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AP Chemistry : Oxidation-Reduction Reactions

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Example Questions

Example Question #1 : Reactions And Equilibrium

What is the balanced equation when heptane is combusted?

Possible Answers:

2 C₇H₁₄ + 5 O₂ → 3 H₂O + 6 CO₂

C₅H₁₂ + 8 O₂ → 6 H₂O + 5 CO₂

C₃H₈ + 5 O₂ → 4 H₂O + 3 CO₂

C₇H₁₆ + O₂ → H₂O + CO₂

C₇H₁₆ + 11 O₂ → 8 H₂O + 7 CO₂

Correct answer:

C₇H₁₆ + 11 O₂ → 8 H₂O + 7 CO₂

Explanation:

Heptane: C₇H₁₆. Combustion is when a molecule reacts with O₂ and the products are CO₂ and H₂O. Balancing gives 7 CO₂, 8 H₂O, 1 heptane, and 11 O₂

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Example Question #1 : Oxidation Reduction Reactions

Which of the following reactions has the most exothermic heat of reaction?

Possible Answers:

The combustion of propane

The combustion of butane

The combustion of isobutane

The combustion of methane

The combustion of isopropane

Correct answer: The combustion of butane

Explanation:

The longer the hydrocarbon chain, the greater the amount of combustion products (CO2 and H20) generated. Branched molecules such as isopropane and isobutane are more difficult to combust than their straight-chain counterparts.

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Example Question #1 : Reaction Types

Which of the following conditions would describe a combustion reaction?

Possible Answers:

Endergonic with a negative $\Delta H$

Exothermic with a positive $\Delta H$

Exothermic with a negative $\Delta H$

Endothermic with a positive $\Delta H$

Exergonic with a positive $\Delta H$

Correct answer:

Exothermic with a negative $\Delta H$

Explanation:

An exothermic reaction will have a negative $\Delta H$ value, indicating that it releases heat. Conversely, an endothermic reaction will have a positive $\Delta H$ value, indicating a consumption of heat.

A combustion reaction releases heat; thus it must have a negative $\Delta H$ value and be exothermic.

Exergonic reactions have a negative $\Delta G$ value, indicating spontaneity, while endergonic reactions are non-spontaneous. While most combustion reactions will be non-spontaneous, it is impossible to draw this conclusion for certain without knowing more about the reaction. The only thing we know for certain is that heat is released, and the reaction is exothermic.

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Example Question #52 : Reaction Types

The following ReDox reaction takes place in acidic solution:

Fe²⁺ + Cr₂O₇^2– → Fe³⁺ + Cr³⁺

What is the sum of coefficients in this redox reaction?

Possible Answers:

Correct answer:

Explanation:

When you balance the redox reaction in acidic conditons, there are 6Fe²⁺, 1 Cr₂O₇^2–, 14 H⁺, 6 Fe³⁺, 2 Cr³⁺, and 7 H₂O. Don't forget to add the 1 in front of the Cr₂O₇^2–

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Example Question #31 : Chemical 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; Hg is oxidized

One electron is transferred; Hg is oxidized

Two electrons are transferred; P 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:

4 e^-

2 e^-

3 e^-

5 e^-

1 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 #2 : Oxidation Reduction Reactions

How many electrons are involved in the following reaction?

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

Possible Answers:

10 e^-

2 e^-

5 e^-

4 e^-

1 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 #2 : 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 #62 : Reaction Types

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

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

Possible Answers:

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

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

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

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

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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AP Chemistry : Oxidation-Reduction Reactions

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Reactions And Equilibrium

Example Question #1 : Oxidation Reduction Reactions

Example Question #1 : Reaction Types

Example Question #52 : Reaction Types

Example Question #31 : Chemical Reactions

Example Question #1 : Balancing Oxidation Reduction Reactions

Example Question #2 : Oxidation Reduction Reactions

Example Question #2 : Balancing Oxidation Reduction Reactions

Example Question #4 : Oxidation Reduction Reactions

Example Question #62 : Reaction Types

All AP Chemistry Resources

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