6.5: Classifying Chemical Reactions (Redox) (Problems)

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1. Contributors

Some of these problems represent a culmination of all the reaction types we've covered in Unit 6!

PROBLEM \(\PageIndex{1}\)

Indicate what type, or types, of reaction each of the following represents:

1. \(\ce{Ca}(s)+\ce{Br2}(l)\rightarrow \ce{CaBr2}(s)\)
2. \(\ce{Ca(OH)2}(aq)+\ce{2HBr}(aq)\rightarrow \ce{CaBr2}(aq)+\ce{2H2O}(l)\)
3. \(\ce{C6H12}(l)+\ce{9O2}(g)\rightarrow \ce{6CO2}(g)+\ce{6H2O}(g)\)

Answer a

oxidation-reduction (addition)

Answer b

acid-base (neutralization)

Answer c

oxidation-reduction (combustion)

PROBLEM \(\PageIndex{2}\)

Indicate what type, or types, of reaction each of the following represents:

1. \(\ce{H2O}(g)+\ce{C}(s)\rightarrow \ce{CO}(g)+\ce{H2}(g)\)
2. \(\ce{2KClO3}(s)\rightarrow \ce{2KCl}(s)+\ce{3O2}(g)\)
3. \(\ce{Al(OH)3}(aq)+\ce{3HCl}(aq)\rightarrow \ce{AlBr3}(aq)+\ce{3H2O}(l)\)
4. \(\ce{Pb(NO3)2}(aq)+\ce{H2SO4}(aq)\rightarrow \ce{PbSO4}(s)+\ce{2HNO3}(aq)\)

Answer a

oxidation-reduction (single displacement)

Answer b

oxidation-reduction (dissociation)

Answer c

acid-base (neutralization)

Answer d

precipitation (double replacement)

PROBLEM \(\PageIndex{3}\)

Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.

Answer

It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.

PROBLEM \(\PageIndex{4}\)

Determine the oxidation states of the elements in the following compounds:

1. NaI
2. GdCl3
3. LiNO3
4. H2Se
5. Mg2Si
6. RbO2 (rubidium superoxide)
7. HF

Answer a

Na +1, I -1

Answer b

Gd +3, Cl -1

Answer c

Li +1, N +5, O -2

Answer d

H +1, Se -2

Answer e

Mg +2, Si -4

Answer f

Rb +1, O -1/2

Answer g

H +1, F -1

Click here to see a video of the solution

PROBLEM \(\PageIndex{5}\)

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

1. H3PO4
2. Al(OH)3
3. SeO2
4. KNO2
5. In2S3
6. P4O6

Answer a

H +1, P +5, O −2

Answer b

Al +3, H +1, O −2

Answer c

Se +4, O −2

Answer d

K +1, N +3, O −2

Answer e

In +3, S −2

Answer f

P +3, O −2

PROBLEM \(\PageIndex{6}\)

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

1. H2SO4
2. Ca(OH)2
3. BrOH
4. ClNO2
5. TiCl4
6. NaH

Answer a

H +1, O -2, S +6

Answer b

H +1, O -2, Ca +2

Answer c

H +1, O -2, Br +1,

Answer d

O -2, Cl -1, N +5

Answer e

Cl -1, Ti +4

Answer f

H +1, Na -1

PROBLEM \(\PageIndex{7}\)

Classify the following as acid-base reactions or oxidation-reduction reactions:

1. \(\ce{Na2S}(aq)+\ce{2HCl}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{H2S}(g)\)
2. \(\ce{2Na}(s)+\ce{2HCl}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{H2}(g)\)
3. \(\ce{Mg}(s)+\ce{Cl2}(g)\rightarrow \ce{MgCl2}(s)\)
4. \(\ce{MgO}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{H2O}(l)\)
5. \(\ce{K3P}(s)+\ce{2O2}(g)\rightarrow \ce{K3PO4}(s)\)
6. \(\ce{3KOH}(aq)+\ce{H3PO4}(aq)\rightarrow \ce{K3PO4}(aq)+\ce{3H2O}(l)\)

Answer a

acid-base

Answer b

oxidation-reduction: Na is oxidized, H+ is reduced

Answer c

oxidation-reduction: Mg is oxidized, Cl2 is reduced

Answer d

acid-base

Answer e

oxidation-reduction: P3− is oxidized, O2 is reduced

Answer f

acid-base

PROBLEM \(\PageIndex{8}\)

Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:

1. \(\ce{Mg}(s)+\ce{NiCl2}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{Ni}(s)\)
2. \(\ce{PCl3}(l)+\ce{Cl2}(g)\rightarrow \ce{PCl5}(s)\)
3. \(\ce{C2H4}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{2H2O}(g)\)
4. \(\ce{Zn}(s)+\ce{H2SO4}(aq)\rightarrow \ce{ZnSO4}(aq)+\ce{H2}(g)\)
5. \(\ce{2K2S2O3}(s)+\ce{I2}(s)\rightarrow \ce{K2S4O6}(s)+\ce{2KI}(s)\)
6. \(\ce{3Cu}(s)+\ce{8HNO3}(aq)\rightarrow \ce{3Cu(NO3)2}(aq)+\ce{2NO}(g)+\ce{4H2O}(l)\)

Answer a

Mg: 0 \(\rightarrow\) +2; loses electrons; oxidized; reducing agent

Ni: +2 \(\rightarrow\) 0; gains electrons; reduced; oxidizing agent

Answer b

P: +3 \(\rightarrow\) +5; loses electrons; oxidized; reducing agent

Cl: 0 \(\rightarrow\) -1; gains electrons; reduced; oxidizing agent

Answer c

C: -2 \(\rightarrow\) +4; loses electrons; oxidized; reducing agent

O: 0 \(\rightarrow\) -2; gains electrons; reduced; oxidizing agent

Answer d

Zn: 0 \(\rightarrow\) +2; loses electrons; oxidized; reducing agent

H: +1 \(\rightarrow\) 0; gains electrons; reduced; oxidizing agent

Answer e

S: +2 \(\rightarrow\) +5/2; loses electrons; oxidized; reducing agent

I: 0 \(\rightarrow\) -1; gains electrons; reduced; oxidizing agent

Answer f

Cu: 0 \(\rightarrow\) +2; loses electrons; oxidized; reducing agent

N: +5 \(\rightarrow\) +2; gains electrons; reduced; oxidizing agent

Click here to see a video of the solution

PROBLEM \(\PageIndex{10}\)

When heated to 700–800 °C, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction.

Answer

\(\ce{C_{diamond}}(s)+\ce{O2}(g)\rightarrow \ce{CO2}(g)\)

PROBLEM \(\PageIndex{11}\)

The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?

Answer

\(\ce{H2}(g)+\ce{F2}(g)\rightarrow \ce{2HF}(g)\)

PROBLEM \(\PageIndex{12}\)

In a common experiment in the general chemistry laboratory (that you will do if you haven't already), magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of Mg3N2, a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.

Answer

\(\ce{2Mg}(s)+\ce{O2}(g)\rightarrow \ce{2MgO}(s)\)

\(\ce{3Mg}(s)+\ce{N2}(g)\rightarrow \ce{Mg3N2}(s)\)

PROBLEM \(\PageIndex{13}\)

Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid hydrogen sulfate as the only product. Write the two balanced equations which represent these reactions.

Answer

\(\ce{CuS}(s)+\ce{2O2}(g)\rightarrow \ce{SO3}(g)+\ce{CuO}(s)\)

\(\ce{SO3}(g)+\ce{H2O}(l)\rightarrow \ce{H2SO4}(l)\)

Contributors

• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]).

• Adelaide Clark, Oregon Institute of Technology