Transcript 2nd lect., anal. che..

Reduction- Oxidation
Reactions
2nd lecture
1
Learning Objectives
What are some of the key things we learned from this lecture?
• Types of electrochemical cells. √
• How to measure Standard electrode potential. √
•
reference electrode. √
• How to measure the electrode potential from the Nernst
equation. √
• Factors affecting the oxidation potential. √
2
Electrochemical cell
An electrochemical cell is a device used for generating an
electromotive force (voltage) and current from chemical reactions,
(galvanic, voltaic cell).
or the reverse, inducing a chemical reaction by a flow of current,
(electrolytic).
The current is caused by the reactions releasing and accepting
electrons at the different ends of a conductor.
There are two types of electrochemical cells:
• Electrolytic cell,
(electrical Energy
chem. Energy)
• Galvanic cell (voltaic),
(chemical energy)
electrical energy)
In an electrochemical cell, an electric potential is created
between two dissimilar metals. This potential is a measure
of the energy per unit charge which is available from the
oxidation/reduction reactions to drive the reaction. It is
customary to visualize the cell reaction in terms of two halfreactions, an oxidation half-reaction and a reduction halfreaction.
Reduced species
oxidized species + ne-
Oxidation at anode
Oxidized species + ne-
reduced species
Reduction at cathode
Electrolytic cell
An electrolytic cell decomposes chemical compounds by means of electrical energy, in
a process called electrolysis.
The result is that the chemical energy is increased.
An electrolytic cell has three component parts: an electrolyte and two electrodes (a
cathode and an anode). The electrolyte is usually a solution of water or other solvents
in which ions are dissolved.
Molten salts such as sodium chloride are also electrolytes.
When driven by an external voltage applied to the
electrodes, the electrolyte provides ions that flow to
and from the electrodes, where charge-transferring, or
redox, reactions can take place.
Only for an external electrical potential (i.e. voltage) of
the correct polarity and large enough magnitude can
an electrolytic cell decompose a normally stable, or
inert chemical compound in the solution.
At anode:
2Cl-
At cathode: 2Na+ + 2eCell reaction: 2Na+ + 2Cl-
Cl2 +2e-
2Na
(oxidation)
(reduction)
2Na + Cl2 (redox reaction)
2- Galvanic Cells
• In redox reaction, electrons are transferred from the oxidized species to the reduced
species.
• Imagine separating the two 1/2 cells physically, then providing a conduit through
which the electrons travel from one cell to the other.
8H+ + MnO4- + 5e
Fe2+
Mn2+ + 4H2O
Fe3+ + e-
x5
We need to “complete the circuit” by allowing positive ions to flow as well.
We do this using a “salt bridge” which will allow charge neutrality in each cell to be
maintained.
Salt bridge/porous disk: allows for ion migration such
that the solutions will remain neutral.
Galvanic Cell
Electrochemical cell in which chemical reactions are used to
create spontaneous current (electron) flow
Standard electrode potential (E0 )
• If the concentration of the components of the cell
are unity (1M), measured at 250c
• The potential is called standard potential.
• It is not possible to measure the absolute
potential of an electrode, but it could be
measured relative to a standard reference
electrode.
How to measure E0 standard electrode potential?
 Suppose we have Cu rod
immersed in its ions (half cells).
 In order to measure potential of
an electrode, it must be
connected to another electrode
(half cell) having known constant
potential called reference
electrode and we use
“Normal (standard) Hydrogen
Electrode“.
So an electric current will flow from that having a higher potential
to the other one, this current is called e.m.f. Which is the algebric
difference of the electrode potential of the two half cells.
If the reference electrode has a potential equal to zero,
therefore, so the e.m.f. measured will be equal to E0
Reference Electrodes
What is Normal Hydrogen Electrode (NHE)?
 It consists of a piece of platinium foil coated electrolytically
with platinum black and immersed in a solution of HCl
(normal solu).
 Hydrogen gas at pressure of 1 atmosphere is passed.
 Platinum black layer absorbs a large amount of H2
Pt wire
and can be considered as a bar of H2.
H+ + e-
½ H2
Glass tube to
contain H2 (g)
 Under fixed conditions:
1 atm pressure
Normal solu of HCl
 The potential is assumed to be zero.
 By connecting the electrode of unknown
potential to “NHE“, the E0 can be measured
by a potentiometer.
(1 atm)
(1 N)
Bubbles of H2 (g)
Standard electrode potential E0: Is the e.m.f. produced when
a half cell consisiting of an element immersed in one molar
solution of its ions is coupled with NHE.
By measuring E0 for all elements and arrange into a series
known as “Electrochemical Series “:Starting from
System
E0
E0 = -ve value, to
Zn/Zn2+ -0.76
hydrogen where
E0 = zero, then to
Fe/Fe2+ -0.44
E0 = +ve value
Sn/Sn2+ -0.13
H2(pt)/H+
0
Cu/Cu2+ +0.34
From this series we can conclude that:
1- The sign of E0 is similar to the charge on the metal electrode.
2- The greater the –ve value of E0, the greater the tendency of the
metal to pass into ionic state (oxidized) i.e. E0 is a quantitative
measure of the ability of the element to lose electrons giving its
ions.
3- Metal with more negative potential will displace any other metal
below it in the series from its solution.
Reduction Potentials
Electrode Couple
Na+ + e- --> Na
Mg2+ + 2e- --> Mg
Al3+ + 3e- --> Al
Zn2+ + 2e- --> Zn
Fe2+ + 2e- --> Fe
Cd2+ + 2e- --> Cd
Tl+ + e- --> Tl
Sn2+ + 2e- --> Sn
Pb2+ + 2e- --> Pb
2H+ + 2e- --> H2(SHE)
S4O62- + 2e- --> 2S2O32Sn4+ + 2e- --> Sn2+
SO42- + 4H+ + 2e- --> H2O + H2SO3(aq)
Cu2+ + e- --> Cu+
S + 2H+ + 2e- --> H2S
AgCl + e- --> Ag + ClSaturated Calomel (SCE)
UO22+ + 4H+ + 2e- --> U4+ + 4H2O
"E0, V"
-2.7144
-2.3568
-1.676
-0.7621
-0.4089
-0.4022
-0.3358
-0.141
-0.1266
0
0.0238
0.1539
0.1576
0.1607
0.1739
0.2221
0.2412
0.2682
Reduction Potentials
Hg2Cl2 + 2e- --> 2Cl- + 2Hg
Bi3+ + 3e- --> Bi
Cu2+ + 2e- --> Cu
Fe(CN)63- + e- --> Fe(CN)64Cu+ + e- --> Cu
I2 + 2e- --> 2II3- + 2e- --> 3IH3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O
2HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2ClHg2SO4 + 2e- --> 2Hg + SO42I2(aq) + 2e- --> 2IO2 + 2H+ + 2e- --> H2O2(l)
O2 + 2H+ + 2e- --> H2O2(aq)
Fe3+ + e- --> Fe2+
Hg22+ + 2e- --> Hg
Ag+ + e- --> Ag
Hg2+ + 2e- --> Hg
2Hg2+ + 2e- --> Hg22+
NO3- + 3H+ + 2e- -->HNO2(aq) + H2O
0.268
0.286
0.3394
0.3557
0.518
0.5345
0.5354
0.5748
0.6011
0.6152
0.6195
0.6237
0.6945
0.769
0.7955
0.7991
0.8519
0.9083
0.9275
If an inert electrode such as platinum is placed in a
redox system e.g. Fe3+/Fe2+ it will assume a definite
potential (oxidation potential) indicative of the position
of equilibrium.
Oxidation potential is the pot. difference arround an
inert electrode immersed in solu. of oxidized and
reduced form.
If the oxidizing
tendency predominate,
the system will take
electrons from platinum
leaving it positively
charged
If the system has
reducing properties,
electrons will be given
up to platinum and it
will acquire a negative
charge.
The magnitude of the potential will be a measure of oxidizing or
reducing character of the system. If +ve oxidizing and if –ve reducing.
To measure the oxidation potential on pt wire, the system should be
connected to NHE.
Standard Oxidation potential (SOP): Is the e.m.f. produced when
a half cell consisting of an inert electrode (pt) dipped in equal
concentration of both oxidized and reduced form of a system is
connected with NHE.
Depending on the SOP the systems were arranged in a series from
this series we can conclude that:
1- The higher the SOP of the system, the stronger the oxidizing
power of its oxidized form and the weaker the reducing power of
its reduced form.
2- the most powerful oxidizing agents are those at the top of the
series with higher +ve potential, and the most powerful reducing
agents occupy the bottom with higher –ve potential.
There should be adifference in oxidation potential between the two
systems of about 0.4 v for oxidation to occur.
Nernst Equation FOR OXIDATION POTENTIAL
• Compensated for non unit activity (not 1 M)
• Relationship between cell potential and activities
• aA + bB +ne- <--> cC + dD
   
2. 30 RT
nF
log
c
d
a
b
[C ] [ D ]
[ A ] [ B]
• At 298K, 2.3RT/F = 0.0592
• What is potential of an electrode of Zn(s) and 0.01 M Zn2+
• Zn2+ +2e- <--> Zn °= -0.763 V
• activity of metal is 1
  0. 763 
0.0592
2
log
1
0. 01
 0.822 V
• Eind.electrode = E0 - 0.059
n
• Eind.electrode = E0M/M+ -
log
0.059
n
[products]
[reactants]
log
1
[M+]
Factors affecting the oxidation potential:
1- Common ion effect:
If we determine a sample of Fe2+ with KMnO4 in presence of Clions we have 3 systems:
MnO4-/Mn2+ (1.52)
Cl2/Cl(1.36)
Fe3+/Fe2+ (0.77)
KMnO4 will oxidize both Cl- and Fe2+ so more KMnO4 will be
consumed.
To overcome this problem the oxidation potential of MnO4-/Mn2+
should be reduced below Cl2/Cl- HOW?
By adding Zimmermann´s reagent
It contains [MnSO4, H3PO4, H2SO4]
E25 = E0 - 0.059/n log [Mn2+] / [MnO4-]
 Presence of MnSO4 increases the conc of Mn2+ ion.
 So the oxidation potential will decrease below that Cl2/Cl-
2-Effect of H+ ion concentration:
[H+] has a deciding effect on the oxidation potential of oxidizing
agents containing oxygen and so [H+] should be included in
Nernest equation.
E increases with increasing [H+] i.e.increasing acidity.
MnO4- + 8H+ + 5e ↔ Mn2++ 4H2O
EMnO
2+
4 /Mn
= E0 - 0.059/5 log [Mn2+] / [MnO4-] [H+]8
At pH 5: KMnO4 oxidizes I- but not Br- and ClAt pH 3: KMnO4 oxidizes I- and Br- but not ClAt lower pH : KMnO4 oxidizes I- , Br- and Cl-
AsO43- + 2H+ + 2e  AsO33- + H2O
EAsO
334 / AsO3
= E0 - 0.059/2 log [ASO33-] / [AsO43-] [H+]2
If [H+] increases: The oxidation potential of AsO43- increases.
If [H+] decreases: The oxidation potential of AsO43-decreases
and the reducing power of AsO33- increases.
Therefore: AsO43- oxidizes I- to I2 in acid medium.
While:
I2 oxidizes AsO3-3 in alkaline medium with NaHCO3
acid
AsO4-3 + 2 I- + 2H+
AsO3-3 + I2 + H2O
alkaline
3-Effect of complexing agents:
E.g.1- Oxidation potential of I2/2I- system increases in
presence of HgCl2 WHY?
Because HgCl2 forms a complex with the I- ions [HgI4]2-
I2 + 2 e  2I-
EI
2 /2I
-=
E0 -
0 . 059
2
log
[ I  ]2
[ I 2]
 removing the I- ions from the share of the reaction.
 minimizing its concentration.
 decreasing the ratio of [I-]2 / I2 .
 increasing the oxidation potential of I2 /2I- system .
E.g.2- Oxidation potential of Fe3+/ Fe2+ system
decreases on addition of F- or PO43- WHY?
Because F- and PO43- form stable complexs with Fe3+
which are [FeF6]3- or [Fe (PO4)2]-3
Fe3+ + e  Fe2+
2
0 . 059
[ Fe ]
log
EFe3+/Fe2+ = E0 3
1
[ Fe
]
 removing the Fe3+ ions from the share of the reaction.
 minimizing its concentration.
 Increasing the ratio of Fe2+/ Fe3+ .
 lowering the oxidation potential of Fe3+/ Fe2+ system .
4-Effect of precipitating agents:
Oxidation potential of Ferricyanide/ Ferrocyanide system
increases on addition of Zn2+ WHY?
Because Zn2+ will precipitate Zn2[Fe(CN)6]
Fe (CN)63- + e  Fe(CN)64E = Eo -
0 . 059
1
log
[ Fe (CN)
[Fe(CN)
6
]
6
]
4-
3-
 removing the Ferrocyanide ions from the share of the
reaction.
 minimizing conc of ferrocyanide.
 decreasing the ratio of [Fe(CN)6]4-/ [Fe (CN)6]3- .
 increasing the oxidation potential of [Fe(CN)6]3-/ [Fe (CN)6]4-