AP Chemistry : Balancing Oxidation-Reduction Reactions

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

Example Question #1 : Balancing Oxidation Reduction Reactions

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

Possible Answers:

37

33

34

35

36

Correct answer:

36

Explanation:

When you balance the redox reaction in acidic conditons, there are 6Fe2+, 1 Cr2O72–, 14 H+, 6 Fe3+, 2 Cr3+, and 7 H2O. Don't forget to add the 1 in front of the Cr2O72–

Example Question #1 : Balancing Oxidation Reduction Reactions

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

Possible Answers:

One electron is transferred; P is oxidized

One electron is transferred; Hg is oxidized

Two electrons are transferred; Hg is oxidized

Two electrons are 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 \displaystyle +1. Phosphorus begins with an oxidation number of \displaystyle +3 and ends with an oxidation number of \displaystyle +2. Note that the oxidation numbers for fluorine and iodine reamain constant at \displaystyle -1 for each.

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

Balance the atoms.

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

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.

Example Question #1 : Balancing Oxidation Reduction Reactions

How many electrons are involved in the following reaction?

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

Possible Answers:

2 e-

5 e-

1 e- 

3 e-

4 e-

Correct answer:

5 e-

Explanation:

02

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

Example Question #111 : Reactions And Equilibrium

How many electrons are involved in the following reaction?

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

Possible Answers:

10 e- 

1 e-

2 e-

4 e-

5 e-

Correct answer:

10 e- 

Explanation:

01

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

Example Question #1 : Balancing Oxidation Reduction Reactions

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

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

Possible Answers:

5

4

1

3

2

Correct answer:

2

Explanation:

03

 

Add them together:

\displaystyle 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: 

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

 

Add Hydroxides to each side to counter H+.

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

 

Simplify:

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

Example Question #112 : Reactions And Equilibrium

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

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

Possible Answers:

12

14

16

8

10

Correct answer:

14

Explanation:

04

Add the equations together

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

 

Simplify

\displaystyle 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+.

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

 

Simplify:

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

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