Name: _______________________
Block: _______
Chemistry 12 Unit 5- Electrochemistry
Lesson 1: Redox Reactions
Oxidation – Reduction
Definitions:
(species means atom, ion or molecule)
Oxidation – a species undergoing oxidation loses electrons
(charge becomes more positive)
Reduction – a species undergoing reduction gains electrons
(charge becomes more negative)
Oxidizing agent – The species being reduced
(gains electrons, causes the other one to be oxidized)
Reducing agent – The species being oxidized
(loses electrons, causes the other one to be reduced)
2 e-
E.g.)
Cu2+ (aq) + Zn (s)
 Cu(s) + Zn2+(aq)
Oxidizing
agent
LEO says GER
Losing
Electrons is
Oxidization
Gaining
Electrons is
Reduction
Redox – Short for Oxidation – Reduction
Charge on neutral atom or molecule = 0
Oxidation – Charge gets more + (loses electrons)
Reduction – Charge gets more – (gains electrons)
Reduction (charge decreases)
E.g.) Pb2+(aq) + Mg0(s)  Pb0(s) + Mg2+(aq)
Oxidation (Charge increases)
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Question
In the reaction:
2Fe2+ + Cl2  2Fe3+ + 2ClIdentify:
a)
b)
c)
d)
e)
f)
g)
h)
The Oxidizing Agent: _____________________
The species being oxidized:________________
The reducing agent:______________________
The species being reduced:________________
The species gaining electrons:______________
The species losing electrons:_______________
The product of oxidation___________________
The product of reduction___________________
Do Ex. 1 (a-e) pp. 192 SW
Half-Reactions
-Redox reactions can be broken up into oxidation & reduction half reactions.
e.g.) Redox rx:
Pb2+(aq) + Zn(s)  Pb(s) + Zn2+(aq)
The Pb2+ (loses/gains) __________ 2 electrons.
Reduction Half-rxn:
Pb2+(aq) + 2e-  Pb(s)
Oxidation Half-rxn:
Note: Half-rx’s always have e-‘s, redox (oxidation-reduction) reactions never show e-‘s!
Given the redox reaction:
F2(g) + Sn2+(aq)  2F-(aq) + Sn4+(aq)
Write the oxidation half-rxn:
__________________________
Write the reduction half-rxn:
__________________________
Do ex. 2 a-c on p. 192
SW
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Block: _______
Oxidation Numbers
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Block: _______
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Name: _______________________
Block: _______
Practice
1. Given the redox reaction:
2MnO4- + 3C2O42- + 4H2O  2MnO2 + 6CO2 + 8OHFind:
a) The species being reduced: _____________.
b) The species undergoing oxidation: _____________.
c) The oxidizing agent: ________________.
d) The reducing agent: ________________.
e) The species gaining electrons: ______________.
f) The species losing electrons: ______________.
2. Given the balanced redox reaction:
3S + 4HNO3  3SO2 + 4NO + 2H2O
Find:
a)
b)
c)
d)
e)
f)
g)
h)
The oxidizing agent: _____________.
The reducing agent: _____________.
The species being reduced: ___________.
The species being oxidized: ___________.
The species losing electrons: ____________.
The species gaining electrons: ____________.
The product of oxidation: ____________.
The product of reduction: ____________.
3. Given the following:
6Br2 + 12KOH  10KBr + 2KBrO3 + 6H2O
Find:
a)
b)
c)
d)
e)
f)
The oxidizing agent: ______________.
The reducing agent: ______________.
The species undergoing oxidation: _____________.
The species being reduced: ________________.
The product of oxidation: _______________.
The product of reduction: _______________.
Read p. 193-195 of SW.
Do Exercise 3 on p. 194 of SW.
Do Exercises 4, 5 and 6 on p. 194-195 of SW.
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Name: _______________________
Block: _______
Lesson 2: Predicting Spontaneous Reactions
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Block: _______
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Block: _______
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Name: _______________________
Block: _______
Assignment: Using data to make your own simple Redox table
Example problem:
1) Four metals A, B, C, & D were tested with separate solutions of A2+, B2+, C2+ & D2+.
Some of the results are summarized in the following table:
Solution
2+
Metal
A
B2+
C2+
D2+
(1) no reaction
(2) reaction
A
(4) no reaction
B
(3) reaction
D
List the ions in order from the strongest to weakest oxidizing agent.
Using data
NOTE: For the same element:
The more positive species is
always the Oxidizing Agent.
Eg.) A2+ A
OA
RA
Another example –
Four non-metal oxidizing agents X2, Y2, Z2 and W2 are combined with solutions of ions:
X-, Y-, Z- and W-.
The following results were obtained;
(1) X2 reacts with W- and Y- only.
(2) Y- will reduce W2
List the reducing agents from strongest to weakest
Do Exercises 14,15,16 & 18 on p. 200 of SW.
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Name: _______________________
Block: _______
Lesson 3: Balancing Half-Reactions
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Name: _______________________
Block: _______
Balancing Redox Reactions Using Half-Reactions
(1) Break up Rx into 2 half-rx’s.
(2) Balance each one (in acidic or basic as given)
(3) Multiply each half rx by whatever is needed to cancel out e-‘s
(Note: balanced half-rx have e-‘s (on left reduction on right oxidation) Balanced redox don’t
have e-‘s)
(4) Add the 2 half-rx’s canceling e-‘s and anything else (usually H2O’s, H+’s or OH-‘s) in order
to simplify.
Example: U4+ + MnO4-  Mn2+ + UO22+ (acidic)
Balance each ½ rx
U4+  UO22+
(Major (U) balanced already)
Oxygen  U4+ + 2H2O  UO22+
Hydrogen  U4+ + 2H2O  UO22+ + 4H+
Charge  U4+ + 2H2O  UO22+ + 4H+ + 2e-
MnO4-  Mn2+
(Major (Mn) balanced already)
MnO4-  Mn2+ + 4H2O
MnO4- + 8H+  Mn2+ + 4H2O
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
Try this one:
SO2 + IO3-  SO42- + I2 (basic solution)
-See examples p.205-207 in SW
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Name: _______________________
Block: _______
Quick notes
-Some redox equations have just one reactant
- Use this as the reactant in both half-rx’s.
- These are called “self-oxidation-reduction” or Disproportionation reactions.
Eg)
Br2  Br-- + BrO3- (basic) (found in some hot tubs)
Half rx’s are:
Br2  Br--
Br2  BrO3-
Answer: ___________________________________________________
Do Ex 24 a-m p. 207
Extension: Balancing using Oxidation Numbers
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Name: _______________________
Block: _______
Practice
1.
U4+ + MnO4-  Mn2+ + UO22+ (acidic)
2.
SO2 + IO3-  SO42- + I2 (basic solution)
3.
Br2  Br-- + BrO3- (basic)
-This is optional
- As long as one method (not guessing!) works for you that’s fine. (This method
or half-rx method.)
- Read examples p. 271-272 SW
- Do any ex 10 a-d, f, j, m & check with key
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Name: _______________________
Block: _______
Lesson 4: Redox Titrations
Read p. 210-212 carefully – go over the examples! Do ex 26 & 29 p. 213-214 SW.
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Name: _______________________
Block: _______
Lesson 5: Electrochemical Cells
3.
Demo: Cu/Zn Cell
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Block: _______
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Name: _______________________
Block: _______
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Name: _______________________
Block: _______
Consider the following cell:
e-
Metal
“X”
Cr
1M X(NO3)2
1M Cr(NO3)3
The voltage on the voltmeter is 0.45 volts.
a) Write the equation for the half-reaction taking place at the anode. Include the Eo.
______________________________________________ Eo: _________v
b) Write the equation for the half-reaction taking place at the cathode.
______________________________________________ Eo: _________v
c) Write the balanced equation for the redox reaction taking place as this cell operates.
Include the Eo.
______________________________________________Eo: __________
d) Determine the reduction potential of the ion X2+.
Eo:
__________v
e) Toward which beaker (X(NO3)2) or (Cr(NO3)3) do NO3- ions migrate?
_______________________ _
f)
Name the actual metal “X” ________________________________
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Name: _______________________
Block: _______
Consider the following cell:
e-
Ag
Cd
AgNO3
Cd(NO3)2
The initial cell voltage is 1.20 Volts
a) Write the equation for the half-reaction which takes place at the cathode. Include the Eo
____________________________________________________ Eo= ____ ___v
b) Write the equation for the half-reaction taking place at the anode:
____________________________________________________ Eo= ____ ___v
c) Write the balanced equation for the overall redox reaction taking place. Include the Eo.
____________________________________________________ Eo= ____ ___v
d) Find the oxidation potential for Cd: Eo= ____ ___v
e)
Find the reduction potential for Cd2+: Eo= ____ ___v
f)
Which electrode gains mass as the cell operates? _______
g) Toward which beaker (AgNO3 or Cd(NO3)2) do K+ ions move? _______
h) The silver electrode and AgNO3 solution is replaced by Zn metal and Zn(NO3)2 solution.
What is the cell voltage now? __________Which metal now is the cathode? _________
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Name: _______________________
Block: _______
Consider the following electrochemical cell:
Ni
Cu
Ni(NO3)2
Cu(NO3)2
a) Write the equation for the half-reaction taking place at the nickel electrode. Include the Eo
____________________________________________________ Eo= ____ ___v
b) Write the equation for the half-reaction taking place at the Cu electrode. Include the Eo.
____________________________________________________ Eo= ____ ___v
c) Write the balanced equation for the redox reaction taking place.
____________________________________________________ Eo= ____ ___v
d) What is the initial cell voltage? _________________ _
e) Show the direction of electron flow on the diagram above with an arrow with an “e-“ written above it.
f) Show the direction of flow of cations in the salt bridge using an arrow with “Cations” written above
it.
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Name: _______________________
Block: _______
Challenge:
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Block: _______
Lesson 6: Standard Reduction Potentials
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Block: _______
Read SW. p. 215-224
Do Ex. 35 p. 217 and Ex. 36 a-d & 37-45 on p. 224-226 of SW
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Name: _______________________
Block: _______
Lesson 7: Electrolysis
3 Types
Type 1:
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Block: _______
Type 2:
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Block: _______
Overpotential Effect
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Block: _______
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Block: _______
Type 3:
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Block: _______
Lesson 8: Applications of Electrochemistry- Part 1: Electroplating and Electrorefining
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Block: _______
Demo: Copper Plating
Anode Half Reaction:
Cathode Half Reaction:
Overall Reaction:
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Block: _______
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Block: _______
Lesson 9: Applications of Electrochemistry- Part 2: CORROSION
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Block: _______
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