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AP Chemistry : Reaction Types

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Example Questions

Example Question #2 : Oxidation Reduction Reactions

For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?

$PF_2I_{(l)}+Hg_{(l)}\rightarrow P_2F_4_{(g)}+Hg_2I_2_{(s)}$

Possible Answers:

Two electrons are transferred; P is oxidized

One electron is transferred; Hg is oxidized

Two electrons are transferred; Hg is oxidized

One electron is transferred; P is oxidized

Correct answer:

Two electrons are transferred; Hg is oxidized

Explanation:

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.

Now we can begin to look at the half-reactions.

$PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

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Example Question #1 : Balancing Oxidation Reduction Reactions

How many electrons are involved in the following reaction?

$MnO_4^-+ Fe^{2+} \rightarrow Fe^{3+}+Mn^{2+}$

Possible Answers:

5 e^-

3 e^-

2 e^-

1 e^-

4 e^-

Correct answer:

5 e^-

Explanation:

The common factor between 1 e^- and 5 e^- is 5. Therefore the number of electrons involved is 5 e^-.

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Example Question #1 : Balancing Oxidation Reduction Reactions

How many electrons are involved in the following reaction?

$MnO_4^-+ I^-\rightarrow I_2+Mn^{2+}$

Possible Answers:

4 e^-

10 e^-

5 e^-

1 e^-

2 e^-

Correct answer:

10 e^-

Explanation:

The common factor between 2 e^- and 5 e^- is 10. Therefore the number of electrons involved is 10 e^-.

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Example Question #1 : Balancing Oxidation Reduction Reactions

What is the balanced coefficient on OH^- for the following reaction:

$Br^- + MnO_4^- \rightarrow MnO_2 + BrO_3^-$ (under basic conditions)

Possible Answers:

Correct answer:

Explanation:

Add them together:

$3 H_2 O + Br^- + 2 MnO_4^- + 8 H^+ + 6 e^- \rightarrow 2 MnO_2 + BrO_3^- + 6 H^+ + 4 H_2 O + 6 e^-$

Simplify:

$Br^- + 2 MnO_4^- + 2 H^+ \rightarrow 2 MnO_2 + BrO_3^- + H_2 O$

Add Hydroxides to each side to counter H⁺_.

$Br^- + 2 MnO_4^- + 2 H_2 O \rightarrow 2 MnO_2 + BrO_3^- + H_2 O+2 OH^-$

Simplify:

$Br^- + 2 MnO_4^- + H_2 O \rightarrow 2 MnO_2 + BrO_3^- +2 OH^-$

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Example Question #4 : Oxidation Reduction Reactions

What is the sum of all the balanced coefficients in the following reaction:

$O_2 + Sb \rightarrow H_2 O_2 + SbO_2^-$ (basic conditions)

Possible Answers:

Correct answer:

Explanation:

Add the equations together

$2Sb+4H_2 O + 3 O_2 + 6 H^+ + 6e^- \rightarrow 2SbO_2^-+8H^++3 H_2 O_2 + 6e^-$

Simplify

$2Sb+4H_2 O+3 O_2 \rightarrow 2SbO_2^-+2H^++3H_2 O_2$

Add 2 OH^- to each side to cancel out the H⁺_.

$2Sb+4H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+2H_2 O+3H_2 O_2$

Simplify:

$2Sb+2 H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+3H_2 O_2$

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Example Question #1 : Half Reactions

$Fe_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_{2(g)}$

For the redox reaction shown, which of the following half reactions occurs in the anode?

Possible Answers:

$2H_{2} + 2e^{-} \rightarrow H_{2}$

$Fe \rightarrow Fe^{3+} + 3e^{-}$

$Fe \rightarrow Fe^{2+} + 2e^{-}$

$Fe^{2+} + 2e^{-} \rightarrow Fe$

Correct answer:

$Fe \rightarrow Fe^{2+} + 2e^{-}$

Explanation:

Recall that oxidation always occurs at the anode (in both the electrochemical and galvanic cells). loses two electrons in this case to become $Fe^{2+}$ . The presence of $Fe^{2+}$ is hinted by the ionic compound FeCl_2 _.

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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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AP Chemistry : Reaction Types

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #2 : Oxidation Reduction Reactions

Example Question #1 : Balancing Oxidation Reduction Reactions

Example Question #1 : Balancing Oxidation Reduction Reactions

Example Question #1 : Balancing Oxidation Reduction Reactions

Example Question #4 : Oxidation Reduction Reactions

Example Question #1 : Half Reactions

Example Question #1 : Oxidation State

Example Question #1 : Oxidation State

Example Question #111 : Reactions And Equilibrium

Example Question #1 : Oxidation State

All AP Chemistry Resources

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