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AP Chemistry : Oxidation State

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Example Questions

Example Question #1 : Oxidation State

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

Possible Answers:

Only when it is bonded to Chlorine

Only when it is bonded to Flourine

Never

When it is bonded to a less electronegative element

When it is bonded to a more electronegative element

Correct answer: When it is bonded to a less electronegative element

Explanation:

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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Example Question #1 : Oxidation State

What is the oxidation number of manganese in KMnO_4 ?

Possible Answers:

Correct answer:

Explanation:

KMnO_4

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 (+1)+4(-2)+Mn=0 .

(+1)-8+Mn=0

-7+Mn=0

Mn=+7

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

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Example Question #111 : Reactions And Equilibrium

What is the oxidation number of chlorine in perchlorate?

Possible Answers:

Correct answer:

Explanation:

Perchlorate is a complex ion with the formula ClO_4^- .

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

(Cl)+4(O)=-1

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

(Cl)+4(-2)=-1

(Cl)-8=-1

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

Cl=+7

Chlorine must be .

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Example Question #1 : Oxidation State

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?

Possible Answers:

No difference

Correct answer:

Explanation:

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

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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Example Question #1 : Oxidation State

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

Possible Answers:

6, 6, 3

3, 6, 3

3, 6, 6

3, 3, 3

6, 6, 6

Correct answer:

6, 6, 3

Explanation:

(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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Example Question #1 : Oxidation State

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

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

Possible Answers:

$+3,\:0$

There is no change in oxidation state

$0,\:+3$

+3,+2

$-3,\:0$

Correct answer:

$+3,\:0$

Explanation:

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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Example Question #1 : Oxidation State

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:

Possible Answers:

This is a red-ox reaction

$S^{-2}$ gets oxidizes to $S_{8}$

$SO_{2}$ is the oxidant

All of them are correct

Correct answer:

All of them are correct

Explanation:

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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AP Chemistry : Oxidation State

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Oxidation State

Example Question #1 : Oxidation State

Example Question #111 : Reactions And Equilibrium

Example Question #1 : Oxidation State

Example Question #1 : Oxidation State

Example Question #1 : Oxidation State

Example Question #1 : Oxidation State

All AP Chemistry Resources

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