 oxidation number and K has a +1 oxidation number (1 valence
electron it gives up), that means that Cr must have an oxidation number of +6. (b) Since H
has a +1 oxidation number and O has a –2 oxidation number, S has a +6 oxidation number.
(c) Fe has an oxidation number of +3 in order for it to have a net 0 oxidation state.

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Question

Consider the following balanced equation:

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

What is the difference between the oxidation state of aluminum on the right side of the equation versus the left?

Answer

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

On the left side of the equation, $Al_{(s)}$ is a solid, so its oxidation state is zero, but on the right side it is in a salt, so it is not in its zero state.

Sulfate, $SO_4^{-2}$ , is an anionic salt, and there are three sulfate ions in each complex, yielding a net charge of -6. The two aluminum ions must have a net charge of +6, which, divided over two aluminum ions, yields an oxidation state of +3 for each aluminum ion.

The difference comes from simple subtraction: $3-0=3$ .

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Question

What is the oxidation number of manganese in ?

Answer

Potassium always has an oxidation state of , while oxygen always has an oxidation state of . Since we have four oxygens, there is a total charge of from them.

The most important rule of oxidation numbers is that their sum must equal the charge on the molecule. In this compound, we have .

Manganese needs to cancel out the charge from potassium and oxygen in order to give us a neutral compound.

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Question

$Fe_{2}O_{3}+2Al \rightarrow Al_{2}O_{3}+2Fe$

In the above reaction, what are the initial and final oxidation states of ?

Answer

To determine the initial oxidation state of , we first must realize that $Fe_{2}O_{3}$ has 3 oxygen atoms, each with a charge of , for a total charge contribution of . Furthermore, since $Fe_{2}O_{3}$ has no net charge, the atoms must contribute a total charge of to balance out the charge coming from the oxygens. And since there are two atoms, then each must have a charge of .

On the product side of the reaction, notice that is all by itself without any charge. The oxidation state of any individual atom is .

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Question

The Claus process is used in the petroleum industry to convert sulfur containing gases into solid (rhombic) sulfur. One of the reactions that takes place in this process is:

$16H_{2}S; +; ; 8SO_{2}; \rightarrow 3{S_{8}}; +; 16H_{2}O$

Which of the following statements is correct:

Answer

We have an oxidizing $\left ( SO_{2}\right )$ and a reducing $\left ( H_{2}S\right )$ agent, then is a red-ox reaction and the two semi reactions are:

$2\left ( H_{2}S; \rightarrow ; S_{8}+16H^{+}+16H_{2}O+16e^{-}\right )$ (oxidation, $S^{-2}$ losses electrons)

$32e^{-}+32H^{+}+8SO_{2}: \rightarrow :S_{8}, +,16H_{2}O$ (reduction, $S^{+4}$ gains electrons)

In total 32 electrons are transferred from the $S^{-2}$ to $S^{+4}$ .

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Question

What is the oxidation number of nitrogen in ?

Answer

First, note that the molecule does not have a charge, meaning that the oxidation numbers of each atom must add up to zero. Hydrogen has an oxidation number of and oxygen has an oxidation number of . Thus, if we call the oxidation number of nitrogen , we can get the equation $1+x+3\times(-2)=0$ . Solving this gives , so the oxidation number of nitrogen is in this molecule.

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Question

The following is a modified true/false question. In it, you must decide if each individual statement is true (T) or false (F). If both are true, then you must also decide if the second statement is a correct explanation (CE) of the first statement.

I: An $\text{Al}^{3+}$ ion would undergo reduction to form $\text{Al}$ metal

BECAUSE

II: reduction is a loss of electrons

Answer

Statement II is false: reduction is a gain of electrons. Looking at statement I, the ion's charge would decrease by 3 in becoming a metal, which suggests that the ion gains 3 electrons. Thus, it is undergoing reduction.

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Question

When is the oxidation number of H (-1)?

Answer

We typically think of Hydrogen as having an oxidation number of +1. However when it is bonded to a less electronegative element such as Na it is actually assigned an oxidation number of -1.

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Question

What is the oxidation number of Cr, S, and Fe in the following substances: (a) K2Cr2O7 (b) H2SO4 (c) Fe2O3.

Answer

(a) Since O has a –2 oxidation number and K has a +1 oxidation number (1 valence
electron it gives up), that means that Cr must have an oxidation number of +6. (b) Since H
has a +1 oxidation number and O has a –2 oxidation number, S has a +6 oxidation number.
(c) Fe has an oxidation number of +3 in order for it to have a net 0 oxidation state.

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Question

Consider the following balanced equation:

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

What is the difference between the oxidation state of aluminum on the right side of the equation versus the left?

Answer

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

On the left side of the equation, $Al_{(s)}$ is a solid, so its oxidation state is zero, but on the right side it is in a salt, so it is not in its zero state.

Sulfate, $SO_4^{-2}$ , is an anionic salt, and there are three sulfate ions in each complex, yielding a net charge of -6. The two aluminum ions must have a net charge of +6, which, divided over two aluminum ions, yields an oxidation state of +3 for each aluminum ion.

The difference comes from simple subtraction: $3-0=3$ .

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Question

What is the oxidation number of manganese in ?

Answer

Potassium always has an oxidation state of , while oxygen always has an oxidation state of . Since we have four oxygens, there is a total charge of from them.

The most important rule of oxidation numbers is that their sum must equal the charge on the molecule. In this compound, we have .

Manganese needs to cancel out the charge from potassium and oxygen in order to give us a neutral compound.

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Question

What is the oxidation number of chlorine in perchlorate?

Answer

Perchlorate is a complex ion with the formula .

The most important rule for oxidation number is that the sum of the oxidation states of the atoms must equal the overall molecular charge.

This compound has four oxygens, which always have a oxidation state.

Chlorine usually has an oxidation state of , but in this case it must balance out the oxygens.

Chlorine must be .

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Question

$Fe_{2}O_{3}+2Al \rightarrow Al_{2}O_{3}+2Fe$

In the above reaction, what are the initial and final oxidation states of ?

Answer

To determine the initial oxidation state of , we first must realize that $Fe_{2}O_{3}$ has 3 oxygen atoms, each with a charge of , for a total charge contribution of . Furthermore, since $Fe_{2}O_{3}$ has no net charge, the atoms must contribute a total charge of to balance out the charge coming from the oxygens. And since there are two atoms, then each must have a charge of .

On the product side of the reaction, notice that is all by itself without any charge. The oxidation state of any individual atom is .

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Question

The Claus process is used in the petroleum industry to convert sulfur containing gases into solid (rhombic) sulfur. One of the reactions that takes place in this process is:

$16H_{2}S; +; ; 8SO_{2}; \rightarrow 3{S_{8}}; +; 16H_{2}O$

Which of the following statements is correct:

Answer

We have an oxidizing $\left ( SO_{2}\right )$ and a reducing $\left ( H_{2}S\right )$ agent, then is a red-ox reaction and the two semi reactions are:

$2\left ( H_{2}S; \rightarrow ; S_{8}+16H^{+}+16H_{2}O+16e^{-}\right )$ (oxidation, $S^{-2}$ losses electrons)

$32e^{-}+32H^{+}+8SO_{2}: \rightarrow :S_{8}, +,16H_{2}O$ (reduction, $S^{+4}$ gains electrons)

In total 32 electrons are transferred from the $S^{-2}$ to $S^{+4}$ .

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Question

When is the oxidation number of H (-1)?

Answer

We typically think of Hydrogen as having an oxidation number of +1. However when it is bonded to a less electronegative element such as Na it is actually assigned an oxidation number of -1.

Compare your answer with the correct one above

Question

What is the oxidation number of Cr, S, and Fe in the following substances: (a) K2Cr2O7 (b) H2SO4 (c) Fe2O3.

Answer

(a) Since O has a –2 oxidation number and K has a +1 oxidation number (1 valence
electron it gives up), that means that Cr must have an oxidation number of +6. (b) Since H
has a +1 oxidation number and O has a –2 oxidation number, S has a +6 oxidation number.
(c) Fe has an oxidation number of +3 in order for it to have a net 0 oxidation state.

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Question

Consider the following balanced equation:

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

What is the difference between the oxidation state of aluminum on the right side of the equation versus the left?

Answer

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

On the left side of the equation, $Al_{(s)}$ is a solid, so its oxidation state is zero, but on the right side it is in a salt, so it is not in its zero state.

Sulfate, $SO_4^{-2}$ , is an anionic salt, and there are three sulfate ions in each complex, yielding a net charge of -6. The two aluminum ions must have a net charge of +6, which, divided over two aluminum ions, yields an oxidation state of +3 for each aluminum ion.

The difference comes from simple subtraction: $3-0=3$ .

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Question

What is the oxidation number of manganese in ?

Answer

Potassium always has an oxidation state of , while oxygen always has an oxidation state of . Since we have four oxygens, there is a total charge of from them.

The most important rule of oxidation numbers is that their sum must equal the charge on the molecule. In this compound, we have .

Manganese needs to cancel out the charge from potassium and oxygen in order to give us a neutral compound.

Compare your answer with the correct one above

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