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Electrochemical Cells
There are two types: Galvanic and Electrolytic
NOTES:
Galvanic Cell: a cell in which a _______________________________________ is used to
produce electrical energy,
i.e., Chemical energy is transformed into Electrical energy.
Electrolytic Cell: a cell in which electrical energy is used to bring about
__________________________________________,
i.e., electrical energy is transformed into chemical energy.
will be dealt with later on.)
(This type
Refer to the cover of the CHEM*1050 lab manual which shows the Zinc–Copper cell as
well as Figures 19.2 and 19.3.
Consider the Zinc–Copper Cell:
If you place Zn metal in an aqueous CuSO4 solution:
1.
Cu2+ ions plate out on the Zn surface;
2.
At the same time the Zn metal dissolves to yield Zn2+ ions;
3.
The solution also becomes warmer.
4.
The blue colour fades.
(You will be studying this in Experiment 5: Galvanic Cells.)
What is happening? Reaction?
If we arrange things properly we can generate electrical energy. To "see" how this cell
operates in detail, view the animated Zn/Cu galvanic cell:
http://www.chemistry.uoguelph.ca/educmat/chm19105/galvanic/galvanic1.htm
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NOTES:
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-7 (Pg 871)
What does this process tell us?
1. Zn(s) has a greater tendency to ______________________________ than Cu(s)
2. Cu2+(aq) has a greater tendency to _________________________ than Zn2+(aq).
3. Zn(s) is a stronger ___________________________________ than Cu(s).
4. Cu2+(aq) is a stronger _________________________________ than Zn2+(aq).
NOTE, the cell is constructed such that
• have a Zn/Zn2+(aq) and a Cu/Cu2+(aq) _______________________________
• metal electrodes are connected via a _______________________________.
• _____________________________________________ links the solutions.
• Oxidation occurs at the _______________________ (connected to negative terminal
of voltmeter)
• Reduction occurs at the _______________________ (connected to positive terminal
of voltmeter)
Note: If the anode is connected to the negative terminal of the voltmeter, the voltage
reading is always positive. Thus, the anode and cathode can be determined experimentally
by using a voltmeter (i.e., if hooked up backwards, the voltage reading will be negative).
Why a Salt Bridge?
Zn2+(aq) is produced in the _______________ compartment
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Cu2+(aq) is consumed in the ______________ compartment
NOTES:
To maintain electroneutrality, we need to get _________________ into the anode (to
counter the build up of positive charge) and ________________ into the cathode (to counter
the loss of positive charge).
This is where the SALT BRIDGE comes in (e.g., KCl and gel in a glass tube).
It completes the electrical circuit by allowing ____________________________ (anions to
anode and cations to cathode); maintains electroneutrality.
It also prevents _____________________________ between Cu2+(aq) and Zn(s).
At the ANODE,
The Zn(s) goes into solution and 2e– go along the wire to the Cu plate.
Anode compartment ___________________________________________
Cl–(aq) ions pop out of the Salt Bridge to maintain ___________________.
At the CATHODE,
The 2 electrons entice a Cu2+ ion to plate out of the solution
Cathode compartment ______________________________________________
________________________ pop out of the Salt Bridge to maintain neutrality.
Note: A porous separator can also be used to separate the two solutions (as shown above).
OTHER POINTS:
•
If we have 1.0 M Cu2+(aq) and 1.0 M Zn2+(aq), we get a voltage of exactly +1.10
volts, no matter how ______________________________________ (the anode is
connected to the negative terminal of the voltmeter and the cathode is connected to
the positive terminal).
•
The "products", Zn2+(aq) and Cu(s), are ____________________________; they
will be made as soon as the current flows.
•
The concentrations of BOTH cations [Cu2+] and [Zn2+] DO affect the cell voltage
– more on this later.
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NOTES:
Another Voltaic cell:
Hydrogen - Copper Cell
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-15 (page 878).
Anode: An inert platinum surface is placed in H+(aq) with H2(g) bubbling past it.
(When we have 1.0 M solutions, this is a standard hydrogen electrode-see below.)
H2(g) → 2H+(aq) + 2e–
(______________ occurs at the anode)
(______________ occurs at the cathode)
Cu2+(aq) + 2e– → Cu(s)
___________________________________________________________
H2(g) + Cu2+(aq) → 2H+(aq) + Cu(s)
If we have 1.0 M solutions and H2(g) at 1 atm, we get a cell voltage
of exactly +0.34 volts.
Other types of electrodes and half–cells:
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NOTES:
Reference: Olmsted & Williams, Chemistry, 3rd ed., Figure 18-9 (page 872)
Reaction: H2(g) + 2Fe2+(aq) → 2H+(aq) + 2Fe3+(aq)
Oxidation Rxn:
Reduction Rxn:
Note: Within the two half cell reactions there are no metals. This creates a problem because
we need an electrode in each half cell to allow for the flow of electrons from the anode
compartment to the cathode compartment.
(Another example: Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)
The two half cells are made up of only ions in solution – no metals present
electrodes.)
to act as
ANSWER: INERT ELECTRODES
Use a solid, inert conductor in the solution, e.g., ____________________________, and let
the ions in solution transfer electrons to or from the surface of the inert electrode.
The inert electrode does not participate in the redox reaction taking place within the halfcell, but it is included in the design of the galvanic cell since it completes the electrical
circuit.
Question : The galvanic cell below uses the half-cells Z+/Z and Y2+/Y,
bridge containing MX(aq). The voltmeter gives a positive voltage reading.
and a salt
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Identify C.
NOTES:
A.
B.
C.
D.
E.
Z+
Z
M+
X–
Y
Cell Diagrams & Conventional Shorthand Notation
(Refer to section 19.3 )
Instead of cumbersome cell drawings we usually use a shorthand notation
specify the structure of the cell.
e.g. Zinc - Copper Cell:
to
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
Salt
Cathode
Bridge
• The Anode is ALWAYS on the _____________ (and the Cathode on the
________________).
• "|" - indicates a junction between different ___________________________, e.g., sol’
& metal, sol’n & gas or metal & compound as in Ag(s) & AgCl(s).
• "||" – indicates the ______________________ that connects the two sol’ns.
• States are indicated, but ____________________________ in the eq’ns are not.
• Usual convention is _______________________________ for each half rxn.
Examples:
Zinc - Chlorine Cell:
Tin (II) - Iron (III) Cell:
Zn(s) | Zn2+(aq) || Cl–(aq) | Cl2(g) | Pt(s)
Au(s) | Sn4+(aq), Sn2+(aq) || Fe3+(aq), Fe2+(aq) | C(s, graphite)
Note: Tin(IV) and tin(II), as well as iron (III) and iron (II), are separated by a commas
since there is no phase change (i.e., both are in aqueous solution).
1. Write the cell notation for: 2Ag(s) + Hg2Cl2(s) → 2AgCl(s) + 2Hg(l)
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NOTES:
2. Write the cell notation for:
b)
a)
Reference: Olmsted & Williams,
Chemistry, 3rd ed., Figure 18-17 (pg.
884)
Ref: Jones & Atkins, Chemistry: Molecules,
Matter and Change, 4th ed., Fig. 18.10 (p.
803).
a)
b)
Question #3: The following reaction occurs in basic solution,
2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)
The cell notation for this voltaic cell is
A) Ag+(aq) | Cu(s) || Ag(s) | Cu2+(aq).
B) Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s).
C) Cu(s) | Cu2+(aq) || 2Ag+(aq) | 2Ag(s).
D) Ag(s) | Ag+(aq) || Cu2+(aq) | Cu(s).
E) Cu(s) | Cu2+(aq) || Ag(s) | Ag+(aq) .
Question #4: Consider the following cell notation:
Pt(s) | H2(g) | H+(aq) || Cl2(g) | Cl−(aq) | Pt(s).
The balanced reaction for this voltaic cell is
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A) Pt(s) + H2(g) + Cl2(g) → H+(aq) + Cl−(aq) + Pt(s).
+
NOTES:
−
B) Pt(s) + H2(g) + Cl2(g) → 2H (aq) + 2Cl (aq) + Pt(s).
C) H2(g) + Cl2(g) → 2H+(aq) + 2Cl−(aq).
D) 2H+(aq) + 2Cl−(aq) → H2(g) + Cl2(g).
E) Pt(s) + 2H+(aq) + 2Cl−(aq) → H2(g) + Cl2(g) + Pt(s).
Standard Cell Potentials, E°
_______________________ is a measure of the driving force, or spontaneity, of a chemical
reaction.
STANDARD conditions are the same as in thermodynamics, i.e., solutes ________, gases
___________, pure _____________________________________.
We cannot measure potentials of individual half–cells, we can only measure the potential
of two half cells coupled together.
To solve this problem, we arbitrarily assign a voltage of 0 to the
___________________________________________, then measure/report all other
standard half cells relative to it.
Standard Hydrogen Electrode :
2H+ + 2e– → H2(g), E° = 0.000 V
In a Galvanic Cell, the standard cell potential (E°) is the difference between the standard
reduction potentials of the two electrodes:
E°(cell) = E°(cathode) – E°(anode)
where the E° values at the cathode (where reduction takes place) and the anode (where
oxidation takes place) are both written as ____________________.
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NOTES:
EXAMPLE: Zinc-Hydrogen Cell:
Zn(s) | Zn2+(aq) || H+(aq) | H2(g) | Pt(s) for which E°(cell) = _____________ V
For this cell, reduction takes place in the H+/H2 half-cell and oxidation
in the Zn2+/Zn half-cell.
takes place
_______________________________
Zn(s) + 2H (aq) → Zn2+(aq) + H2(g) E° = 0.76 (measured)
+
This gives,
E°(cell) = E°(H+/H2)– E°(Zn2+/Zn)
+0.76 = 0.00 – E°(Zn2+/Zn)
∴ E°(Zn2+/Zn) = –0.76 V.
Thus, the standard reduction potential for Zn2+(aq) + 2e– → Zn(s) is – 0.76 V.
CONVENTION: Always report half–cell potentials in the form:
Ox + ne– → Red
i.e., Standard Reduction Potentials.
Now look back at the Standard Zn(s)/Zn2+(aq) and Cu(s)/Cu2+(aq) cell:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) for which E°(cell) = 1.10 V.
takes place
For this cell, reduction takes place in the Cu2+/Cu half-cell and oxidation
2+
in the Zn /Zn half-cell.
Ox:
Red:
_______________________________
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s); E° = +1.10 V
E°(Cu2+/Cu) = +0.34 V.
Therefore, Cu2+(aq) + 2e– → Cu(s)
E° = +0.34 V
By proceeding in this way we can build up a Table of Standard Electrode Potentials or
Standard Reduction Potentials. These tables are VERY USEFUL!
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Sample of a Standard Reduction Potential Table
Ox + ne– → Red
Potential in Volts
F2(g) + 2e– → 2F–(aq)
+2.87
Au3+(aq) + 3e– → Au(s)
+1.50
Cl2(g) + 2e– → 2Cl–(aq)
+1.36
Ag+(aq) + e– → Ag(s)
+0.80
Fe3+(aq) + e– → Fe2+(aq)
+0.77
Cu2+(aq) + 2e– → Cu(s)
+0.34
Sn4+(aq) + 2e– → Sn2+(aq)
+0.15
2H+(aq) + 2e– → H2(g)
0.000 (by convention)
Cd2+(aq) + 2e– → Cd(s)
–0.40
Zn2+(aq) + 2e– → Zn(s)
–0.76
Na+(aq) + e– → Na(s)
–2.71
Li+(aq) + e– → Li(s)
–3.05
Check out the table of standard reduction potentials in Appendix I. In this table the Standard Potentials
are listed in electrochemical order – very useful when comparing reactivity.
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Reference: Appendix 2B, page A14 from the text Chemistry - Molecules, Matter and Change,
P. Atkins & L. Jones, 4th ed., W.H. Freeman and Company, New York, 2000.
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OBSERVATIONS from the table above:
__________ is the BEST oxidizing agent (and the most easily reduced)
__________ is the WORST oxidizing agent (and the most difficult to reduce)
__________ is the BEST reducing agent (and the most easily oxidized)
__________ is the WORST reducing agent (and the most difficult to oxidize)
Consider the following: Which metals were first known and used by early civilizations? Why?
Comment on their relative positions in the table.
How do we obtain Standard Cell Potentials from
Standard Reduction Potentials?
NOTES:
What would the Standard Cell Potential be for a Cu|Cu2+ & Zn|Zn2+ cell?
Step 1: Write out the two standard reduction potentials (SRP) equations:
Step 2: The half–reaction with the more positive SRP goes as written (is the
reduction reaction) and thus the lower SRP (the oxidation reaction) is always
subtracted from the higher SRP.
E°(cell) = E°(cathode) – E°(anode)
E°(cell) = E°(
/
) – E°(
/
)
Step 3: Now,
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq), E°(cell) = +0.34 – (–0.76) = 1.10V
Comments:
Note that the E° value is POSITIVE (this is true for all voltaic or galvanic cells),
i.e., the reaction is spontaneous as we have written it (under Standard Conditions).
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NOTES:
Now look at the Ni|Ni2+ and Ag|Ag+ cell.
What is the spontaneous cell reaction, anode, cathode, E°?
Step 1: Get SRPs from Tables
Ni2+(aq) + 2e– → Ni(s)
E° =
Ag+(aq) + e– → Ag(s)
E° =
Step 2: The half–reaction with the more positive SRP goes as written and thus the
lower SRP is always subtracted from the higher SRP.
E°(cell) =
Step 3: Write the overall reaction equation and the cell potential
Does anything strike you as odd about this??
Although we multiplied the coefficients in the silver half–cell by 2, we did NOT
double the half–cell voltage.
Why NOT ??
Compare:
1) Ag+(aq) + e– → Ag(s)
2) 2Ag+(aq) + 2e– → 2Ag(s)
The ENERGY release by equation (2) is certainly twice that of (1), but remember
that Voltage = Joules/Coulombs (energy per coulomb of charge transferred)
Equation (2) involves twice the energy, but it also involves twice the coulombs – so
the VOLTAGE does not change. Voltage is an Intensive property.
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Question: Would the reaction: Al3+(aq) + Mg(s) → Al(s) + Mg2+(aq) go as
written under standard sonditions, or would the reverse reaction occur?
NOTES:
Step 1: Get SRP equations
Step 2: The half–reaction with the more positive SRP goes as written and thus the
lower SRP is always subtracted from the higher SRP.
E°(cell) =
Step 3: Write the overall equation and cell potential:
Answer: (Yes/No) the reaction (WOULD/WOULD NOT) occur as written.
Question #5: Consider the following standard electrode potentials:
Ag+(aq) + e– → Ag(s)
E° = 0.80 V
2+
–
Mn (aq) + 2e → Mn(s) E° = –1.18 V
For the standard cell:
Mn(s)⏐Mn2+(aq)⏐⏐Ag+(aq)⏐Ag(s)
which of the following statements are true?
A. The cell voltage will be 1.98 V.
B. The half-cell reaction is Mn(s) → Mn2+(aq) + 2e–.
C. The overall cell potential will decrease with time.
Question #6: For a galvanic cell using Fe⏐Fe2+(1.0 M) and Pb⏐Pb2+(1.0 M) half-cells, which of
the following statements is correct?
Fe2+ + 2e– → Fe; E° = – 0.44 V
Pb2+ + 2e– → Pb; E° = – 0.13 V
A) The mass of the iron electrode increases during discharge.
B) Electrons leave the Pb electrode to pass through the external circuit during discharge.
C) The concentration of Pb2+ decreases during discharge.
D) The iron electrode is the cathode.
E) When the cell has completely discharged (to 0 V), the concentration of Pb2+ is zero.
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Problem 1: With access to an SRP table, deduce what reaction (if any) will occur when:
(a) Metallic Sn is added to 1 M HCl(aq) solution.
(b) A solution of 1.0 M Sn2+(aq) is treated with
(i) metallic Zn:
(ii) 1.0 M FeCl3(aq) sol’n (no solids are formed):
(c) A solution of 1.0 M Sn4+(aq) is treated with metallic Zn.
Reduction Potentials:
Fe3+(aq) + e– → Fe2+(aq); E° = +0.77 V
Sn4+(aq) + 2e– → Sn2+(aq); E° = +0.15 V
2H+(aq) + 2e– → H2(g); E° = 0.0 V
Fe3+(aq) + 3e– → Fe(s); E° = –0.04 V
Sn2+(aq) + 2e– → Sn(s); E° = –0.14 V
Zn2+(aq) + 2e– → Zn(s); E° = –0.76 V
Problem 2: You are told the following:
1. Solid iodine and solid sodium DO react.
4. Au(s) and Ag+(aq) do NOT react
2. Au+(aq) and iodide ion DO react.
5. Ag(s) and I2(s) do NOT react
+
3. Ag (aq) and sodium DO react.
Deduce from these data alone WITHOUT access to SRP tables the correct ORDER of the
relevant SRPs for the processes involved.
Problem 3: You are told the following:
1. Br2(l) and K(s) DO react.
4. Hg(l) and Co2+(aq) do NOT react
2. Hg2+(aq) and solid potassium DO react.
5. Hg(l) and Br2(l) DO react
+
3. Co(s) and K (aq) do NOT react.
Deduce from these data alone without access to SRP tables the correct ORDER of the
relevant SRPs for the processes involved.
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Problem #4: All the following statements are true EXCEPT
A) most metal oxides are basic.
B) most metals have positive reduction potentials.
C) most metals are dense solids at 500 K.
D) metals are good heat conductors.
E) metals are good electrical conductors.
Problem #5: The overall reaction for the lead storage battery is:
Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq) → 2PbSO4(s) + 2H2O(l)
When the battery discharges, which of the following is (are) true?
1. PbSO4 is formed at both electrodes.
2. PbO2 is the oxidizing agent.
3. The pH increases.
A) a only
B) b only
C) c only
D) a & b only
E) a, b, & c
Problem #6: A piece of iron half-immersed in a sodium chloride solution will corrode more
rapidly than a piece of iron half-immersed in pure water because
A) the sodium ions oxidize the iron atoms.
B) the chloride ions oxidize the iron atoms.
C) the chloride ions form a precipitate with iron.
D) the chloride ions increase the pH of the solution.
E) the sodium ions and chloride ions carry a current through the solution.
Answers to Problem #1:
a) Metallic Sn is added to 1 M HCl(aq) solution:
Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g), E°(reaction) = E°(H+/H2) – E°(Sn2+/Sn) = +0.14 V
(b) A solution of 1.0 M Sn2+(aq) is treated with
(i) metallic Zn: Sn2+(aq) + Zn(s) → Sn(s) + Zn2+(aq), E° = +0.62 V
(ii) FeCl3(aq) solution: Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq), E° = +0.62 V
(c) A solution of 1.0 M Sn4+(aq) is treated with metallic Zn:
Sn4+(aq) + Zn(s) → Sn2+(aq) + Zn2+(aq), E° = +0.91 V
Answer to Problem 2:
Based on the information given, write chemical equations for the spontaneous reactions.
(iii) Au+(aq) + e– → Au(s), E°(iii)
(i) I2(s) + 2e– → 2I–(aq), E°(i)
(ii) Na+(aq) + e– → Na(s), E°(ii)
(iv) Ag+(aq) + e– → Ag(s), E°(iv)
From observation 1, E°(i) is more positive than E°(ii). From observation 2, E°(iii) is
more positive than E°(i). From observation 3, E°(iv) is more positive than E°(ii). From
observation 4, E°(iii) is more positive than E°(iv). From observation 5, E°(iv) is more
positive than E°(i). Therefore, E°(iii) > E°(iv) > E°(i) > E°(ii).
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