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The highlights are:
1. Electrochemistry
2. Half reactions
3. Balancing redox reactions
4. Galvanic cells
5. Cell potential
6. Electrochemical series
7. Standard potential, free energy, and equilibrium constant
8. Nernst equation
9. Electrolysis
2.
What is Electrochemistry
The branch of chemistry that deals with
the use of spontaneous chemical reaction
to produce electricity and the use of
electricity to bring about non spontaneous
chemical change.
3.
What are Half reactions?
Oxidation occurs together with redox
reaction but it is helpful to consider each
process separately by writing two half
reactions.
Mg(s) Mg2+(aq) + 2e-
Fe3+(aq) + 3e- Fe(s)
4.
Class Practice
Write the equation for the half reaction for
(a) the oxidation of iron (II) ions to iron(III)
ions in aqueous solution; (b) the reduction
of copper (II) ions in aqueous solution to
copper metal.
5.
Balancing redox reactions
The chemical equation of a reduction
half reaction is added to that of an
oxidation half reaction to form the
chemical equation for the overall redox
reaction.
You can balance a redox reaction in
acidic and basic reaction.
6.
Class example
Permanganate ions, MnO4-, react with
oxalic acid, H2C2O4, in acidic aqueous
solution to produce manganese(II)ions
and carbon dioxide gas. The partial
skeletal equation is
MnO4- (aq) + H2C2O4(aq) Mn2+(aq)
+CO2(g)
7.
In basic solution
The products of the reaction of bromide
ions with permanganate ions in basic
aqueous solution are solid manganese
(IV) oxide, MnO2, and bromate ions, BrO3-
.
Balance the chemical equation for the
reaction.
9.
Galvanic cell
The Galvanic cell, named after Luigi Galvani,
consists of two different metals connected by a
salt bridge or a porous disk between the
individual half-cells. It is also known as a voltaic
cell and an electrochemical cell.
10.
What occurs in a galvanic
The zinc electrode is losing mass as Zn metal is
oxidized to Zn2+ ions which go into solution.
The concentration of the Zn2+ solution is
increasing. Anions, negative ions (e.g. SO42-), are
flowing from the salt bridge toward the anode to
balance the positive charge of the Zn2+ ions
produced.
11.
The copper electrode is gaining mass as Cu2+ ions in the
solution are reduced to Cu metal.
The concentration of the Cu2+ solution is decreasing.
Cations, positive ions (e.g. Na+), are flowing from the salt
bridge toward the cathode to replace the positive charge of
the Cu2+ ions that consumed.
A reaction may start at standard-state conditions, but as
the reaction proceeds, the concentrations of the solutions
change, the driving behind the reaction becomes weaker,
and the cell potential eventually reaches zero.
12.
In a galvanic cell, a spontaneous
chemical reaction tends to draw
electrons into the cell through the
cathode, the site of reduction, and to
release them at the anode, the site of
oxidation.
13.
Notation For Cells
Instead of drawing a cell diagram,
chemists have devised a shorthand way
of completely describing a cell called line
notation. This notation scheme places
the constituents of the cathode on the
right and the anode components on the
left. For example, a half-cell containing
1M solutions of CuO and HCl and a Pt
electrode for the reduction of Cu2+ would
be written as: Pt (s) | Cu2+ (aq), H+ (aq)
14.
Note that the spectator ions, oxide and
chloride, have been left out of the notation
and that the anode will be written to the
far left.
The salt bridge or porous disk is shown in
the notation as a double line ( || ).
Therefore, a cell that undergoes the
oxidation of magnesium by Al3+ could have
the following cell notation if the anode is
magnesium and the cathode is aluminum:
Mg (s) | Mg2+ (aq) || Al3+ (aq) | Al (s)
15.
An electrode is designated by
representing the interfaces between
phases by Ι.
A cell diagram depicts the physical
arrangement of species and inter faces
with a salt bridge denoted by II.
16.
Class Practice
Write the diagram for a cell that has a
hydrogen electrode on the left, an iron
(III)/iron(II) electrode on the right, and
includes a salt bridge. Both electrode
contacts are platinum.
17.
Cell Potential
An idealized cell for the electrolysis of sodium chloride is shown in
the figure below. A source of direct current is connected to a pair
of inert electrodes immersed in molten sodium chloride. Because
the salt has been heated until it melts, the Na + ions flow toward
the negative electrode and the Cl- ions flow toward the positive
electrode.
When Na+ ions collide with the negative electrode, the battery carries a
large enough potential to force these ions to pick up electrons to form
sodium metal.
18.
Negative electrode (cathode): Na+ + e-- Na
Cl- ions that collide with the positive electrode
are oxidized to Cl2 gas, which bubbles off at this
electrode.
Positive electrode (anode): 2 Cl- Cl2 + 2 e-
The net effect of passing an electric current
through the molten salt in this cell is to
decompose sodium chloride into its elements,
sodium metal and chlorine gas.
19.
Electrolysis of NaCl:
Cathode (-): Na+ + e- Na
Anode (+): 2 Cl- Cl2 + 2 e-
The potential required to oxidize Cl- ions to Cl2
is -1.36 volts and the potential needed to
reduce Na+ ions to sodium metal is -2.71 volts.
The battery used to drive this reaction must
therefore have a potential of at least 4.07 volts.
This example explains why the process is called
electrolysis. Electrolysis uses an electric
current to split a compound into its elements.
Electrolysis: 2 NaCl(l) 2 Na(l) + Cl2(g)
20.
Home work
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21.
Standard cell potential
A cell's standard state potential is the potential
of the cell under standard state conditions,
which is approximated with concentrations of 1
mole per liter (1 M) and pressures of 1
atmosphere at 25oC.
To calculate the standard cell potential for a
reaction:
Write the oxidation and reduction half-
reactions for the cell.
Look up the reduction potential, E oreduction, for
the reduction half-reaction in a table of
reduction potentials
22.
Look up the reduction potential for the reverse of
the oxidation half-reaction and reverse the sign
to obtain the oxidation potential. For the
oxidation half-reaction, Eooxidation = - Eoreduction.
Add the potentials of the half-cells to get the
overall standard cell potential.
Eocell = Eoreduction + Eooxidation
Example:
Find the standard cell potential for an
electrochemical cell with the following cell
reaction.
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
23.
Write the half-reactions for each process.
Zn(s) Zn2+(aq) + 2 e- Cu2+(aq) + 2 e- Cu(s)
Look up the standard potentials for the reduction half-reaction.
Eoreduction of Cu2+ = + 0.339 V
Look up the standard reduction potential for the reverse of the
oxidation reaction and change the sign.
Eoreduction of Zn2+ = - 0.762 V Eooxidation of Zn = - ( - 0.762 V) = + 0.762 V
Add the cell potentials together to get the overall standard cell
potential.
oxidation: Zn(s) Zn2+(aq) + 2 e- Eoox. = - Eored. = - (- 0.762 V) =
+ 0.762 V
reduction: Cu2+(aq) + 2 e- Cu(s) Eored. = + 0.339 V overall:
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Eocell = + 1.101 V
24.
Class Practice
The standard potential of a Zn2+/Zn
electrode is – 0.76 V, and the standard
potential of the cell Zn(s)IZn2+(aq)IICu2+
(aq)ICu(s) is 1.10 V. What is the standard
potential of the Cu2+/Cu electrode?
25.
Home work
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26.
The Electrochemical Series
The oxidizing and reducing power of a
substance can be determined by its
position in the electrochemical series.
The strongest oxidizing agents are at the
top of the table as reactants, the
strongest reducing agents are at the top
of the table as products.
27.
Class practice
Can aqueous potassium permanganate
be used to oxidize iron(II) to iron(III)
under standard conditions in acidic
solution.
28.
To find a standard cell potential arising from
a spontaneous reaction we must combine
the standard potential of the cathode half
reaction(reduction) with that of the anode
half reaction (oxidation) in such a way so as
to obtain a positive value. The overall
potential must be positive because that
corresponds to a spontaneous process, and
only a spontaneous process can generate a
potential. If the calculation results in a
negative value, this means that the reverse
reaction is spontaneous.
29.
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30.
Standard potential, free
energy and equilibrium
The free energy change is a measure of the
change in the total entropy of a system and
its surroundings at constant pressure;
spontaneous processes are accompanied
by a decrease in free energy.
G=H−TS
H is the enthalpy
T is the temperature
S is the entropy
31.
When the reaction free energy ∆Gf is
negative , the cell reaction is spontaneous
and the cell generates a positive potential.
When ∆Gf is is large as well as negative,
the cell potential is high as well as positive.
The relationship suggests that ∆G f =−nFE
n is the number of moles of electrons that
are transferred between the electrodes for
the cell reaction as written in the chemical
equation.
32.
The Faraday constant, F is the
magnitude of the charge per mole of
electrons: F=Nae=9.6485 X104 C/mol.
1C.V=1J
We can write F= 9.6485 X104 J/V.mol
33.
Class Practice
The cell Cr(s) ΙCr2+(aq)Ι Cu (s) was found to
have E⁰ = +1.08V at 298K.(a) Write the balanced
net equation for the cell reaction;(b) determine n;
and© calculate the standard reaction free energy
at 298K.
34.
The following cell was set up: Hg(l)
ΙHg2Cl₂(s)ΙHCl (aq)ΙΙHg₂(NO₃)₂(aq)ΙHg(l), E⁰ =+ 0.52V
at 298.(a) Write the equation for the cell reaction.
(b) determine n, and © calculate the standard
reaction free energy at 298K.
35.
The equilibrium constant of a reaction
can be calculated from standard
potentials by combining the equation for
the half reactions to give the reaction of
interest and determining the standard
potential of the corresponding cell.
36.
Class Practice
The reaction between zinc metal and
iodine in water generates 1.30 v under
standard conditions. Determine a)n and
b) ∆Gf ⁰ for the cell reaction Zn(s)+I2(aq) Zn2+ (aq)
+ 2I- (aq)
The solubility product is the equilibrium constant
for the dissolution of a salt. Calculate the solubility
product of silver chloride.
37.
The Nernst Equation
The variation of cell potential with
composition is expressed by the Nernst
equation:
E=E⁰ ─(RT/nF) ln Q
38.
Class Practice
Calculate the potential at 25⁰C of a Daniel
cell in which the concentration of Zn²+ ions is 0.10
mol/L and that of the Cu2+ ions is 0.0010 mol/L
39.
Home work
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40.
Electrolysis is a method of separating
chemically bonded elements and
compounds by passing an electric
current through them.
41.
In an electrolytic cell, current supplied
by an external source is used to drive a
nonspontaneous redox reaction.
The Galvanic cell, consists of two
different metals connected by a salt
bridge or a porous disk between the
individual half-cells. It is also known as a
voltaic cell and an electrochemical cell.
42.
The porous bridge is substituted by a
Salt bridge for a galvanic cell.
43.
Oxidation-reduction or redox reactions take
place in electrochemical cells. There are two
types of electrochemical cells. Spontaneous
reactions occur in galvanic (voltaic) cells;
nonspontaneous reactions occur in electrolytic
cells.
Both types of cells contain electrodes where
the oxidation and reduction reactions occur.
Oxidation occurs at the electrode termed the
anode and reduction occurs at the electrode
called the cathode.
44.
The anode of an electrolytic cell is positive
(cathode is negative), since the anode
attracts anions from the solution. However,
the anode of a galvanic cell is negatively
charged, since the spontaneous oxidation at
the anode is the source of the cell's
electrons or negative charge. The cathode of
a galvanic cell is its positive terminal. In
both galvanic and electrolytic cells,
oxidation takes place at the anode and
electrons flow from the anode to the
cathode.
45.
The redox reaction in a galvanic cell is a
spontaneous reaction. Hence, galvanic
cells are commonly used as batteries.
Galvanic cell reactions supply energy
which is used to perform work. The
energy is harnessed by situating the
oxidation and reduction reactions in
separate containers, joined by an
apparatus that allows electrons to flow. A
common galvanic cell is the Daniell cell.
46.
Class practice
Predict the products resulting from the
electrolysis of 1M ZnNO2(aq) at pH=7
47.
Home work
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