Chemistry 12
Unit 8 ~ Learning Guide Name:
Instructions:
Using a pencil, complete the following notes as you work through the related lessons. Show ALL
work as is explained in the lessons. You are required to have this package completed BEFORE
you write your unit test. Do your best and ask questions if you don’t understand anything!
Please print out the Formula Sheet and found at the start of the course. You will need this
to do assignments and tests!
The Electrochemical Cell:
1. What does ULIRBLRWIO stand for?
Upper Left Is Reduced By Lower Right Which Is Oxidised
2. Will you ever double or triple an Eo value? Explain.
E0 is the potential difference of a particular cell and is an intensive property (does not depend
on amount). Thus, it cannot be doubled or tripled.
3. What cell was used to measure the Eo values of all other cells? What is the Eo value for this cell?
The standard Hydrogen Reference Cell. The E0 value is zero.
4. If a reduction half reaction had Eo = -4.52, what would be the E0 value of the same reaction in
reverse (oxidation)?
The reduction half-reaction E0 = -4.52 and E0reduction = -E0oxidation thus the oxidation half
reaction would be E0= +4.52.
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Chemistry 12
5. Determine if the reaction below is spontaneous by calculating the net Eo.
3 Hg(l)
+ 2 NO3- + 8 H+  3 Hg2+ + 2 NO(g)
E0 Hg2+/Hg = +0.85V (anode)
E0NO3-/NO = +0.96V (cathode)
E0 = E0cathode – E0anode
E0 = 0.96 – 0.85 = 0.11V
Hence, the reaction is spontaneous.
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+ 4 H2O
Chemistry 12
6. What would the net Eo be for the reverse reaction of the one shown in question 5. Would it be
spontaneous?
For the reverse reaction, E0 = -0.11V, thus it would be non-spontaneous.
7. What process (oxidation or reduction) occurs at the anode in an electrochemical cell?
Oxidation
8. What is the purpose of the salt bridge in an electrochemical cell?
A salt bridge is necessary to keep the charge flowing through the cell. Without a salt
bridge, the electrons produced at the anode would build up at the cathode and the
reaction would stop running.
9. Will Hg2+ ions react with water to produce Hg metal? Explain. Be careful, the overpotential
effect comes into play here!
When one dissolves a mercuric salt in water, the mercuric ions do not react with the water to
form metallic mercury. However, if the water is impure, and full of contaminants, you might
see some reaction overnight. Fehling's solution, for example, will reduce Hg+ to HgO, which,
under the right conditions, can cause the mercury metal to precipitate out as a silvery mirror
on the walls of a test tube. When made with Ag+ instead of Hg+ and placed on a the cleaned
surface of flat glass.
10. What are two common terms for the oxidation of a metal?
Rusting and/or corrosion.
11. What are the only two metals that can be used to cathodically protect Rubidium? Bonus: Whyis
this unlikely to work very well?
Magnesium And Zinc are the two metals that are used for the cathodic protection of
Rubidium. It prevents the corrosion by forming metals surface as cathode in the
electrochemical cell.
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Chemistry 12
12.A very expensive electrochemical cell is set up with a silver electrode in a solution of silver
nitrate in the first beaker and a gold electrode in a solution of gold nitrate in the second beaker.
The electrodes are connected with a wire and a salt bridge is used to connect the two beakers.
a) Write the two half reactions with their correct Eo. Indicate which is undergoing reduction
and which is undergoing oxidation. Determine the net reaction and the net Eo.
The two half-reactions for the given cell are
Gold: Au3+ + 3 e- → Au0 Here gold ions in gold nitrate solution is undergoing reduction
reaction with an E0= +1.50 V
Silver: Ag0 - 1 e- → Ag1+ Here silver electrode is undergoing oxidation reaction with an
Eo= -0.80 V
The net reaction is: Ag + AuNO3 → AgNO3 + Au
Net EMF= E0 oxidation + E0 reduction = 1.50-0.80 = 0.7 V
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Chemistry 12
b) Identify the cathode and the anode in this cell.
Gold acts as the cathode and Silver acts as the anode.
c) Identify the species reduced and the species oxidized.
Gold(III) cation is reduced to Gold(0) and Silver (0) is oxidized to Silver (I) cation
d) In this cell (as in all electrochemical cells) electrons move from the :
Anode to Cathode
13. Consider the electrochemical cell shown below?
a) Write the two half reactions with their correct Eo. Indicate which is undergoing reduction
and which is undergoing oxidation. Determine the net reaction and the net Eo.
E0 for Pb2+ +2e- --> Pb is = -0.13V (reduced)
E0 for Mn --> Mn2+ +2e- is = 1.18V (oxidised)
E0cell = 1.18 – 0.13 = 1.05V
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Chemistry 12
b) As the reaction proceeds which ion will decrease in concentration and which will increase?
As the reaction proceeds, the Pb2+ ion concentration will decrease and the Mn2+ ion
concentration will increase.
c) Which electrode will increase in mass as the reaction proceeds, the cathode or the anode?
Cathode mass will increase.
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Chemistry 12
Electrolytic Cells:
1. What are the two most common materials used to make inert electrodes?
Carbon and Platinum.
2. Platinum electrodes are used in a molten solution of AlBr3 in an electrolytic cell. Write the two
half reactions that occur with their correct Eo. Indicate which is undergoing reduction and which
is undergoing oxidation. Determine the net reaction and the net Eo.
Oxidation (anode): 6Br- --> 3Br2 + 6eReduction (cathode): 2Al3+ +6e- --> 2Al(s)
(E0= +1.09)
(E0= -1.66)
Overall:
6Br-(l) + 2Al3+(l) --> 3Br2(s) + 2Al(s)
E0 = -1.66 – 1.09 = -2.75V
3. What is the complication added when comparing a type II electrolytic cell to a type I electrolytic
cell?
In a type-II electrolytic cell, aqueous solutions are used for electrolysis. Thus, oxidation and
reduction happen in addition with dissolved cations or anions.
In a type-I electrolytic cell, molten salts are used. Thus, only anion and cation
oxidation/reduction occurs (there is no water present.)
Also the polarity of the electrodes are swaped between the types of electrolytic cells.
4. How do you determine which species will be reduced in an electrolytic cell?
The species with the higher oxidation potential.
5. How do you determine which species will be oxidized in an electrolytic cell?
The species with the higher reduction potential.
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Chemistry 12
6. Direct current is passed through inert electrodes in a 1.00 M solution of CuF. Identify which
species will be oxidized and reduced and then write the two half reactions and the net reaction for
the events which occur. Also calculate the net E° value.
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Chemistry 12
7. For a Tin ion | Tin || Silver | Silver ion cell (this could be abbreviated as Sn2+ | Sn || Ag | Ag+ )
a) Write two half reactions and the net reaction. (Water will not be oxidized or reduced in
thiscell). Calculate the net Eo value.
Sn2+(aq) + 2e- --> Sn(s) (cathode) E0 = -0.14V
2Ag(s) --> 2Ag+(aq) +2e- (anode) E0 = -0.80V
Net reaction:
Sn2+(aq) + 2Ag(s) --> 2Ag+(aq) + Sn(s)
E0cell = -0.14 – 0.80 = -0.940V
b) Identify the oxidizing agent.
Sn2+(aq) is the oxidation agent
c) Identify the reducing agent.
Ag(s) is the reducing agent
d) Identify this as an electrolytic or an electrochemical cell based on your Net Eo calculation.
Electrolytic cell
e) Is this reaction spontaneous?
Non-spontaneous
8. What are the two common purposes type III electrolytic cells are used for?
9. If you want to plate something with a metal, does it need to be the anode or the cathode? Explain
The cathode because that is where reduction occurs.
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Chemistry 12
Answers:
5. Yes, it would be spontaneous
12.
a) Au3+
15.
2 Al3+ + 6 I-
+ 3 Ag(s)
 Au(s) + 3 Ag+
 2 Al(g) + 3 I2(g)
20. e. No, it is not spontaneous
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6
Net Eo = +0.70
Net Eo = -2.20