Transcript Slide 1

Electrochemistry
Physical Chemistry
Daniel Cell
1.
2.
3.
Electrochemistry is the study of the interconversion of electrical and
chemistry energy.
Voltaic cell, in which a spontaneous reaction generates electrical energy.
Electrolytic cell: electrical energy is used to bring about a non spontaneous
reaction.
Oxidation-reduction (redox) reactions
occur when electrons are given up by the
substance being oxidized (the reducing
agent) and simultaneously gained by the
substance being reduced (the oxidizing
agent).
2 Ag+ + Zn  Zn2+ + 2 Ag
At Anode : Zn  Zn2+ + 2e
At Cathode: Ag+ + e  Ag
Daniel Cell
1.
2.
3.
4.
The two half-cells are connected by a salt bridge, which
prevents free mixing of both electrolyte solutions but permits
the proper movement of ions to maintain electrical neutrality.
When Zn is oxidized some anions must enter (or cations must
leave) the Zn half-cell to compensate for the added positive
charge of the Zn2+ produced. Also when Ag+ reduced cations
must enter (or anions must leave) the Ag half-cell.
Let’s say the salt bridge used is: KNO3 , K+ ions move from the
salt bridge into cathode half bridge. At the same time NO3- ions
move into the anode half cell.
The Daniel Cell Diagram:
Zn(s)/ZnSO4 (aq)
AgSO4(aq)/Ag(s)
Practice
1.
a)
b)
c)
When chlorine gas is bubbled through a water solution of NaBr, a
spontaneous redox reaction occurs:
Cl2 (g) + 2Br-(aq)  2Cl-(aq) + Br2 (g)
What is the cathode and anode reaction?
Which way do electrons move in the external circuit?
Which way do anions and cations move within the cell?
Solution:
a)
Cathode: Cl2 + 2e  2Cl- Anode: 2Br-  Br2 + 2e
b)
From anode to cathode
c)
Anion to anode, cations to cathode
Writing Redox Equation
Oxidation Number

a)
b)
c)
d)
e)
f)
Determine the oxidation number of the underlined atom:
V2O5
SO42-SO32OClHCO3MnO4-
Changes in Oxidation Number
1.
xA + yB  Product
x(+a) + y(-b) = 0
Practice:
1.
What are the oxidation numbers of KIO3, KIO, Na2SO3 and Na2SO4. Thus,
balance the equation.
xKIO3 + y Na2SO3  cKIO + d Na2SO4
2.
a)
Use the oxidation number method to balance the equation:
Sn +HNO3  SnO2 + NO2 + H2O
Writing Half Equation
ClO3- + I-  Cl- + I2
1.
Determine change of oxidation number:
ClO3-  Cl- : +5 to -1 = -6….. Reduction
I-  I2 : -1 to 0 = +1…. Oxidation
2.
3.
ClO3- + 6e  Cl2I I2 + 2e
H2O, H+ and OH- can be added to balance up the half equatons.
ClO3- + 6H+ + 6e  Cl- + 3H2O
2I I2 + 2e
)x3
--------------------------------------------------------ClO3- + 6H+ + 6I-  Cl- + 3H2O + 3I2
Practice 2
1.
2.
3.
4.
5.
Cr2O72- + H+ + NO2-  Cr3+ + NO3Cl2 + SO2  Cl- + SO42C + H2SO4  CO2 + SO2
H2O2 + Mn2+  MnO2 + H2O
Cu + HNO3  Cu2+ + NO
Electrode Potential
The Electrode Potential of Metal
1.
Suppose you have a piece of magnesium in a beaker of water. There
will be some tendency for the magnesium atoms to shed electrons
and go into solution as magnesium ions. The electrons will be left
behind on the magnesium
In a very short time,
there will be a build-up
of electrons on the
magnesium, and it will
be surrounded in the
solution by a layer of
positive ions
(Helmholtz double
layer). This produces
a potential defference
between the metal and
the solution.
The Electrode Potential of Metal
1.
Dynamic equilibrium will be established when the rate at
which ions are leaving the surface is exactly equal to the rate at
which they are joining it again. At that point there will be a
constant negative charge on the magnesium, and a constant
number of magnesium ions present in the solution around it.
The potential difference produced between
the electrode and the solution in half cell is
called electrode potential. The magnitude of
the electrode potential depends on the
tendency of the metal to lose electrons and
to form hydrated metal ions
The Electrode Potential of Metal
1. Copper is less reactive and so forms its ions less readily.

The position of the magnesium equilibrium . . .


. . . lies further to the left than that of the copper equilibrium.
The Electrode Potential of Metal

All we need to know is that the magnesium equilibrium lies
further to the left than the copper one. We need to know that
magnesium sheds electrons and forms ions more readily than
copper does.

That means that we don't need to be able to measure the
absolute voltage between the metal and the solution. It is
enough to compare the voltage with a standardized system
called a reference electrode.

The system used is called a standard hydrogen electrode.
The Electrode Potential of Non Metal


The standard hydrogen electrode
The standard hydrogen electrode looks like this:
As the hydrogen gas flows over the porous
platinum, an equilibrium is set up between
hydrogen molecules and hydrogen ions in
solution. The reaction is catalyzed by the
platinum.
• The ions in the solution have a
concentrated of 1.0 M
• All gases involved have a pressure of 1
atm.
•A temperature of 25oC
Using the standard hydrogen electrode
1.
The standard hydrogen electrode is attached to the electrode system you are
investigating - for example, a piece of magnesium in a solution containing
magnesium ions.
Using the standard hydrogen electrode
1.
The emf for the cell is 2.37 volt.
2.
According to IUPAC, the standard potential for the magnesium
electrode is -2.37. The negative sign is used to show that the
electrode is the negative pole.
3.
The voltmeter shows that the electrons flow from the
magnesium electrode to the hydrogen electrode in the external
circuit. This means that at Mg electrode:
Mg  Mg2+ + 2e
3.
At hydrogen electrode: 2H+ + 2e  H2
4.
The overall cell reaction: Mg + 2H+  Mg2+ + H2
5.
The cell diagram:
Mg(s)/Mg2+(aq, 1M)
H+ (aq, 1M)/ H2 (g) /Pt(s)
EӨ = -2.37 V
Using the standard hydrogen electrode
1.
2.
3.
a)
b)
Half equation is written as a reduction process:
Oxidized species + ne  reduced species.
Thus, Ag+(aq)/Ag(s) EӨ = +0.80 V means that the following half cell
reaction has a standard electrode potential of +0.80
Ag+ + e  Ag
EӨ = +0.80 V
The cell diagram is written with:
The half cell undergoing oxidation ( negative electrode) on the left of the
diagram.
The half cell undergoing reduction ( positive electrode) on the right of the
diagram
metal / metal ion combination
E° (volts)
Mg2+ / Mg
-2.37
Zn2+ / Zn
-0.76
Cu2+ / Cu
+0.34
Ag+ / Ag
+0.80
Practice
1. Consider the following equilibrium involving I2 and Fe3+:
I2 + 2e  2IEӨ = +0.54 V
Fe3+ + e  Fe2+
EӨ = +0.77 V
a)
Write the ionic equation for the overall cell reaction.
b)
State the observation after the half cell have been connected for a short
while.
c)
Determine the polarity of the electrode in the electrochemical cell.
Solutions
a)
2Fe3+ + 2I-  2Fe2+ + I2
b)
The beaker on the right (Fe3+) : yellow to green
The beaker on the left (I-)
: yellow to brown
c)
I2 + 2e  2I- is anode cause oxidation happen.
Fe3+ + e  Fe2+ is cathode cause reduction occur.


We could therefore write down the emf of the above cell (under
standard conditions) as:
emf of cell = potential of Cu - potential of Zn
or conventionally
For example, one could use the reaction


to construct the cell
The electrochemical Series
1.
2.
3.
4.
5.
6.
The electrochemical series shows the ease of a species accept or loses
electrons.
The higher a species the easier it loses electron (oxidized)
The EӨ value shows the possibility of a reaction but not the quantity of
substances involved.
If EӨ for I2 + 2e  2I- +0.54 V, then ½ I2 + e  I- not (1/2 x 0.54V) but is
+0.54V.
The reactivity of metals increases as the electrode potential become more
negative while the reactivity if non metal increases as the electrode
potential become more positive.
The strength of oxidizing agents increases on descending the series while
the strength of reducing agents increases on ascending the series.
Practice
1.
a)
b)
c)
Arrange the following substances in the ascending order of their
oxidizing strength:
Na, Cl2, Cu
Zn, Ag+, Ag
Cl2, Br2, Br-
2. The EӨ value for three elements, X, Y and Z are given below:
X  X+ + e
EӨ = -0.34V
Y  Y+ + e
EӨ = -0.80V
Z  Z+ + e
EӨ = +2.46V
Predicting the Possibility of Redox Reaction.
1.
The possibility of a spontaneous reaction occurring can be predicted by
using:
a)
Electrochemical series
b)
EӨ value for the overall reaction
2.
An oxidizing agent on the left of the electrochemical series will oxidize a
reducing agent on the right if the reducing agent is above the oxidizing
agent in the series.
I2 + 2e  2IEӨ = +0.54 V
Br2 + 2e  2BrEӨ = +1.07 V
Cl2 + 2e  2ClEӨ = +1.33 V
Br2 is able to oxidize I- into I2 but cannot oxidize Cl- into Cl2
Predicting the Possibility of Redox Reaction.
1.
The possibility of a spontaneous reaction occurring can be predicted by
using:
a)
EӨ value for the overall reaction
If EӨ for a combined reaction is positive, then the reaction will occur
spontaneously.
Sn2+ + 2Fe3+  Sn4+ + 2Fe2+
Sn2+  Sn4+ + 2e
EӨ = -0.15 V
2Fe3+ + 2e  2Fe2+
EӨ = +0.77V
--------------------------------------------------------------EӨ = +0.62V
--------------------------------------------------------------2.
Practice
1.
Predict whether the following reactions can take place under
standard condition.
a)
Reduction of Sn2+ to Sn by Mg
b)
Oxidation of Cl- to Cl2 acidified Cr2O72Hints:
Mg2+ + 2e  Mg
EӨ = -2.38 V
Sn2+ + 2e  Sn
EӨ = -0.14 V
-------------------------------------------------------------------Cr2O72- + 14H+ + 6e  2Cr3+ + 7H2O EӨ = +1.33 V
Cl2 + 2e  ClEӨ = +1.36 V
Predicting the Stability of Aqueous Ions
The relative stability of Cr and Cr2+ can be explain as follow:
Cr2+ + 2e  Cr
EӨ = -0.91 V
2H+ + 2e  H2
EӨ = 0.00 V
------------------------------------------------------Cr + 2H+  Cr2+ + H2
EӨ = + 0.91V
------------------------------------------------------2.
This implies that Cr2+ is more stable than Cr because the forward reaction
can occur spontaneously.
3.
The Co3+ ion can be stabilized in aqueous solution by complexing it with
other ligands like ammonia.
1.
The Nernst Equations
1.
The value of the electrode potential depends on:
a)
The concentration of the ion and the temperature of the
solution
b)
The pH value (acidity) of the solution
c)
The effect of complex formation.
For the electrode system:
oxidized species + ne  reduced species
E = EӨ + RT/nF {ln [oxidized species]/[reduced species]}
or
E = EӨ + 0.059/n {lg [oxidized species]/[reduced species]}
The Nernst Equations

1.
2.
aA+bB=cC+dD
Q = [C]c[D]d/ [A]a[B]b
E = EӨ - 0.059/n {lg Q}
Sn2+(aq) + Br2 (l)
Sn4+(aq) + 2 Br-(aq)
Q = [Sn4+][Br-]2 / [Sn2+]
The Nernst equation for this reaction is:
E = 0.092 – 0.059/2 lg[ (0.0001)(0.0001)2/ 0.05]
E = 0.92 V - (-0.316 V)
E = 1.24 V
Effect of the Concentration of an ion on the emf of a cell
1.
Cu(s) / Cu2+ (aq) // Ag+ (aq) / Ag (s)
Cu + 2Ag+  2Ag + Cu2+
E
= EӨ
- 0.059/n {lg Q}
If the concentration of Cu2+ ions in the copper half cell is reduced, the ECu is
less than EӨ and the emf of the electrochemical cell increases.
If the concentration of Cu2+ ions in the copper half cell is reduced, the
equilibrium will shift to the right.
The Nernst Equation and Solubility Product
1.
When a silver electrode is immersed into a saturated silver chloride
solution, the electrode potential at 298 K is 0.51 V. Calculate the solubility
product of silver chloride at 298 K. EӨ Ag+/Ag = +0.80 V.
E = EӨ + 0.059/1 lg[Ag+]
0.051 = 0.80 + 0.059 lg[Ag+]
[Ag+] = 1.22 x 10-5 moldm-3
Ksp = [Ag+][Cl-]
= (1.22 x 10-5 )2
= 1.49 x 10 -10 mol2dm-6
The Nernst Equation and Solubility Product
1.
Consider the cell:
Pt(s) /H2 (g, 1.0 atm)/HCl(0.01M)// AgCl(aq)/Ag(s)
The measured cell emf is 0.046 V at 25oC and EӨ Ag+/Ag = +0.80 V
a)
Write equation for the reaction occurring at the electrodes.
Write an equation for the overall cell reaction.
Calculate the emf of the cell under standard conditions.
Calculate the [Ag+] in equilibrium with AgCl and Cl-.
What is the concentration of Cl-?
Find the Ksp of AgCl.
b)
c)
d)
e)
f)
The Nernst Equation and Equilibrium Constant
1.
When the redox reaction in a Daniel cell has achieved equilibrium, Ecell =
0.0 V. This is because at equilibrium, there is no overall change taking place
in a chemical reaction.
0 = EӨ - 0.059/n lg Kc
EӨ = 0.059/n [lg Kc]
2. Consider the following reaction:
2Fe3+ + Cu  2Fe2+ + Cu2+
a)
Calculate the EӨ the system.
b)
Calculate the equilibrium constant Kc for the reaction. [ 0.43V, 3.8 x1014]
The Nernst Equation and pH
1.
The cell reaction:
Zn + 2H+  Zn2+ + H2
EӨcell = 0.00-(-0.76)
= + 0.76 V
Using Nernst Equation:
Ecell = EӨcell + 0.059/n lg [H+]2/[Zn2+]PH2
Ecell = 0.76 + 0.059/2 lg [H+]2 / (1)(1)
pH = (0.76 – Ecell ) / 0.059
If the emf for the electrochemical cell is + 0.54 V, the pH would be
pH = (0.76 – 0.54)/0.059
= 3.7