This document provides an overview of electrochemistry. It discusses electron transfer reactions, oxidation and reduction, voltaic cells, cell potentials, and balancing redox reactions. Key points include: - Electron transfer reactions are oxidation-reduction or redox reactions that result in the generation of an electric current. - Oxidation is the loss of electrons and reduction is the gain of electrons. Reduction cannot occur without oxidation providing electrons. - A voltaic cell uses the energy from a spontaneous redox reaction to generate an electric current by transferring electrons through an external circuit between the anode and cathode.

1. 1
ELECTROCHEMISTRY
Chapter 18

2. 2
Why Study Electrochemistry?
• It’s important because it involves many of the
devices that we use every day that are battery
powered.
• Corrosion
• Industrial production of chemicals such as
Cl2, NaOH, F2 and Al.

3. 3
Electron Transfer Reactions
• Electron transfer reactions are oxidation-
reduction or redox reactions.
• it Results in the generation of an electric
current (electricity)
• ELECTROCHEMISTRY is the connection
between chemistry and electricity.

4. Electrochemistry
Oxidation Numbers
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.

5. Electrochemistry
Oxidation and Reduction
• A species is oxidized when it loses electrons.
➢Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.

6. Electrochemistry
Oxidation and Reduction
• A species is reduced when it gains electrons.
➢Here, each of the H+ gains an electron and they
combine to form H2.

7. Electrochemistry
Oxidation and Reduction
• What is reduced is the oxidizing agent.
➢H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
➢Zn reduces H+ by giving it electrons.

8. 8
You can’t have one… without the
other!
• Reduction (gaining electrons) can’t happen
without an oxidation to provide the electrons.
• You can’t have 2 oxidations or 2 reductions in
the same equation. Reduction has to occur
at the cost of oxidation
LEO the lion says GER!
GER!

9. 9
Another way to remember
•OIL RIG
Oxidation is loss of
electrons and
Reduction is gain of
electrons.

10. Electrochemistry
The Big Idea!
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and
energy is released.

11. Electrochemistry
Voltaic Cells
• We can use the energy
from the Redox Rxn to
do work if we make the
electrons flow through an
external device.
• We call such a setup a
voltaic cell or a Galvanic
cell or even an
electrochemical cell..

12. Electrochemistry
Voltaic Cells
• A typical cell looks
like this.
• The oxidation
occurs at the anode.
• The reduction
occurs at the
cathode.

13. Electrochemistry
Voltaic Cells
Once even one
electron flows from
the anode to the
cathode, the
charges in each
beaker would not be
balanced and the
flow of electrons
would stop.

14. Electrochemistry
Voltaic Cells
• Therefore, we use a
salt bridge, usually a
U-shaped tube that
contains a salt
solution, to keep the
charges balanced.
➢Cations move toward
the cathode.
➢Anions move toward
the anode.

15. Electrochemistry
Voltaic Cells
• In the cell, then,
electrons leave the
anode and flow
through the wire to the
cathode.
• As the electrons leave
the anode, the cations
formed dissolve into
the solution in the
anode compartment.

16. Electrochemistry
Voltaic Cells
• As the electrons
reach the cathode,
cations in the
cathode are
attracted to the now
negative cathode.
• The electrons are
taken by the cation,
and the neutral
metal is deposited
on the cathode.

17. 17
Anode Cathode
Basic Concepts of Electrochemical Cells
Zn is oxidized and is the
reducing agent
Zn(s) ---> Zn2+(aq) + 2e-
Cu2+ is reduced and is the
oxidizing agent
Cu2+(aq) + 2e- ---> Cu(s)

18. 18
•Electrons travel thru external wire.
Salt bridge allows anions and cations to move
between electrode compartments.
Zn --> Zn2+ + 2e- Cu2+ + 2e- --> Cu
<--Anions
Cations-->
Anode
Oxidation
Negative
Reduction
Cathode
Positive

19. Electrochemistry
Cell Potential • Water only
spontaneously
flows one way in a
waterfall.
• Likewise,
electrons only
spontaneously
flow one way in a
redox reaction—
from higher to
lower potential
energy.

20. Electrochemistry
Cell Potential
•The potential difference between the
anode and cathode in a cell is called the
cell potential, and is designated Ecell.
Cell potential is measured in volts (V).
1 V = 1
J
C

21. 21
Calculating Cell Voltage (Eo)
Balanced half-reactions can be added
together to get overall, balanced equation.
Zn(s) ---> Zn2+(aq) + 2e-
Cu2+(aq) + 2e- ---> Cu(s)
--------------------------------------------
Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
If we know Eo for each half-reaction, we
could get Eo for net reaction.

22. 22

23. Electrochemistry
Standard Cell Potentials
The cell potential at standard conditions
can be found through this equation:
Ecel
l
° = Ered (cathode) − Ered
(anode)
° °
Because cell potential is based on
the potential energy per unit of
charge, it is an intensive property.

24. Electrochemistry
Cell Potentials
• For the oxidation in this cell,
• For the reduction,
Ered = −0.76
V
°
Ered = +0.34
V
°

25. Electrochemistry
Cell Potentials
Ecel
l
° = Ere
d
° (cathode) − Ere
d
°
(anode)
= +0.34 V − (−0.76 V)
= +1.10 V
E ° > 0 Spontaneous reaction
E° ＜ 0 Not Spontaneous ; Doesn’t “GO”
E° = 0 Equilibrium

26. Electrochemistry
Oxidizing and Reducing Agents
• The strongest
oxidizers have the
most positive
reduction potentials.
• The strongest
reducers have the
most negative
reduction potentials.

27. Electrochemistry
Oxidizing and Reducing Agents
The greater the
difference between
the two, the greater
the voltage of the
cell.

28. What is Electroplating?
An electrochemical process where metal ions are transferred
from a solution and are deposited as a thin layer onto surface of
a cathode.
The setup is composed DC circuit with an anode and a cathode
sitting in a bath of solution that has the metal ions necessary for
coating or plating
Electrolysis is the process of making a non-spontaneous chemical
redox reaction occur by passing electricity through a substance.
(electrolytic cell)

29. History
◼ In 1805, Italian chemist Luigi Brugnatelli,
successfully electroplated silver medals with gold
◼ 1940 first patent for electroplating awarded, and
soon factories in England were mass producing
silver plated items, including teapots, brushes and
utensils
Wikienergia.com

30. Benefits of Electroplating
Electroplating can enhance:
▪ Chemical properties---increase corrosion resistance
▪ Physical properties---increase thickness of part
▪ Mechanical properties---increase tensile strength &
hardness
Commercial Examples:
◼ Coating jewelry with thin layer of expensive metal.
◼ Coating chromium over steel to make rust resistant.

31. Electrolytic VS Electrochemical
Electrochemical
◼ 2 half-cells with 1
electrode each.
◼ Spontaneous redox
reactions produce
electric current
◼ REDCAT +, cathode
grows.
◼ ANOX -, anode shrinks
◼ External circuit carries e-
‘s
◼ Salt bridge neutralizes
charge build-up
Electrolytic
● 1 container(cell)
● 2 electrodes in same
container
● electricity from outside
source supplies e-’s for the
non-spontaneous rxn.
● REDCAT -,item to be plated
goes here.
● ANOX +, metal to do the
plating here.
● Ions transfer e-’s from
electrode to electrode.

32. How it Works
◼In electroplating, the anode is made up of the metal
you want to coat the surface of another metal with.
◼There is also a salt solution present of the anode
metal.
◼While electrolysis is taking place, the anode metal is
oxidized and goes into solution as positive ions.
◼These positive ions are then reduced on the surface
of the cathode (the metal you wish to coat).

33. How it works

34. Sample Problem
A spoon is to be plated with
silver, Ag.
1. Identify the anode and
cathode.
2. Write an equation for the
reaction taking place at the
anode and at the cathode
and indicate whether it is
oxidation or reduction.
3. What electrolyte is used?

35. Answers
1. Cathode = spoon (metal object to be coated), Anode =
silver electrode
2. As electrolysis takes place, the silver anode is oxidized,
Ag(s) → Ag+(aq) + 1e-
◼ The Ag+ (aq) ions in solution travel to the spoon cathode
and are reduced to form neutral Ag(s) on the surface of
the spoon (cathode):
Ag+ (aq) + 1e- → Ag(s)
3. Electrolyte solution is made of AgNO3

36. electric Current
An ampere is a unit of measure of the rate of electron
flow or current in an electrical conductor. One ampere of
current represents one coulomb of electrical charge
(6.24 x 1018charge carriers) moving past a specific
point in one second.
The coulomb (symbolized C) is the standard unit of
electric charge in the International System of Units (SI).
It is a dimensionless quantity, sharing this aspect with
the mole.
1 mole = 96 500 Coulombs
1 mole = 6.02 x 1023 electrons

37. Stoichiometry of Electrolysis problems
To determine the quantity of substance either produced or
consumed during electrolysis given the time a known current
flowed:
Write the balanced half-reactions involved.
Calculate the number of moles of electrons that were
transferred.
Calculate the number of moles of substance that was
produced/consumed at the electrode.
Convert the moles of substance to desired units of measure.

38. Sample Problem
A 40.0 amp current flowed through a solution of iron(III) chloride for 10.0 hours.
Determine the mass of iron (measured at 25oC and 1 atm) that is produced
during this time.

39. Calculating the time required
To determine the quantity of time required to produce a
known quantity of a substance given the amount of
current that flowed:
● Find the quantity of substance produced/consumed in
moles.
● Write the balanced half-reaction involved.
● Calculate the number of moles of electrons required.
● Convert the moles of electrons into coulombs.
● Calculate the time required.

40. Sample Problem
How long must a 20.0 amp current flow through a solution of ZnSO4 in order
to produce 25.00 g of Zn metal.

41. Electrochemistry
Free Energy
ΔG for a redox reaction can be found by
using the equation
ΔG = −nFE
n is the number of moles of electrons
transferred, and F is a constant, the Faraday.
1 F = 96,485 C/mol = 96,485 J/V-mol

42. Electrochemistry
Nernst Equation
Dividing both sides by −nF, we get the
Nernst equation:
E = E°
−
RT
nF
ln Q
or, using base-10 logarithms,
E = E°
−
2.303 RT
nF
ln Q
•The value of the cell potentials change if conditions are nonstandard.
•The Nernst equation describes the electrode potentials at nonstandard
conditions.

43. Electrochemistry
Nernst Equation
At room temperature (298 K),
Thus the equation becomes
E = E°
−
0.0592
n
ln Q
2.303 RT
F
= 0.0592 V

44. 44
Dry Cell Battery
Anode (-)
Zn ---> Zn2+ + 2e-
Cathode (+)
2 NH4
+ + 2e- --->
2 NH3 + H2

45. 45
Alkaline Battery
Nearly same reactions as
in common dry cell, but
under basic conditions.
Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2e-
Cathode (+): 2 MnO2 + H2O + 2e- --->
Mn2O3 + 2 OH-

46. 46
Mercury Battery
Anode:
Zn is reducing agent under basic conditions
Cathode:
HgO + H2O + 2e- ---> Hg + 2 OH-

47. 47
Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4
- ---> PbSO4 + H+ + 2e-
Cathode (+) Eo = +1.68 V
PbO2 + HSO4
- + 3 H+ + 2e-
---> PbSO4 + 2 H2O

48. 48
Ni-Cad Battery
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2e-
Cathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-

49. 49
H2 as a Fuel
Cars can use electricity generated by H2/O2
fuel cells.
H2 carried in tanks or generated from
hydrocarbons

50. 50
Balancing Equations
Step 1: Divide the reaction into half-reactions, one
for oxidation and the other for reduction.
Ox Cu ---> Cu2+
Red Ag+ ---> Ag
Step 2: Balance each element for mass. Already
done in this case.
Step 3: Balance each half-reaction for charge by
adding electrons.
Ox Cu ---> Cu2+ + 2e-
Red Ag+ + e- ---> Ag

51. 51
Balancing Equations
Step 4: Multiply each half-reaction by a factor so
that the reducing agent supplies as many electrons
as the oxidizing agent requires.
Reducing agent Cu ---> Cu2+ + 2e-
Oxidizing agent 2 Ag+ + 2 e- ---> 2 Ag
Step 5: Add half-reactions to give the overall
equation.
Cu + 2 Ag+ ---> Cu2+ + 2Ag
The equation is now balanced for both
charge and mass.

52. 52
Balancing Equations
Balance the following in acid solution—
VO2
+ + Zn ---> VO2+ + Zn2+
Step 1: Write the half-reactions
Ox Zn ---> Zn2+
Red VO2
+ ---> VO2+
Step 2: Balance each half-reaction for
mass.
Ox Zn ---> Zn2+
Red
VO2
+ ---> VO2+ + H2O
2 H+ +
Add H2O on O-deficient side and add H+
on other side for H-balance.

53. 53
Balancing Equations
Step 3: Balance half-reactions for charge.
Ox Zn ---> Zn2+ + 2e-
Red e- + 2 H+ + VO2
+ ---> VO2+ + H2O
Step 4: Multiply by an appropriate factor.
Ox Zn ---> Zn2+ + 2e-
Red 2e- + 4 H+ + 2 VO2
+
---> 2 VO2+ + 2 H2O
Step 5: Add balanced half-reactions
Zn + 4 H+ + 2 VO2
+
---> Zn2+ + 2 VO2+ + 2 H2O

54. 54
Tips on Balancing Equations
• Never add O2, O atoms, or
O2- to balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end
to make sure mass and
charge are balanced.
• PRACTICE!