Electrons move
from anode to
cathode in the wire.
Anions & cations
move thru the salt
bridge.
Terms Used for Galvanic Cells
Galvanic Cell
CELL POTENTIAL, E
wire V
???
e le c t ro ns
• We can calculate the potential of a
Galvanic cell using one of the following
¾Ecell = Ecathode + Eanode (use Red/Ox potentials)
¾Ecell = - [Eanode – Ecathode] (use Red potentials)
¾Ecell = Ecathode – Eanode (use Red potentials
Zn
Zn and Zn2+,
anode
Cu
salt
bridge
Zn 2+ ions
1.0 M
Cu 2+ ions
1.0 M
Cu and Cu2+,
cathode
• Electrons are “driven” from anode to
cathode by an electromotive force or emf.
• For Zn/Cu cell, this is indicated by a voltage
of ??? V at 25 ˚C and when [Zn2+] and [Cu2+]
= 1.0 M.
wire
elect rons
Zn
salt
bridge
Zn2+ ions
Cu
CELL POTENTIAL, E
Cu2+ ions
• For Zn/Cu cell, potential is +1.10 V at 25 ˚C
and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL
POTENTIAL, Eo
–a quantitative measure of the tendency of
reactants to proceed to products when all are
in their standard states at 25 ˚C.
Uses of Eo Values
• Organize halfreactions by
relative ability to act
as oxidizing agents
• Table 17.1 / A5.5
• Use this to predict
cell potentials and
direction of redox
reactions.
wire
elect rons
Zn
Zn2+ ions
salt
bridge
Cu
Cu2+ ions
1
Standard Redox Potentials, Eo
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
Any substance on the right
will reduce any substance
higher than it on the left.
• Zn can reduce H+ and
Cu2+.
• H2 can reduce Cu2+ but
not Zn2+
• Cu cannot reduce H+ or
Zn2+.
Using Standard Potentials, Eo
Think of it this way…
• What happens when you place Mg(s) in a
solution of Cu2+(aq)? (with a little acid)
– Mg goes into solution
– Cu plates out.
• In terms of the Reduction Potentials
– The more negative the value the more active the
substance.
– So, they tend to want to go into solution.
– Therefore those substances are the Anode
Eo for a Voltaic Cell
Volts
• In which direction do the following
Cd
reactions go?
Fe
Salt Bridge
• Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
• 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)
• What is Eonet for the overall reaction?
Cd2+
Cd --> Cd2+ + 2eor
Cd2+ + 2e- --> Cd
Fe2+
Fe --> Fe2+ + 2eor
Fe2+ + 2e- --> Fe
All ingredients are present. Which way does
reaction proceed?
Eo for a Voltaic Cell
Volts
Cd
Fe
Salt Bridge
Cd2+
Fe2+
Overall reaction
Fe + Cd2+ ---> Cd + Fe2+
Eo = E˚cathode - E˚anode
= (-0.40 V) - (-0.44 V)
= +0.04 V
From the table,
you see
• Fe is a better
reducing agent
than Cd
• Cd2+ is a better
oxidizing agent
than Fe2+
More About
Calculating Cell Voltage
Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OHCathode
2 I- ---> I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚net = E˚cathode - E˚anode
= (-0.828 V) - (+0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction
2
CELL POTENTIALS, Eo
Can’t measure 1/2 reaction Eo
directly. Therefore, measure it
relative to a STANDARD
HYDROGEN CELL, SHE.
2
H+(aq,
1 M) + 2e- <----> H2(g, 1 atm)
Eo = 0.0 V
Which is the Anode/Cathode?
• Standard notation has the Anode on
the left and the Cathode on the right.
• In a Galvanic cell, you always want to
have a positive Ecell.
• Out of the two components (in terms of
the standard reduction potential)
Zn/Zn2+ half-cell hooked to a SHE.
Eo for the cell = +0.76 V
Negative
electrode
Zn
Supplier
of
electrons
Volts
-
+
Positive
electrode
Salt Bridge
H+
Zn2+
2+ + 2eZn -->Zn
Zn
2+
Zn
+ 2eOxidation
OXIDATION
Anode
ANODE
H2
Acceptor
of
electrons
2 H+ + 2e- --> H2
2 H+ + 2eH
Reduction 2
REDUCTION
Cathode
CATHODE
Is E˚ related to ∆G?
YES!
– The more negative one is the ANODE
– The more positive one is the CATHODE
Michael Faraday
1791-1867
Originated the terms anode,
cathode, anion, cation,
electrode.
Discoverer of
• electrolysis
• magnetic props. of matter
• electromagnetic induction
• benzene and other organic
chemicals
Was a popular lecturer.
Eo and ∆Go
Eo is related to ∆Go, the free energy
change for the reaction.
∆Go = - n F Eo
where F = Faraday constant
= 9.6485 x 104 J/V•mol
and n is the number of moles of
electrons transferred
Michael Faraday
1791-1867
3
E at Nonstandard
Conditions
RT
E = Eo −
ln(Q)
nF
Eo and ∆Go
∆Go = - n F Eo
For a product-favored reaction
Reactants ----> Products
∆Go < 0 and so Eo > 0
Eo is positive
For a reactant-favored reaction
Reactants <---- Products
∆Go > 0 and so Eo < 0
Eo is negative
• The NERNST
EQUATION
• E = potential under nonstandard conditions
• n = no. of electrons exchanged
Electrolysis of Aqueous NaOH
Electrolysis
Using electrical energy to produce
chemical change.
Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)
Electric Energy ----> Chemical Change
Anode (+)
4 OH- ---> O2(g) + 2 H2O + 4eCathode (-)
SnCl2(aq)
4 H2O + 4e- --->2 H2 + 4 OHEo for cell = -1.23 V
Cl2
Sn
Electrolysis
Electric Energy ---> Chemical Change
• Electrolysis of
molten NaCl.
• Here a battery
+
“pumps” electrons
Anode
from Cl- to Na+.
• Polarity of
electrodes is
reversed from
batteries.
Electrolysis of Molten NaCl
electrons
BATTERY
Cathode
Cl-
Na+
4
Electrolysis of Molten NaCl
BATTERY
2 Cl- ---> Cl2(g) + 2e-
+
Anode
Cathode
Cl-
Anode (+)
2 Cl- --->
Cl2(g) + 2eCathode (-)
2 H2O + 2e- --->
H2 + 2 OHo
E for cell = -2.19 V
Note that H2O is
more easily
reduced than Na+.
Anode (+)
electrons
Na+
Cathode (-)
Na+
Electrolysis of Aqueous NaCl
+ e- ---> Na
Eo
for cell (in water) = E˚c - E˚a
= - 2.71 V – (+1.36 V)
= - 4.07 V (in water)
External energy needed because Eo is (-).
Electrolysis of Aqueous NaCl
electrons
BATTERY
+
Anode
Cathode
Cl- Na +
H2O
Also, Cl- is oxidized in
preference to H2O because of
kinetics.
Electrolysis of Aqueous NaI
• Cells like these are the source of NaOH and Cl2.
• In 1995: 25.1 x 109 lb Cl2 and 26.1 x 109 lb
NaOH
Also the source of NaOCl for use in bleach.
Electrolysis of Aqueous CuCl2
Anode (+)
2 Cl- ---> Cl2(g) + 2e-
Eo
for cell = -1.02 V
Note that Cu is more
easily reduced than
either H2O or Na+.
Electrolytic Refining of Copper
electrons
Cathode (-)
Cu2+ + 2e- ---> Cu
Anode (+):
2 I- ---> I2(g) + 2eCathode (-): 2 H2O + 2e- ---> H2 + 2 OHEo for cell = -1.36 V
BATTERY
+
Anode
Cathode
Cl- Cu2+
H2O
Impure copper is oxidized to Cu2+ at the anode.
The aqueous Cu2+ ions are reduced to Cu metal
at the cathode.
5
Alcoa
http://www.alcoa.com
Producing Aluminum
2 Al2O3 + 3 C ---> 4 Al + 3 CO2
Charles Hall (1863-1914) developed
electrolysis process. Founded Alcoa.
• World's leading producer of primary
aluminum, fabricated aluminum, and
alumina. Active in all major aspects
of the industry — technology,
mining, refining, smelting,
fabricating and recycling.
• Alcoa's aluminum products and
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aircraft, automobiles, beverage cans,
buildings, chemicals, sports and
recreation, and a wide variety of
industrial and consumer
applications, including such Alcoa
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Baco® household wraps.
• Vital statistics: 120,000 Alcoans in 41
countries, $21.5 billion in revenues
Quantitative Aspects of
Electrochemistry
Consider electrolysis of aqueous silver ion.
Ag+ (aq) + e- ---> Ag(s)
1 mol e- ---> 1 mol Ag
If we could measure the moles of e-, we could
know the quantity of Ag formed.
But how to measure moles of e-?
Current =
I (amps)
Quantitative Aspects of
Electrochemistry
I (amps)
charge passing
time

C 
e- 
=  1.60 x 10-19
  6.02 x 1023


e - 
mol
= 96,500
C/mol e- = 1 Faraday
Quantitative Aspects of Electrochemistry
coulombs
I (amps) =
seconds
solution
1.50 amps flow thru a
for 15.0 min. What mass of Ag metal
is deposited?
Solution
(a) Calc. charge
Charge (C) = current (A) x time (t)
= (1.5 amps)(15.0 min)(60 s/min) =
1350 C
Current =
Ch arge on 1 mol e -
coulombs
seconds
Quantitative Aspects of Electrochemistry
coulombs
seconds
But how is charge related to moles of
electrons?
charge passing
time
=
=
I (amps) =
coulombs
seconds
1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What
mass of Ag metal is deposited?
Ag+(aq)
Solution
(a) Charge = 1350 C
(b) Calculate moles of e- used
1350 C •
(c)
1 mol e = 0.0140 mol e 96, 500 C
Calc. quantity of Ag
0.0140 mol e - •
1 mol Ag
= 0.0140 mol Ag or 1.51 g Ag
1 mol e -
6
Quantitative Aspects of Electrochemistry
Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454
g of Pb, how long will the battery last?
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454
g of Pb, how long will the battery last?
Solution
a) 454 g Pb = 2.19 mol Pb
b) Calculate moles of e-
Time (s)
Solution
a) 454 g Pb = 2.19 mol Pb
b) Mol of e- = 4.38 mol
c) Charge = 423,000 C
2.19 mol Pb •
c)
2 mol e = 4.38 mol e 1 mol Pb
Calculate charge
4.38 mol e- • 96,500 C/mol e- = 423,000 C
d)
Calculate time
Time (s) =
423, 000 C
1.50 amp
=
Charge (C)
I (amps)
About 78 hours
= 282, 000 s
7