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Electrochemistry
Electrochemistry
Electrochemistry involves either…
– generating electricity by harnessing a spontaneous chemical
reaction (one with K > 1)
OR
– using electricity to force a chemical reaction to occur (one
that is non-spontaneous, K<1)
Understanding electrochemistry requires a basic understanding of
the processes called “oxidation” and “reduction”.
Oxidation and Reduction
Oxidation is…
– the loss of electrons
– an increase in oxidation
state
– the addition of oxygen
– the loss of hydrogen
Reduction is…
– the gain of electrons
– a decrease in oxidation
state
– the loss of oxygen
– the addition of hydrogen
2 Mg + O2  2 MgO
MgO + H2  Mg + H2O
notice the magnesium is losing
electrons
notice the Mg2+ in MgO is gaining
electrons
Oxidation States
Oxidation states are numbers assigned to atoms that
reflect the net charge an atom would have if the
electrons in the chemical bonds involving that atom
were assigned to the more electronegative atoms.
Oxidation states can be thought of as “imaginary”
charges. They are assigned according to the
following set of rules:
Assigning Oxidation States

Each atom in a pure element
has an oxidation state of “0”.
e.g. F2

K
N2
Al3+
Fe2+
The oxidation state of
hydrogen in a compound
(e.g. CH3OH) is almost
always “+1”.

The sum of the oxidation
states in a neutral compound
must equal “0”.

The sum of the oxidation
states in a complex ion must
equal the charge on the
complex ion.
Fe
For mono-atomic ions, the
oxidation state is equal to the
charge on the ion.
e.g. F-

O3

O2-
The oxidation state of oxygen
in a compound (e.g. CH3OH)
is almost always “-2”.
Examples - assigning oxidation numbers
Assign oxidation states to all elements:
H2
K
+
Cr2O72ClO3
SO42-
SO3
-
-
NH3
MnO4
CH3OH
PO43-
HSO3
-
Cu
Identifying Redox Reactions
Oxidation and reduction always occur together in a
chemical reaction. For this reason, these reactions are
called “redox” reactions.
Although there are different ways of identifying a redox
reaction, the best is to look for a change in oxidation
state:
2 Fe3+ + 2 I-  2 Fe2+ + I2
2 H2O  2 H2 + O2
2 AgNO3 + Cu  2 Ag + Cu(NO3)2
HCl + AgNO3  AgCl + HNO3
More Definitions

Oxidizing Agent
– the substance in a chemical reaction which
causes another species to be oxidized.
– the oxidizing agent always gets reduced in the
reaction.

Reducing Agent
– the substance in a chemical reaction that causes
another species to be reduced.
– the reducing agent always gets oxidized!
Examples - labeling redox reactions



In each reaction, look for changes in oxidation state.
If changes occur, identify the substance being
reduced, and the substance being oxidized.
Identify the oxidizing agent and the reducing agent.
5 Fe2+ + MnO4- + 8 H+  5 Fe3+ + Mn2+ + 4 H2O
H2 + CuO  Cu + H2O
Zn + 2 HCl  ZnCl2 + H2
Balancing Redox Equations
Redox reactions are often quite complicated and difficult
to balance. For this reason, you’ll learn a step-by-step
method for balancing these types of reactions, when
they occur in acidic or in basic solutions.
The procedure is called the “Half-Reactions Method”
of balancing redox equations. It starts by identifying the
substances being reduced and oxidized in the reaction,
and then following the steps below:
Balancing Redox Equations
Half-Reactions Method

First split the original equation into two half-reactions,
one “reduction” and the other “oxidation”.
In each half-reaction, follow these steps:

Balance all elements except “H” and “O”.

Balance the “O’s” by adding water, H2O.

Balance the “H’s” by adding hydrogen ions, H+.

Balance the electric charge by adding electrons, e-.

Multiply the two equations by appropriate coefficients
to make the # of electrons in the equations equal.

Re-combine the two equations, canceling if needed.
Example -
Balancing with Half-Reactions
Fe2+ + Cr2O72-  Fe3+ + Cr3+
Fe2+  Fe3+
Cr2O72-  Cr3+
Fe2+ Fe3+ + e-
Cr2O72-  2 Cr3+
6 Fe2+ Fe3+ + 6 e-
Cr2O72-  2 Cr3+ + 7 H2O
Cr2O72- + 14 H+  2 Cr3+ + 7 H2O
Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
Cr2O72- + 6 Fe2+ + 14 H+  2 Cr3+ + 6 Fe3+ + 7 H2O
What if the solution was basic?
Notice that the method has assumed the solution was acidic
- we added H+ to balance the equation. The [H+] in a basic
solution is very small. The [OH-] is much greater.
For this reason, we will add enough OH- ions to both sides of
the equation to neutralize the H+ in the overall reaction.
The hydrogen and hydroxide ions will combine to make
water, and you may have to do some canceling before you’re
done.
Cr2O72- + 6 Fe2+ + 7 H2O  2 Cr3+ + 6 Fe3+ + 14 OHWhy is this reaction unlikely in basic solution?
Balancing Redox Equations Practice

Balance in acidic solution:
H2C2O4 + MnO4-  Mn2+ + CO2
5 H2C2O4 + 2 MnO4- +16 H+  2 Mn2+ + 10 CO2 + 8 H2O

Balance in basic solution:
CN- + MnO4-  CNO- + MnO2
3 CN- + 2 MnO4- + H2O  3 CNO- + 2 MnO2 + 2 OH-
Redox Reactions - What’s Happening?

Zinc is added to a
blue solution of
copper(II) sulfate
Zn (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu (s)

The blue colour
disappears…the zinc
metal “dissolves”, and
solid copper metal
precipitates on the
zinc strip

The zinc is oxidized
(loses electrons)
The copper ions are
reduced (gain
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
electrons)

Copper ions (Cu2+)
collide with the zinc
metal surface
A zinc atom (Zn)
gives up two of its
electrons to the
copper ion
The result is a
neutral atom of Cu
deposited on the
zinc strip, and a
Zn2+ ion released
into the solution
Harnessing the Electricity

During the spontaneous redox reaction, electrons
flow from the zinc atoms to the copper ions.
Zn  Zn2+ + 2 eCu2+ + 2 e-  Cu



(oxidation)
(reduction)
Electricity can be thought of as the “flow of electric
charge”.
How can we make use of the flow of electrons from
the zinc atoms to the copper ions??
Answer: SEPARATE the copper ions from the zinc
atoms. This will force the electrons from the zinc
atoms to travel through an external path to reach the
copper ions.

Electrochemical Cell
– a device that uses a
spontaneous redox
reaction to produce
electricity

Anode
– the electrode where
oxidation occurs

Cathode
– the electrode where
reduction occurs

Salt Bridge
– connects two “half-cells”
to complete the electric
circuit.
– for example, a U-tube
filled with salt solution
Electrochemical Cells
Definitions
Electrochemical Cells
Cell Potential

Water flows spontaneously over a waterfall because of a
difference in potential energy between the top of the falls and
the stream below.

In a similar fashion, electrons flow from the anode of a voltaic
cell to the cathode because of a difference in potential energy.

The potential energy of electrons is higher in the anode than in
the cathode, and they spontaneously flow through an external
circuit from the anode to the cathode.

The potential difference between the two electrodes of an
electrochemical cell provides the driving force that pushes
electrons through the external circuit.

We call this potential difference, denoted Ecell, the cell potential.

Because Ecell is measured in volts, we often refer to it as the cell
voltage. For any cell reaction that proceeds spontaneously, such
as that in a voltaic cell, the cell voltage will be positive.
Standard Hydrogen Electrode (SHE)




It is impossible to measure
the potential of a single
electrode.
Using a voltmeter, we can
measure the difference in
potential between two
electrodes.
Chemists arbitrarily assigned
a potential of 0 V for the
SHE.
By measuring the difference
in potential between the SHE
and other electrodes,
potentials can then be
assigned to other electrodes.
2 H+ + 2 e-  H2
E° = 0 V



When a zinc electrode is connected to the SHE as shown
above, the voltmeter reads: E°cell = + 0.76 V
The positive sign means electrons are flowing from zinc to the
hydrogen cell. The zinc is being oxidized.
Since E°SHE= 0 V, we can conclude that E°Zn = + 0.76 V . This is
the “oxidation potential” for the reaction: Zn  Zn2+ + 2 e-
Table of Standard Reduction Potentials
Calculating a Cell Potential
Step 1:
Which electrode is more
likely to be reduced? This is
the cathode.
Step 2:
Write half-reactions each
half-cell - one reduction and
one oxidation reaction.
Include the values for E°red
and E°ox from the table of
standard reduction
potentials.
Remember: E°red = -E°ox
Step 3:
Calculate the cell potential:
E°cell = E°red + E°ox
Step 1:
Since Cu has a higher
reduction potential than
Zn, copper forms the
cathode. We could also
argue that Zn has a
higher oxidation
potential…and forms
the anode.
Step 2:
Cathode:
Cu2+ + 2e-  Cu
E°red = +0.34 V
Anode:
Zn  Zn2+ + 2 eE°ox = +0.76 V
Cu
Step 3:
The overall reaction is:
Zn + Cu2+  Zn2+ + Cu
E°cell = E°red + E°ox = 0.34 + 0.76 = 1.10 V
Zn
Line Notation
An electrochemical cell needs to be described in a more
convenient way than drawing a diagram!
We use “line notation” to describe a cell. The zinccopper standard cell is described:
Zn | Zn2+ (1.0 M) || Cu2+ (1.0 M) | Cu
–
–
–
–
The ANODE is described before the CATHODE.
Concentrations of ions are indicated in brackets.
A vertical line represents a phase boundary.
A double vertical line represents the salt bridge.
Line Notation - an example

Calculate the cell potential for the cell described
below:
Mg | Mg2+ (1.0 M) || Ag+ (1.0 M) | Ag
Anode:
Mg  Mg2+ + 2 e-
E° = 2.37 V
(oxidation)
Cathode: Ag+ + e-  Ag
E° = 0.80 V
(reduction)
Overall:
Mg + 2 Ag+  Mg2+ + 2 Ag
E° = 3.17 V
Review: What’s the oxidizing agent? the reducing agent?
What’s E° for the reverse reaction?
The Nernst Equation


So far, the cells we have considered have operated
under standard conditions - 1.0 M solutions and 1.0
atm pressures at 25°C.
Cell potential is dependent on concentration and on
temperature, described by the Nernst Equation:
RT
Ecell  E 
ln Qcell 
nF

where “E°“ is the standard cell potential
“R” is the Ideal Gas Constant, 8.314
“T” is the temperature of the cell
“n” is the number of moles of electrons transferred
“F” is the Faraday, the charge on one mole of electrons, 96 500 C/mol
“Q” is the reaction quotient (remember your equilibrium unit!)
Using the Nernst Equation

Calculate Ecell for the cell described below at 25°C
Mg | Mg2+ (0.0050 M) || Ag+ (2.0 M) | Ag
– According to the balanced equation, Q 
[Mg 2 ]
[ Ag  ]2
– We have already seen that E°cell = 3.17 V
– According to our balanced equation, n = 2 mol e– Substituting into the Nernst Equation:
 (0.0050 M) 
(8.314)(298 K)

Ecell  (3.17 V ) 
ln 

2
(2 mol)(96500 C/mol)  (2.0 M) 
Thus, Ecell = 3.26 V
Calculating Equilibrium Constants



When an electrochemical cell operates, the
concentrations of the ions change until Q = K .
When the cell reaches equilibrium, Ecell = 0.
Combining these facts with the Nernst equation
gives:
Ecell  0  E 

RT
ln K 
nF
Rearranging, we can derive an equation to calculate
the equilibrium constant for a redox reaction:
RT
E 
ln K 
nF

Calculating K for a Redox Reaction


Calculate K for the reaction that occurs when Mg is
added to a solution of AgNO3.
The net reaction is:
Mg + 2 Ag+  Mg2+ + 2 Ag
E 
RT
ln K 
nF
(8.314)(298 K)
3.17 V 
ln K 
(2 mol)(96500 C/mol)
247 = ln (K)
so K = e247 = HUGE
Corrosion



Oxidation of metals with oxygen to form a metal oxide
In some cases a thin layer of oxide coats the metal
surface, preventing further oxidation (Al2O3)
When iron corrodes, the process is called “rusting”.
The iron oxide (Fe2O3) flakes off, exposing more
metal to corrosion.
Galvanized Iron - Preventing Corrosion


Iron can be protected from corrosion by coating it
with zinc metal - a metal that will oxidize first, instead
of the iron
Note that E°oxidation of zinc is greater than that of iron.
Cathodic Protection - Preventing Corrosion
To protect underground
pipelines, a sacrificial
anode is added.
The water pipe is
turned into the cathode
and an active metal is
used as the sacrificial
anode.
Magnesium is used as
the sacrificial anode,
since it is more easily
oxidized than the
iron water pipe.
Dry-Cell Batteries
Zn + 2 MnO2 + 2 NH4+ Zn+2 + Mn2O3 + 2 NH3 + H2O
The Alkaline Battery
(KOH is the electrolyte)
Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s)
Lead Storage Battery (Car Battery)

Anode:
Pb (s) + HSO4- (aq) 
PbSO4 (s) + H+ (aq) + 2 e- Eo = 0.296 V

Cathode:
PbO2 (s) + 3 H+ (aq) + HSO4- (aq) + 2 e- 
PbSO4(s) + 2 H2O ()
Eo = 1.628 V

Overall:
Pb (s) + PbO2 (s) + 3 H+ (aq) + 2HSO4 
2 PbSO4(s) + 2 H2O ()
Eo = 1.924 V
A Picture of a Car Battery
A 12-V battery has six cells
connected in series.
Each cell generates 2 V.
Fuel Cells



Anode: 2 H2 (g) + 4 OH- (aq) 2 H2O () + 4 eCathode: O2 (g) + 2 H2O () + 4 e-  4 OH- (aq)
Overall: 2 H2 (g) + O2 (g)  2 H2O ()
Electrolysis of Molten Sodium Chloride
Anode:
Cathode:
Overall:
Molten sodium
metal is formed
at the cathode.
Chlorine gas
is formed at the
anode.
2 Cl- ()  Cl2 (g) + 2 e2 Na+ () + 2 e-  Na()
2 Cl- () 2 Na+ ()  Cl2 (g) + Na()
Electrolysis of Water
Anode: 2 H2O  O2 + 4 H+ + 2 eCathode: 4 H2O + 4 e-  2 H2 + 4OH-
H 2O
Na2SO4
Overall:
H2
O2
6 H2O  O2 + 2 H2 + 4 OH- + 4 H+
Since 4 OH- + 4 H+  4 H2O
Net:
2 H2O  O2 + 2 H2
cathode
anode
Electrolysis of Aqueous Sodium Bromide

Two possible reactions at the cathode:
2 H2O + 2 e-  H2 +2OHE° = -0.83 V
Na+ + e-  Na
E° = -2.71 V

Two possible reactions at the anode:
2 Br-  Br2 + 2 eE° = -1.09 V
2 H2O  O2 + 4 H+ + 4 eE° = -1.23 V
Overall reaction:
2 Br- + 2 H2O  Br2 + H2 + 2 OHEocell = Eored + Eoox = -0.83 + (-1.09)
Eocell = -1.92 V
Quantitative Electrolysis
Q =It
Charge = Current x time
(Coulombs)
(Amperes)
(seconds)
1 F = 96 500 C/mol e-
What mass of aluminum can be produced in 8.00 min
by passing a constant current of 100 A through a molten
mixture of aluminum oxide, Al2O3?
Al3+ + 3 e-  Al (s)
60 s 100 C 1mol e  1mol Al 26.98 g Al
8.00 min 





1min 1 sec 96 500 C 3 mol e
1mol Al
= 4.47 g of Aluminum will be produced