AP Chemistry Chapter 21 Notes
(Student’s edition)
Chapter 21 problems: suggested - 11, 19, 22, 33, 54, 55, 63, 80, 82, 84, 95 additional 2, 3, 4, 92, 93
Because chemical reactions involve movement of electrons, spontaneous redox reactions produce
electricity in a . This supplies electricity to an external source.
Non-spontaneous reactions are found to happen in an by adding electricity (an
external source).
Cell: the .
Electrodes: provide entrance and exit of .
21.1 Electrical Conduction
This is the through metals, pure ionic liquids, or solutions containing electrolytes.
Metallic conduction – atoms don’t move, no chemical change
Ionic conduction (electrolytic conduction) – motion of ions through a solution toward electrodes
of opposite charge.
21.2 Electrodes
This is the surface upon which oxidation and reduction occur.
Electrodes may or may not participate in the reaction (some electrodes are inert)
“ ”
Anode – oxidation occurs/ electrons are lost
Cathode – reduction occurs/ electrons are gained
***each of these could be positive or negative so…
In electrolysis – the cathode is and the anode is
In voltaic cells – the anode is and the cathode is
Electrolytic cells – non-spontaneous reactions are forced to happen by the input of electricity.
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21.3 Electrolysis of Molten Sodium Chloride
Students should take notes on this section in their notebook
The Downs Cell:
The sodium is and chloride is . Sodium is taking from Chlorine. This
is .
Overall cell reaction: ______
Electrons are produced at the carbon anode (+)…and then used at the iron cathode (-).
21.4 Electrolysis of Aqueous Sodium Chloride
Students should take notes on this section in their notebook
In electrolysis, the cathode is negative.
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Anode: Oxidation: 2 Cl-1(aq) à Cl2(g) + 2 e-1
Cathode: Reduction: 2 H2O(l) + 2 e-1 à OH-1(g) + 2H2(g)
Overall cell reaction: 2 H2O(l) + 2 Cl-1(aq) à Cl2(g) + OH-1(g) + 2H2(g)
The Na+1 produced is considered a spectator ion in the above overall reaction. This Na+1
ion will combine with Cl-1 to produce NaCl or it will combine with OH-1 to
produce NaOH.
This is the most important commercial preparation of H2 , Cl2, and NaOH.
Note that the Na+1 ions are not reduced to metallic Na. Instead gaseous H2 and aqueous
OH-1 are produced by the reduction of water. This is due to the fact that water is more easily reduced than Na+1. Also, the products of H2 and OH-1 are more stable than the products of Na+1 reduction (solid Na).
21.5 Electrolysis of Aqueous Sodium Sulfate
Students should take notes on this section in their notebook
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Anode: Oxidation: 2 H2O(l) à O2(g) + 4 H+1 + 4 e-1
Cathode: Reduction: 4 H2O(l) + 4 e-1 à 4 OH-1(g) + 2 H2(g)
Overall cell reaction: 2 H2O(l) à O2(g) + 2H2(g)
Note that the Na+1 ions are not reduced to metallic Na. Instead gaseous H2 and aqueous
OH-1 are produced by the reduction of water. This is due to the fact that water is more easily reduced than Na+1. Also, water is more easily oxidized than SO4-2.
21.6 Counting Electrons: Coulometry and Faraday’s Law of Electrolysis
Faraday’s Law: The amount of substance oxidized or reduced at an electrode is directly
proportional to the amount of electricity that passes through the cell.
Units of electricity: 1 Faraday = passage of 6.02 x 1023 electrons
1 Coulomb = 1 ampere for 1 second
1 Faraday = 96,500 C
1 mole of electrons = 96,500 C
Faradays’ Law can be used to calculate quantity of material, time or current needed, and
oxidation state of metals.
Example1: Calculate the mass of solid copper produced by passing 2.50 amps of
electricity through a solution of CuSO4 for 50.0 minutes ( seconds).
Step 1: write out the half reaction for Cu…
Cu+2(aq) + 2 e-1 à Cu(s)
Step 2: Calculate the Coulombs…
C = (Amperes)(seconds) à C = ( amps)( sec) = C
Step 3: Calculate the amount of solid copper produced using stoichiometry…
C x 1 mole e-1 x mole Cu x 63.5 g Cu = g Cu
96,500 C mole e-1 1 mole Cu
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Example2: How long will it take a current of 5.00 amps passed through a silver nitrate
solution to produce 10.5 grams of Silver.
Step 1: write out the half reaction for Ag…
Step 2: Calculate the amount of Coulombs produced using stoichiometry…
10.5 g Ag x 1 mole Ag x 1 mole e-1 x 96,500 C = C
107.9 g Ag 1 mole Ag 1 mole e-1
Step 3: Calculate the time in seconds and minutes from the Coulombs…
C = (amp)(sec) à C = (5.00 amp)(x) à x = sec or min
Example3: A current of 3.00 amps for 1.00 hour (3600.00 seconds) passed through a solution of
copper ions produced 3.55 grams of solid copper. Calculate the oxidation number of the
copper ions.
Step 1: Calculate the moles of electrons…
C = (3.00 amps)(3600 sec) = 10,800 C
1 mole e-1 = x moles e-1 à x = moles e-1
96,500 C 10,800 C
Step 2: Calculate the moles of copper…
3.55 g Cu x 1 mole Cu = moles Cu
63.5 g Cu
Step 3: Calculate the oxidation number of the copper ions…
Moles of electrons / moles of copper à
Cu?(aq) + __e-1 à Cu(s)
Thus, Cu is a charge
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21.7 Common Applications of Electrolytic Cells
We’ve already seen how electrolytic cells can produce sodium (not found in nature).
Also used for electroplating. Students should take notes on this section in their notebook. Student should also read this section.
21.8 Construction of Simple Voltaic Cells and 21.9 and 21.10 Examples of Voltaic Cells
Voltaic cells –spontaneous reactions happen without the input of electricity.
Also called Cells. These are cells in which spontaneous redox reactions
produce electricity. Oxidation and reduction occur in separate compartments and electrons flow from one side to the other. This is known as a .
Cell notation: anode (oxidation)║ cathode (reduction). In all voltaic cells, the anode is negative and the cathode is positive. Students should take notes on these sections in their notebook.
Standard Electrode Potentials ( section 21-14): The potential (voltage) of a cell is determined by
the tendency of the reaction to go to completion. A positive value = higher tendency
The standard potential of the zinc copper voltaic cell (diagram above):
Zn(s) + Cu+2(aq) à Zn+2(aq) + Cu(s) à 1.10 volts
The standard potential of the silver copper voltaic cell (diagram above):
Cu(s) + Ag+1(aq) à Cu+1(aq) + Ag(s) à +.46 volts
Zinc has of a tendency to give electrons to copper than copper does to give electrons to silver.
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Whether a given electrode acts as an anode or cathode depends on what the other electrode is.
Zn+2 < Cu+2 < Ag+1 Zn > Cu > Ag
Increasing strength as oxidizing agents Increasing strength as reducing agents
The magnitude of a cell’s potential measures the of its redox reaction.
Higher (more ) cell potentials indicate greater driving force for the reaction as written.
21.11 The Standard Hydrogen Electrode and 21.12 and 21.13 Some metal-SHE cells
Students should take notes on these sections in their notebook.
Because a cell is always made of two half cells, it is impossible to determine the potential
of a single half cell. The SHE (standard hydrogen electrode) is assigned a value
of exactly 0 volts.
SHE as anode: H2 à 2 H+1 + 2 e-1 0.000 Volts
SHE as cathode: 2 H+1 + 2 e-1 à H2 0.000 Volts
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Anode, Oxidation: Zn à Zn+2 + 2 e-1 .763 Volts
Cathode, Reduction: 2 H+1 + 2 e-1 à H2 0.000 Volts
Overall Cell Reaction: Eocell = Volts
Note that Zinc gets and Hydrogen gets
Page 9
The Copper – SHE Cell:
Anode, Oxidation: H2 à 2 H+1 + 2 e-1 0.000 Volts
Cathode, Reduction: 2 Cu+2 + 2 e-1 à Cu 0.337 Volts
Overall Cell Reaction: Eocell = Volts
Note that Hydrogen gets and Cu+2 gets . The H2 electrode has the higher
electron pressure…so, electrons flow from the electrode to the electrode.
21.15 Uses of Standard Electrode Potentials and 21.16 SEP for Other Half-Reactions
The electromotive series (activity series) of the elements – developed from standard
reduction potentials (Look at the SRP table – that is, if you haven’t already…)
Li+1 + e- → Li -3.045 V (not going to happen – actually, most likely to get oxidized)
K+1 + e- → K -2.925 V (not going to happen – actually, it will get oxidized)
Ca+2 + 2e- → Ca -2.87 V (not going to happen – actually, it will get oxidized)
2 H+ + 2 e- → H2 0.000 V (depends on what type of cell it’s connected to)
(reference electrode)
Cu+2 + 2 e- → Cu + .337 V (will get reduced)
Ag+1 + e- → Ag + .799 V (will get reduced)
Au+3 + 3 e- → Au + 1.50 V (of the metals listed above, it is most likely to get reduced)
F2 + 2 e- → 2 F-1(aq) + 2.87 V (fluorine; however, always wins...It is most likely to be reduced)
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Example1: Calculate the cell potential for: Cu+2 + K → K+1 + Cu
Anode, Oxidation: 2 ( K à K+1 + e-1 ) Volts
Cathode, Reduction: Cu +2 + 2 e-1 à Cu Volts
Overall Cell Reaction: 2 K + Cu+2 à 2 K+1 + Cu Eocell = Volts
Example2: Will Cr+3 oxidize Cu to Cu+2?
or
Will Cu+2 oxidize Cr to Cr+3?
or (said another way):
Which reaction is spontaneous? CrCl3 + Cu → CuCl2 + Cr
or
Cr + CuCl2 → Cu + CrCl3
Step 1: Answer… Will Cr+3 oxidize Cu to Cu+2?
Oxidation: à Volts
Reduction: à Volts
Overall Cell Reaction: Eocell = Volts
Thus, Cr+3 ______Cu to Cu+2
Step 2: Answer… Will Cu+2 oxidize Cr to Cr+3?
Oxidation: à Volts
Reduction: à Volts
Overall Cell Reaction: Eocell = Volts
Thus, Cu+2 ______Cr to Cr+3
This is a ______electrode reduction and therefore will be ____ likely than the
______electrode reduction in step #1.
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Example3: Can Cl2 oxidize Mn+2 to MnO4-1 in acidic solution and be reduced to Cl-1?
So… Cl2 + Mn+2 → Cl-1 + MnO4-1
Step 1: Oxidation: balance oxygen by adding H2O to the side that needs oxygen,
then balance the H’s in the H2O by adding H+1 , and finally add electrons to balance the charge.
+7
Mn+2 → MnO4-1
+7
Mn+2 + 4 H2O → MnO4-1 + 8 H+1 + 5 e-1
Step 2: Reduction: balance charge by adding electrons
Cl2 → 2 Cl-1
Cl2 + 2 e-1 → 2 Cl-1
Step 3: Balance the electrons:
2 (Mn+2 + 4 H2O → MnO4-1 + 8 H+1 + 5 e-1 )
5 (Cl2 + 2 e-1 → 2 Cl-1 )
Step 4: Combine the half-reactions and reduce:
2 Mn+2 + 8 H2O + 5 Cl2 + 10 e-1 → 2 MnO4-1 + 16 H+1 + 10 e-1 + 10 Cl-1
2 Mn+2 + 8 H2O + 5 Cl2 → 2 MnO4-1 + 16 H+1 + 10 Cl-1
Now, for this chapter and the reduction potentials…
Oxidation: 2 Mn+2 + 8 H2O → 2 MnO4-1 + 16 H+1 + 10 e-1
Mn+2 + 4 H2O → MnO4-1 + 8 H+1 + 5 e-1
Reduction: 5 Cl2 + 10 e-1 → 10 Cl-1
Cl2 + 2 e-1 → 2 Cl-1
So…. (ox) Mn+2 + 4 H2O → MnO4-1 + 8 H+1 + 5 e-1 V
(red) Cl2 + 2 e-1 → 2 Cl-1 V
Eocell = Volts
Thus, Cl2 ______oxidize Mn+2 to MnO4-1 in acidic solution and be reduced to Cl-1
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Example4: Calculate the cell potential for:
+4
MnO2 + 4 H+1 + Pd → Mn+2 + 2 H2O + Pd+2
Oxidation: à Volts
Reduction: à Volts
Overall Cell Reaction: Eocell = Volts
21.17 Corrosion and 21.18 Corrosion Protection
Students should read these sections on their own.
21.19 The Nernst Equation
For notes on section 12.19, refer to the text – it’s the exact same question, solving for
concentrations instead of Ecell (pluggin’ and chuggin’)
The Nernst Equation is used to determine cell potentials at conditions. All
previous problems assumed equimolar concentrations and standard pressure.
The Nernst Equation: E cell = Eo – (2.303RT/nF) log Q
(R = 8.314 J/molK, T(in K), F = Faraday’s constant)
(Eo = standard potential)(E = nonstandard potential) so….
The simplified Nernst Equation.. E cell = Eo – (0.0592/n) log Q (n = # electrons)
Example1: If [Mn+2] = 0.500 M and [Al+3] = 1.50 M, then calculate E cell for:
2 Al(s) + 3 Mn+2(aq) → 2 Al+3(aq) + 3 Mn(s)
Step 1: Calculate the E at standard conditions
Oxidation: Volts
Reduction: Volts
Eocell = Volts
Step 2: Calculate the Q given [Mn+2] = 0.500 M and [Al+3] = 1.50 M …
Q = [Al+3]2 / [Mn+2]3 à Q = ( )2 / ( )3 à Q =
Step 3: Calculate the E cell at the above non-standard conditions…
E cell = Eo – (0.0592/n) log Q
E cell =
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Example2: Calculate E cell for the following situation…
Zn|Zn+2(0.00100 M)║Cu+2(10.0 M)|Cu
Step 1: Calculate the E at standard conditions…
Oxidation: Zn à Zn+2 + 2 e-1 Volts
Reduction: Cu+2 + 2 e-1 à Cu Volts
Eocell = Volts
Step 2: Calculate the Q given [Zn+2] = 0.00100 M and [Cu+2] = 10.0 M …
Zn(s) + Cu+2(aq) → Zn+2(aq) + Cu(s)
Q = [Zn+2] / [Cu+2] à Q = ( ) / ( ) à Q =
Step 3: Calculate the E cell at the above non-standard conditions…
E cell = Eo – (0.0592/n) log Q
E cell =
Example3: Calculate E cell for:
Cr2O7-2 (aq) + 14 H+1(aq) + 6 I-1(aq) → 2 Cr+3(aq) + 3 I2(s) + 7 H2O(l)
When [dichromate] = 2.00 M, [H+1] = 1.0 M, [I-1] = 1.0 M, [Cr+3] = 1 x 10-5 M
Step 1: Calculate the E at standard conditions…
Oxidation: 6 I-1 à 3 I2 + 6 e-1 -.53 Volts
Reduction: Cr2O7-2 + 14 H+1 + 6 e-1 à 2 Cr+3 + 7 H2O +1.33 Volts
Eocell = Volts
Step 2: Calculate the Q given the above concentrations and reaction…
Q = [Cr+3]2 / [Cr2O7-2][I-1]6[H+1]14
Q = (.00001)2 / (2)(1.0)6(1.0)14 à Q =
Step 3: Calculate the E cell at the above non-standard conditions…