Honors text: Ch 20 & 21
Unit 12
Oxidation-Reduction (Redox) Reactions (4.4)
“Redox” reactions:
Oxidation & reduction always occur simultaneously
We use OXIDATION NUMBERS to keep track of electron transfers
Rules for Assigning Oxidation Numbers:
1) the ox. state of any free (uncombined) element is zero.
Ex:
2) The ox. state of an element in a simple ion is the charge of the ion.
Ex:
3) the ox. # for hydrogen is +1
(unless combined with a metal, then it has an ox. # of –1)
4) the ox. # of fluorine is always –1.
5) the ox. # of oxygen is usually –2.
6) in any neutral compound, the sum of the oxidation #’s = zero.
7) in a polyatomic ion, the sum of the oxidation #’s = the overall charge of the ion.
**use these rules to assign oxidation #’s; assign known #’s first, then fill in the #’s for the
remaining elements:
Examples: Assign oxidation #’s to each element:
a) NaNO3
d) H3PO4
b) SO32-
e) Cr2O72-
c) HCO3-
f) K2Sn(OH) 6
Oxidation-Reduction Reactions: oxidation & reduction always occur together (as one loses electrons,
the other gains them)
• oxidation:
• reduction:
• oxidizing agents:
• reducing agents:
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Honors text: Ch 20 & 21
Unit 12
A reaction is “redox” if a change in oxidation # happens; if no change in oxidation # occurs, the
reaction is nonredox.
Examples:
MgCO3

Zn +
CuSO4 
NaCl
+
AgNO3
CO2
+
H2O
MgO
+
ZnSO4 +
Cu


AgCl
C6H12O6
CO2
+
NaNO3
+
O2

Balancing Redox Equations
In balancing redox equations, the # of electrons lost in oxidation (the increase in ox. #) must equal
the # of electrons gained in reduction (the decrease in ox. #)

There are 2 methods for balancing redox equations:
1. Change in Oxidation-Number Method: based on equal total increases and decreases in oxidation #’s
Steps:
1) Write equation and assign oxidation #’s.
2) Determine which element is oxidized and which is reduced, and determine the change in
oxidation # for each.
3) Connect the atoms that change ox. #’s using a bracket; write the change in ox. # at the
midpoint of each bracket.
4) Choose coefficients that make the total increase in ox. # = the total decrease in ox. #.
5) Balance the remaining elements by inspection.
Example:
S
+
HNO3 
SO2
+
NO
+
H2O
If needed, reactions that take place in acidic or basic solutions can be balanced as follows:
Acidic:
Basic:
add H2O to the side needing oxygen
• add 2 OH- to the side needing
oxygen; and add 1 H2O to the other
side
then add H+ to balance the hydrogen
• then add 1 H2O to the side needing
hydrogen, and 1 OH- to the other side
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Honors text: Ch 20 & 21
Unit 12
Example: Balance the following equation, assuming it takes place in acidic solution.
ClO4- +
I
Cl+
I2
2. The Half-Reaction Method: separate and balance the oxidation and reduction half-reactions.
Steps:
1) write equation and assign oxidation #’s.
2) Determine which element is oxidized and which is reduced, and determine the change in oxidation
# for each.
3) Construct unbalanced oxidation and reduction half reactions.
4) Balance the elements and the charges (by adding electrons as reactants or products) in each halfreaction.
5) Balance the electron transfer by multiplying the balanced half-reaction by appropriate integers.
6) Add the resulting half-reaction and eliminate any common terms to obtain the balanced equation.
Example: Balance the following using the half-reaction method:
HNO3 +
H2S

NO
+
S
+
H2O
Electrochemical Cells (18.1-18.6)
Consider the reaction:
Zn(s) + CuSO4(aq)  Cu(s) + ZnSO4(aq)

If the 2 solutions are physically connected, but are connected by an external wire, electrons can
still be transferred through the wire.

Electrons flowing through a wire generate energy in the form of electricity.
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Honors text: Ch 20 & 21
Unit 12

An electrochemical cell is a container in which chemical reactions produce electricity or an electric
current produces chemical change.

When an electrochemical cell produces electricity, it is also known as a voltaic cell.

In the copper-zinc cell, the Cu and Zn electrodes (strips of each metal) are immersed in sulfate
solutions of their respective ions (CuSO4 and ZnSO4)

The solutions are separated by a porous barrier (prevents solutions from mixing, but allows ions
to pass through), or a salt bridge (any medium through which ions can pass slowly).

Cu2+ ions gain 2 electrons at the surface of the Cu strip, where they are deposited as Cu atoms:

Zn atoms in the Zn strip are losing electrons to become Zn 2+ ions in solution:

Voltaic cells are divided into 2 components called HALF CELLS: consist of a metal electrode in
contact with a solution of its own ions.

The ANODE is the half cell at which oxidation occurs; (a source of electrons)

The CATHODE is the half cell at which reduction occurs (use up electrons)
o electrons “flow” from left to right (anode to cathode)
o we can predict the “direction” of the electron flow in any given cell using the activity series
and reduction potentials for metals (see Table 18.1, p. 520)
o The more negative the Standard Reduction Potential value, the more likely a metal is to
“give up” its electrons (become oxidized) and serve as the anode.

The reaction in the cell will be spontaneous if the E° for the cell is positive; it is calculated as
follows:
E° cell = E° cathode – E° anode
• consider the zinc-copper cell:
Zn2+ + 2e-  Zn
E° =
Cu2+ + 2e-  Cu
E° =
-reduction of zinc has the lower value, so Zn is the anode
-what is the E° for the cell?
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Honors text: Ch 20 & 21
Unit 12
• consider a cell made from Al and Zn:
Zn2+ + 2e-  Zn
E° =
Al3+ + 3e-  Al
E° =
-what is the anode?
-what is the E° for the cell?
• Batteries are voltaic cells: a voltage is generated by a battery only if electrons continue to be
removed from 1 substance and transferred to another; when equilibrium is reached between the 2
half cells, the battery is “dead.”
• Rechargeable batteries: an external voltage source is applied to the battery’s electrodes and
reverses the half-reactions; this restores the electrodes to their original state.
• while the battery is being used, it operates as a voltaic cell (converts chemical energy into electric
energy)
• while the battery is being charged, it operates as an electrolytic cell (converts electric energy into
chemical energy)
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Honors text: Ch 20 & 21
Unit 12
Worksheet 1: Oxidation-Reduction Reactions
Part 1: For each of the compounds listed below, write the oxidation number of each
element in the compound.
1) H2O
11) MnO2
2) NiO2
12) NH2OH
3) Cd(OH) 2
13) H2O2
4) C2O42-
14) PBr3
5) Cr2O72-
15) ClO3-
6) MnO4-
16) HClO2
7) CN-
17) Pb(NO3)2
8) Cl2
18) CuSO4
9) H2CO3
19) N2O7
10) Fe(NO2)3
20) Pb3(PO4)4
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Honors text: Ch 20 & 21
Unit 12
Part 2: For each of the equations below, assign oxidation numbers to each element on
both the reactant and product sides. Determine if the reaction is “redox” or
“nonredox”, and list the reducing agent (RA) and oxidizing agent (OA) for all of the
redox reactions.
1) 2H2(g) + O2(g) 2H2O(g)
2) Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
3) Cd(s) + NiO2(s) + 2H2O(l)  Cd(OH)2(s) + Ni(OH)2(s)
4) 2H2O(l) + Al(s) + MnO4-(aq)  Al(OH) 4-(aq) + MnO2(s)
5) 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)
6) Cu(s) + 4H+(aq) + 2NO3-(aq)  Cu2+(aq) + 2NO2(g) + 2H2O(l)
7) I2O5(s) + 5CO(g)  I2(s) + 5CO2(g)
8) 2Hg2+(aq) + N2H4(aq)  2Hg(l) + N2(g) + 4H+(aq)
9) 3H2S(aq) + 2H+(aq) + 2NO3-(aq)  3S(s) + 2NO(g) + 4H2O(l)
10) Ba2+(aq) + 2OH-(aq) + H2O2(aq) + 2ClO2(aq)  Ba(ClO2)2(s) + 2H2O(l) + O2(g)
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Honors text: Ch 20 & 21
Unit 12
Worksheet 2: Balancing Redox
Balance the following using the half-rxn method…
In acidic sol’n:
a) Cu + NO3-  Cu2+ + NO
b) Cr2O72- + Cl-  Cr3+ + Cl2
c) Pb + PbO2 + H2SO4  PbSO4
In basic sol’n:
a) Al + MnO4-  MnO2 + Al(OH)4-
b) Cl2  Cl- + OCl-
c) NO2- + Al  NH3 + AlO2-
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Honors text: Ch 20 & 21
Unit 12
Wkst 3: Electrochemical Cells – Practice Problems
1. Write half-cell reactions and give standard reduction potentials for these reduction reactions. (use
table)
a) Cu+  Cu
b) Cl2  2Clc) Cu2+  Cu
d) Ba2+  Ba
e) Ag+  Ag
2. Determine which metal will be the anode, and E°cell for the following sets of half-reactions:
a)
Al3+ + 3e- ===> Al
E° =-1.66 V
Au3+ + 3e- ===> Au
E° =+1.50 V
b)
Li+ + e- ===> Li
Ag+ + e- ===> Ag
E° = -3.05 V
E° = +0.80 V
c)
Mg2+ +2e- ===> Mg
Fe3+ + 3e- ===> Fe
E° = -2.37 V
E° = -0.036 V
3. Two half-cells, one containing Fe2+ and Fe and the other containing Ag+ and Ag, are connnected to form
a voltaic cell. Use a Standard Reduction Potential table to determine the direction of spontaneous
reaction and the value for E°cell. Diagram the cell and label its parts. Give equations for the half
reactions.
4. Use a Standard Reduction Potential table to determine the E°cell value for the spontaneous reaction
of each pair of half-cells listed below:
a)
Ag ===> Ag+ + 1e-;
Fe2+ + 2e- ===> Fe
2+
b)
Mg ===> Mg + 2e ;
Sn2+ + 2e- ===> Sn
c)
Li ===> Li+ + 1e-;
Mn2+ + 2e- ===> Mn
3+
d)
Cr ===> Cr + 3e ;
Hg2+ + 2e- ===> Hg
e)
Ni ===> Ni2+ + 2e-;
Cu2+ + 2e- ===> Cu
5. Two half cells, one containing Ca2+ and Ca and the other containing Ag+ and Ag, are connected to form a
voltaic cell. Use a Standard Reduction Potential table to determine the direction of spontaneous reaction
and the value for E° cell. Diagram the cell and label its parts. Give equations for the half reactions.
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Honors text: Ch 20 & 21
Unit 12
6. Explain why a rechargeable battery can be considered a combination voltaic-electrolytic cell.
7. A voltaic cell is composed of the following half-cells:
Ca2+ + 2e-  Ca E° = -2.87 V
Fe3+ + e-  Fe2+ E° = +0.77 V
Write the reaction that takes place at the anode (oxidation) and the reaction that takes place at the
cathode (reduction). Calculate the standard cell potential (E° cell).
8. What is the standard cell potential for a voltaic cell composed of the following half-cells:
Cu2+ + 2e-  Cu E° = +0.34 V
Ag+ + e-  Ag E° = +0.80 V
Write the reaction that occurs at the anode (oxidation) and the cell reaction that occurs at the cathode
(reduction).
9. Is the following redox reaction spontaneous as written? (Hint: write oxidation and reduction halfreactions; look up the standard reduction potentials, and find E° cell; if E° is positive, the reaction is
spontaneous!)
2Ag+ + Ni  2Ag + Ni2+
10. Decide if the following redox reaction is spontaneous as written: (see hint in #9)
Cr3+ + Al  Cr + Al3+
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