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AP Chemistry : Reactions and Equilibrium

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Example Questions

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 #11 : Oxidation Reduction Reactions

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

Possible Answers:

3, 3, 3

3, 6, 3

6, 6, 3

3, 6, 6

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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Example Question #1 : Reduction Potential

The standard reduction potentials for certain metals are listed below:

$Au^3^+ + 3e^- \to Au$ Potential = 1.50V

$Ag^2^+ + 2e^- \to Ag$ Potential = 0.80V

$Cu^+ + e^- \to Cu$ Potential = 0.52V

$Ni^2^+ + 2e^- \to Ni$ Potential = -0.23V

$Fe^2^+ + 2e^- \to Fe$ Potential = -0.44V

What is the standard potential of a cell in which copper is reduced by iron?

Possible Answers:

0.08V

1.48V

0.60 V

0.96V

Correct answer:

0.96V

Explanation:

In order to find the potential of a galvanic cell in which copper is reduced by iron, we need to combine the two half reactions for the metals in question.

Since copper is reduced in the net ionic equation, we can use the reduction potential seen in the table.

$Cu^+ + e^- \to Cu$ Potential = 0.52V

Iron must be oxidized in order to reduce the copper. As a result, we must use the reverse reaction in the table. Remember that the table only gives reduction potentials; you will frequently need to charge the reaction to find an oxidation potential.

$Fe \to Fe^2^+ + 2e^-$ Potential = 0.44V

Notice how the potential for the half reaction has been switched as well. Combining these two reactions gives the following balanced reaction, once the electron transfer is balanced.

$2Cu^+ + Fe \to 2Cu + Fe^2^+$

Reduction potentials are intensive properties, so we do not multiply the potential of the copper half reaction by two. The total potential is simply the addition of both half reaction potentials.

0.52V+0.44V=0.96V

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Example Question #1 : Reduction Potential

The standard reduction potentials for some metals are listed below:

$Au^3^+ + 3e^- \to Au$ E^o=1.50

$Ag^2^+ + 2e^- \to Ag$ E^o=0.80

$Cu^+ + e^- \to Cu$ E^o=0.52

$Ni^2^+ + 2e^- \to Ni$ E^o=-0.23

$Fe^2^+ + 2e^- \to Fe$ E^o=-0.44

Which of the following metals is the strongest reducing agent?

Possible Answers:

The answer cannot be determined

Correct answer:

Explanation:

The reducing agent is the metal that will be oxidized in the reaction. Since all of the half reactions shown are reduction potentials, we need to switch the half reactions in order to determine which results in the greatest cell potential when the metal is oxidized. Iron will result in a cell potential of 0.44 volts when oxidized, which makes it the strongest reducing agent on the list.

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Example Question #2 : Reduction Potential

Standard Reduction Potentials

Cr³⁺_(aq)+ 3e^– → Cr (s)–0.74 V

Cu²⁺_(aq) +2e^– → Cu (s) 0.34 V

Consider the following reaction

Cu(s) + Cr³⁺(aq) ⇌ Cu²⁺(aq) + Cr(s)

What is the E^o Cell for the reaction?

Possible Answers:

–1.08 V

–1.76 V

0.46 V

–0.40 V

1.08 V

Correct answer:

–1.08 V

Explanation:

You do not multiply the coefficents that you need to balance to the Eo cell; you just have to see that the copper is being oxidized, so the sign changes (0.34 → –0.34) and the Cr is being reduced so the sign doesn't change. Then, just add the Eo cells: –0.74 + (–0.34) = –1.08.

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Example Question #1 : Principles Of Oxidation Reduction Reactions

Methane combusts in the presence of oxygen according to the following reaction:

$CH_4 + 2O_2 \to CO_2 + 2H_2O$

Which of the following statements is true concerning the reaction?

Possible Answers:

CH_4 will be the limiting reagent

Oxygen has a charge of $\small -2$ throughout the entire reaction

Carbon has an initial oxidation state of $\small +4$

Oxygen is oxidized in the reaction

Carbon is oxidized in the reaction

Correct answer:

Carbon is oxidized in the reaction

Explanation:

By comparing the oxidation number of an atom as a reactant and its oxidation number as a product, we can determine if the atom has been oxidized or reduced. In increase in oxidation number indicates a loss of electrons, or oxidation. A decrease in oxidation number signals a gain of electrons, or reduction.

For electrochemistry, you should familiarize yourself with the traditional oxidation states of hydrogen $\small (+1)$ , halogens $\small (-1)$ , oxygen $\small (-2)$ , and elemental atoms $\small (0)$ .

Carbon is initially in the form of methane, meaning that it is attached to four hydrogen atoms. The molecule is neutral, and each hydrogen has an oxidation number of $\small +1$ . Carbon must have an initial oxidation state of $\small -4$ in order to balance the molecular charge.

In the a product, carbon is attached to two oxygens, each with a charge of $\small -2$ . Again, the molecule is neutral, so carbon must balance these charges. This means that carbon's final oxidation state is $\small +4$ .

Since carbon went from an oxidation state of $\small -4$ to $\small +4$ , we can conclude that carbon has been oxidized in the reaction.

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Example Question #2 : Principles Of Oxidation Reduction Reactions

Which of the following reactions involves the transfer of electrons?

Possible Answers:

Neutralization

Redox

Decomposition

Combination

Correct answer:

Redox

Explanation:

Redox reactions occur when one species is oxidized—loses electrons—and another is reduced—gains electrons. Thus, this is the only one of the given options that involves the transfer of electrons.

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Example Question #2 : Principles Of Oxidation Reduction Reactions

Which statement about this reaction is FALSE?

$Cr_2O_3(s)+2Al(s)\rightarrow 2Cr(s)+Al_2O_3$

Possible Answers:

Chromium is oxidized during the reaction

This reaction is an example of a single displacement reaction.

Chromium is the oxidizing reagent for this reaction.

Aluminum is oxidized during the reaction.

Correct answer:

Chromium is oxidized during the reaction

Explanation:

In Cr₂O₃, Cr has a charge of +3. The chromium metal product is neutral, and therefore has a charge of 0. So, Cr is reduced from +3 to 0.

Remember, OIL RIG. Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons).

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AP Chemistry : Reactions and Equilibrium

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Oxidation State

Example Question #11 : Oxidation Reduction Reactions

Example Question #1 : Oxidation State

Example Question #1 : Oxidation State

Example Question #1 : Reduction Potential

Example Question #1 : Reduction Potential

Example Question #2 : Reduction Potential

Example Question #1 : Principles Of Oxidation Reduction Reactions

Example Question #2 : Principles Of Oxidation Reduction Reactions

Example Question #2 : Principles Of Oxidation Reduction Reactions

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