Chapter 18 Electrochemistry

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Electrochemistry
Chapter 17
Electrochemistry
Electrochemistry is the study of batteries
and the conversion between chemical and
electrical energy.
 Based on redox (oxidation-reduction)
reactions in which one substance gains
electrons and another loses electrons.

GCC CHM152


Ox # examples
Oxidation Numbers (Chapter 4).


Free elements: ox # = 0 (H2(g), Hg(l), etc.)
Ions in binary ionic compounds: ox # = charge.

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Ex. For Al2S3, ox # for Al = +3, ox # for S = -2.
H: ox # = +1, except when H is with alkali
metals it is -1 (LiH, NaH, etc.)
O: ox # = -2, except in peroxides it is -1 (H2O2,
K2O2)
The sum of the oxidation numbers of all the
atoms in a molecule = 0
For polyatomic ions, the sum of the oxidation
numbers must equal the charge on the ion.
Redox: LEO the lion goes GER
Oxidation: loss of electrons; ox # , more +
 Reduction: gain of electrons; ox # , more −
 oxidizing agent: substance that is reduced; it
caused oxidization of other substace.
 reducing agent: substance that is oxidized; it
caused reduction of other substance.
These two processes MUST happen together.
Single replacement, combustion,
combination and decomposition rxns are all
examples of redox reactions.
Find ox numbers for all the atoms in HCO3 H: ox # = +1
O: ox # = -2
 1H + 1C + 3O = -1
 + 1 + C + 3x(-2) = -1
 Ox # for C = +4
 What is the ox # for Cr in Cr2O72-?

2Cr + 7O = -2
2Cr + 7(-2) = -2
 Cr = +6


Zn(s) + Cu(NO3)2(aq)  Zn(NO3)2(aq) + Cu(s)


Zn(s) + Cu(NO3)2(aq)  Zn(NO3)2(aq) + Cu(s)

Identify atom oxidized, atom reduced,
oxidizing agent and reducing agent.
Zn(s)  Zn2+(aq) + 2e-
Zn(s) dissolves
Cu2+(aq) + 2e-  Cu(s)
Cu plates on electrode
1
Redox Reactions
What is oxidized/reduced, give the specific atom.
Agents: give the whole substance as an answer.


MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) 
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
 For this reaction:

What is oxidized?
What is reduced?
 What is the oxidizing agent?
 What is the reducing agent?


Balancing Redox Reactions
Cr3+(aq) + Be(s)  Cr(s) + Be2+(aq)
 First, break it up into two half reactions,
the oxidation and reduction half reactions.
 Balance charge by adding electrons (e-) to
the more positive side for each reaction.
 Oxidation: Be(s)  Be2+(aq) + 2e Reduction: Cr3+(aq) + 3e-  Cr(s)

Types of Cells
Balancing Redox Rxns contd
Balance electrons by multiplying each half
reaction by an integer so that the # electrons
gained = # electrons lost.
 Oxidation: (Be(s)  Be2+(aq) + 2e-)x3
 Reduction: (Cr3+(aq) + 3e-  Cr(s))x2


2Cr3+(aq) + 3Be(s)  2Cr(s) + 3Be2+(aq)
Batteries
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Galvanic cell (also called voltaic cell)
Electrochemical cell in which a
spontaneous reaction generates electricity.

Electrolytic cell
Electrochemical cell in which an electrical
current is used to drive a nonspontaneous
reaction.
Galvanic Cells: Figure17.2
Batteries are examples of galvanic cells we use
in everyday life.
For a Voltaic cell the voltage keeps dropping as
the spontaneous reaction proceeds right.
Battery is dead when Ecell = 0 (at equilibrium)
Bigger batteries only last longer, they don’t
have more volts. Volts are determined by the
chemical reaction that occurs.
2
Dr. Lisa’s cell Mnemonic
Galvanic Cells

Oxidation occurs at the anode - both vowels

Fat red cat eats electons!

Anorexic ox spits them out!
mass of anode decreases - it is dissolving as metal
atoms lose electrons to form ions in solution

Reduction occurs at the cathode - both consonants
mass of cathode increases as metal ions are reduced
to form atoms that plate onto the cathode.


e-
External Circuit - electrons flow from the anode to
the cathode via an external wire.
Salt bridge - soluble salt solution in a bridge that
connects the two half cells; ions flow through the
bridge to complete the electrical circuit.
e-
Cathode is reduced (gains e-) & mass (plating)
Anode is oxidized (loses e-) & mass (dissolves)
What happens in Salt Bridge?
Short-hand Notation
Migration of ions maintains charge
neutrality in both compartments:
 Anions move into the anode where
excess + charge builds up as metal
cations are formed by oxidation.
 Cations move into the cathode where
excess (-) charge builds up as the metal
cations are reduced to form neutral
metal atoms.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode | anode ion || cathode ion | cathode
| indicates a phase boundary between two
phases in the same cell.
|| denotes the salt bridge between the cells.
anode is always on the left side, and the cathode
on the right side.
Electrodes are always on the two ends


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Inert Electrodes
Shorthand Notation for Galvanic Cells
Anode half-reaction:
Cathode half-reaction:
Zn(s)  Zn2+(aq) + 2eCu2+(aq) + 2e-  Cu(s)
Overall cell reaction: Zn(s) + Cu2+ (aq)  Zn2+(aq) + Cu(s)
Salt bridge
Anode half-cell
Zn(s) |
Zn2+(aq)
cathode half-cell
|| Cu2+(aq) | Cu(s)
Pt and graphite are typically electrodes for gas
phase and aqueous reactions.
 Ex. Standard Hydrogen Electrode (SHE) uses
Pt electrode. (figure 17.4)
 SHE consists of Pt electrode in contact with 1
M H+ solution and H2 gas at 1 atm pressure
 H2(g)  2H+(aq) + 2e
Phase boundary
Chapter 17/17
© 2012 Pearson Education, Inc.
3
Galvanic Cells - Pt electrode

Figure 17.4
Electromotive Force
emf = Ecell = cell voltage:
Electromotive force is the cell potential
measured in volts. This is the driving force that
pushes electrons away from the anode and
towards the cathode.
Joules = Coulombs x Volts
Ecell is measured in volts: V = J/C
coulomb - the quantity of charge that passes a
point in 1 sec when a current of 1 ampere flows.
Free Energy and Emf
DG = -nFE
n = number of moles of e- transferred
E = Emf of cell
F = Faraday's constant
1F 
96500 Coulombs 96500 J

1 mol e 
V  mol e 
spontaneous reaction: DG < 0 and E > 0
nonspontaneous reaction: DG > 0 and E < 0
At equilibrium: DG = 0 and E = 0
Reduction Potentials
We can use the table of Standard Reduction
Potentials for reduction half reactions to
determine the cell potential of a galvanic cell.
 Table 17.1: All potentials are listed as
reduction potentials.


Standard Cell Potential, Eo
Gases at 1 atm, Solutions at 1 M,
Temperature at 298 K (25oC)
 Standard potential for any galvanic cell is
the sum of the half-cell potentials.
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Eocell = Eoox + Eored
All cell potentials are compared to hydrogen
(SHE: standard hydrogen electrode)
2H+ + 2e-  H2(g)
Eored = 0 V
Calculating Cell potentials



Ecell
 Ered
(cathode)  Eox
(anode)
For the oxidation reaction at the anode,
make sure you reverse the reaction and
change the sign for Eox, then add the
reduction and oxidation potentials.
Oxidation potentials: reverse the reaction
and


Eox
(anode)   Ered
(cathode)
4
Strength of Oxidizing Agents
Oxidizing Agents:
F2(g) + 2e-  2F-(aq)
Eored = 2.87 V
 F2 has the most positive Eored value.
 F2 is the easiest to reduce. (It wants ethe most!)
 Thus F2 is strongest oxidizing agent.
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As Eored, strength ox agent 
Strength of Reducing Agents
Li+(aq) + e-  Li(s)
Eored = -3.04 V
 Li has the most negative Eored value, so it
is the easiest to oxidize.
 Li(s) is strongest reducing agent.
 Li(s) wants to lose electrons, so reverse
reaction occurs:
 Li(s)  Li+(aq) + e- Eoox = +3.04 V

Table 17.1 info continued
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Cell Potentials
Based on their Eored values, determine the
best oxidizing agent, best reducing agent,
worst oxidizing agent, and worst reducing
agent.
 Au3+ + 3e-  Au(s)
Eored = 1.50 V
 Br2(l) + 2e  2Br (aq)
Eored = 1.07 V
 Pb2+ + 2e-  Pb(s)
Eored = -0.13 V
2+
 Ni + 2e  Ni(s)
Eored = -0.25 V

Active metals tend to be good reducing
agents. (red agent = ox = lose e-)
Active nonmetals tend to be good oxidizing
agents. (ox agent = red = gain e-)
E° is intensive; it does not depend on # of
moles involved. Don't need to multiply E°
by factor if coefficients of reaction are
changed.
Cell Potentials
Ni(s) is the easiest to oxidize, best
reducing agent.
 Au3+ is the easiest to reduce, best
oxidizing agent.
 Au(s) is the hardest to oxidize, worst
reducing agent.
 Ni2+ is the hardest to reduce, worst
oxidizing agent.

5
Cell Potentials
Cell Potentials in Reactions
Positive Eocell means the reaction is productfavored and spontaneous.


A negative Eocell means the reaction won’t
happen in the forward direction.
Therefore, we want two half reactions that
yield the most positive Eocell value.
 Assign rxn with more + Eored as cathode!
 Eocell = Eored + Eoox

What will Eocell be if we react Ni(s) with Au3+?
What will Eocell be if Pb2+ reacts with Br -?
Calculate Eocell for the following cell
Ni(s) | Ni2+(aq) || Br2(l), Br -(aq) | Pt(s)
Which are spontaneous?
Example: Cl2(g) + Zn(s)
Cell Potentials of Reactions
If we were to make a Galvanic cell from
the following metals, which would act as
the anode and which as the cathode?
Refer to table 17.1. Write the half-cell
reactions and the overall balanced
reaction for each cell. Calculate Eocell for
each one.
 Also calculate Eocell for each one.

For a spontaneous reaction (+ Eocell), write the
appropriate anode and cathode reactions:
Cl2 (g) + 2 e- → 2 Cl- (aq) E°red = 1.36 V
Zn2+ (aq) + 2 e- → Zn (s)
E°red = -0.76 V
Cl2 (g) has more + E°red, so it’s the cathode reaction

Cathode: Cl2 (g) + 2 e- → 2 Cl- (aq) E°red = 1.36 V
Zn reaction is reversed for the anode:
Anode: Zn (s) → Zn2+ (aq) + 2 eE°ox = +0.76 V
 The E°ox reaction is flipped so the sign is the opposite!
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Eocell = 1.36 + .76 = 2.12 V
The Nernst Equation: Calculate emf for
nonstandard conditions
Recall: DG =

DGo
+ RT ln Q
Standard emf & K

Plugging in DG = - nFE and DGo = - nFEo
-nFE = -nFEo + RT ln Q
Nernst Equation:
E  E 
2.303RT
logQ
nF
E  E 
R  8.314
At equilibrium, E = 0 and Q = K (plug these
conditions into Nernst Equation)
0  E 
RT
ln Q
nF
RT
ln K
nF
R  8.314
J
J
, F  96500
K  mol
V  mol e 
0.0592 V
logQ
At 298 K: E  E 
n
Both [ ]’s & P’s can be plugged into Q; V = volts (unit)


Fe(s) and Ni(s)
Cu(s) and Ag(s)

E 
RT
ln K
nF
J
J
, F  96500
K  mol
V  mol e 
At 298 K, E   0.0592 V log K
n
6
Calculations
Cell Potentials Summary
Calculate DGo for the Ni(s) + Au3+ reaction. (#4a)

Positive Eocell value
DGo is negative
K is large
 Reaction is product-favored
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Calculate K at 25oC for Ni(s) + Br2 cell (#4c):
Ni(s) | Ni2+(aq) || Br2(l), Br -(aq) | Pt(s)

Negative Eocell value
DGo is positive
K is small
 Reaction is reactant-favored

K e
 nFE  


 RT 


or
 nE  


 0.0592 V 



K  10
The Nernst Equation
Corrosion
A galvanic cell utilizes the following reaction:
2Ag+ + Pb(s)  Pb2+ + 2Ag(s)
a) Write the half reactions and calculate Eocell.
b) Write the short-hand notation for this cell.
The oxidative deterioration of a metal (i.e.,
solid metal converted to ions).
 Rust formation is the corrosion of iron.
 Metals can be plated with non-reactive
metals to protect them (chromium, tin, or
zinc are common).

c) Calculate the cell potential if [Pb(NO3)2] =
0.88 M and [AgNO3] = 0.14 M.
Electrolysis

Electrolytic Cell: Electrical energy from an external
source (outlet or a battery) is used to force a
nonspontaneous redox reaction to proceed.
Molten salt: 2NaCl(l)  2Na(s) + Cl2(g)

Cathode: 2Na+(l) + 2e-  2Na(s)

Anode: 2Cl-(l)  Cl2(g) + 2eEcell = - 4 V
This reaction naturally wants to run in reverse
direction. We need more than 4 volts to drive this
reaction forward. Salts don’t normally decompose
into elements.



Electrolysis of Water
Water doesn’t naturally decompose into
hydrogen and oxygen.
 2H2O(l)  2H2(g) + O2(g) Ecell = -1.23 V

Anode: 2H2O(l) → O2(g) + 4H+(aq) + 4eCathode: 4H2O(l) + 4e- → 2H2(g) + 4OH-(aq)
The electrodes are the still the same for
electrolysis: (oxidation at anode, reduction at
cathode).
7
Electrolysis Calcs
Used to find mass or volume
of product produced by
passing current through cell.
 Current: measured in Amps
A = C/s
 ampere: unit of electric current;
rate of flow of e-.
 charge = current x time
 coulombs = amps x seconds

Batteries

Lead storage (i.e., car batteries)
9V battery – sum of 1.5 Volts
Electrolysis Calculations




How many grams of Cu can be collected in 1.00
hour by a current of 1.62 A from a CuSO4 solution?
Reduction reaction: Cu2+ + 2 e-  Cu(s)
Calculate coulombs: Current (C/s) x time (s) = C
C  mol e-  mol solid  g solid
3600s 

1.62C 1 mol e  1 mol Cu 63.5g
= 1.92 g Cu



s
96,500C 2 mol e 
mol
Worked Examples 17.10, 17.11; Problems 17.22, 17.23
Household batteries

Dry-cell (or Laclanche) batteries
Ni-Cad Batteries - rechargeable
Positive
electrode:
NiOOH
Negative
electrode:
Cd
8
Fuel Cell
Lithium Ion Batteries

Positive electrode:
Lithium cobalt oxide
Negative electrode:
Carbon
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Used for space travel, and in hydrogen fuel cars
Reactants are stored external to the cell and
introduced to the electrodes as they are
needed.
The reactants are usually gaseous, such as O2
and H2, or O2 and CH4, or O2 and NH3.
H2 + 2OH-  2H2O + 2e- Eo = 0.83 V
O2 + 2H2O + 4e-  4OH- Eo = 0.40 V
2H2 + O2  2H2O
Eo = 1.23 V
Fuel cells in action
Fuel
Cell
9
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