ChemQuest 82
Name: ____________________________
Date: _______________
Hour: _____
Introduction Questions
increases
1. During oxidation, the oxidation number of a substance ___________
due to the loss
increases OR decreases
_________ of electrons.
loss OR gain
2. Let’s assume that substance X gets oxidized. Which equation do you think is the best
way to represent oxidation? [Note: “e-“ stands for an electron.]
A) X+2 + 2 e-  X
OR
Notice that the charge
on X goes from +2 to 0
B) X  X+2 + 2 eNotice that the charge
on X goes from 0 to +2
3. Let’s assume that substance Y gets reduced. Which equation do you think is the best way
to represent reduction?
A) Y+3 + 3 e-  Y
OR
B) Y  Y+3 + 3 e-
4. In the previous two questions, we saw that X+2 had two electrons involved oxidation. Y+3
had three electrons involved in its reduction. Write the equation for the reduction of Z+1.
Z+1 + e-  Z
Information: Half Reactions
A reaction in which substances undergo oxidation and reduction are called “redox reactions.”
Sometimes the reactions are written with ions in them. Here’s a common redox reaction in
which one substance is oxidized and the other is reduced:
Zn (s) + Cu+2 (aq)  Zn+2 (aq) + Cu (s)
Zn  Zn+2
We can take all of the zinc substances to
make a zinc “half reaction”
Cu+2  Cu
We can take all of the copper substances
to make a copper “half reaction”
Critical Thinking Questions
5. The above half reactions need some electrons added to one of the sides so that they will
look similar to questions 2 and 3.
a) Label each half reaction as “oxidation” or “reduction.” Also, follow the pattern of
questions 2 and 3 and add electrons to the correct side of each equation.
oxidation
___________________:
Zn (s)
reduction
___________________:
Cu+2 (aq)

Zn+2 (aq)

Cu (s)
b) How many electrons did you add? Why did you choose this number?
2
6. Split the following reaction into two half reactions—an oxidation and a reduction half
reaction. Write each half reaction with the proper number of electrons in each.
a) Ag+ + Ni  Ag + Ni+2
Oxidation ½ reaction:
Ni  Ni+2 + 2 e-
Reduction ½ reaction:
Ag+ + e-  Ag
b) I2 + Sn  2 I- + Sn+2
Oxidation ½ reaction:
Sn  Sn+2 + 2e-
Reduction ½ reaction (done for you ): I2 + 2e-  2 IQuestion: Why are there 2 electrons instead of one?
Each iodine gains an electron.
c) Mn+2 + 2 Br-  Mn + Br2
Oxidation ½ reaction: 2 Br-  Br2 + 2 eReduction ½ reaction: Mn+2 + 2e-  Mn
7. True or False: A substance is reduced by gaining electrons. TRUE
Information: Reduction Potential
Potential: having the likelihood of occurring.
In electrochemistry there is a term, “reduction potential.” The reduction potential is a measure of
how likely a substance is to be reduced. Reduction potential is measured in volts (V) and is
given the symbol Eo. The more positive the reduction potential (Eo), the more likely a substance
is to be reduced.
Table 1: Some common reduction potential values:
Reduction Half Reaction
F2 + 2e  2F
Co3+ + e-  Co2+
Cl2 + 2e-  2ClO2 + 4H+ + 4e-  2H2O
Br2 + 2e-  2BrHg2+ + 2e-  Hg
Ag+ + e-  Ag
Fe3+ + e-  Fe2+
I2 + 2e-  2ICu2+ + 2e-  Cu
Cu2+ + e-  Cu+
Sn4+ + 2e-  Sn2+
2H+ + 2e-  H2
Fe3+ + 3e-  Fe
Pb2+ + 2e-  Pb
Sn2+ + 2e-  Sn
Ni2+ + 2e-  Ni
Co2+ + 2e-  Co
Cr3+ + e-  Cr2+
Cd2+ + 2e-  Cd
Fe2+ + 2e-  Fe
Cr3+ + 3e-  Cr
Zn2+ + 2e-  Zn
2H2O + 2e-  H2(g) + 2OHV2+ + 2e-  V
Mn2+ + 2e-  Mn
Al3+ + 3e-  Al
Mg2+ + 2e-  Mg
-
-
Eo (in Volts)
+2.87
+1.80
+1.36
+1.23
+1.07
+0.85
+0.80
+0.77
+0.54
+0.34
+0.15
+0.15
0.00
-0.04
-0.13
-0.14
-0.25
-0.29
-0.40
-0.40
-0.41
-0.74
-0.76
-0.83
-1.18
-1.18
-1.66
-2.37
Critical Thinking Questions
8. True or False: Pb2+ is more likely to gain electrons than Br2.
FALSE
9. Use the above table to rank the following in order from most likely to least likely to be
reduced: Ni2+, Cl2, Mn2+, Sn4+
Cl2, Sn+4, Ni+2, Mn+2
10. If oxidation is the opposite of reduction, which do you think is most likely to be oxidized:
Cd2+ or Fe3+?
Cd+2
11. Which of the following is most likely to be the best reducing agent—Pb2+ or F2?
Pb+2 because it is most likely to be oxidized
12. The Eo value for Cl2 + 2e-  2Cl- is +1.36 V. If we reverse the reaction so that Cl- is
being oxidized, the reaction looks like this:
2Cl-  Cl2 + 2e- oxidation potential = ???
To find the oxidation potential for this new oxidation reaction, what should we do to the
+1.36 value?
A) Change the sign to make it negative
B) divide it by two C) square it
Hint: What makes the most sense based on what we did for Hess’s Law?