Redox Equilibria

1
Redox Equilibria
-2
-2
-1
6
-2
-2
6
-1
-2
-2
Peroxodisulphate(VI) ion S2O82-
The oxidation state of S is not 7 as it must
range from -2 to 6
The O-atom attached to sulphur, a less
electronegative element than itself, has an
unusual oxidation number of -1, while the other
bond is to itself. The oxidation number of
sulphur is 6
2
Balancing redox reactions

• In investigating the reaction of ClO4-(aq), it
  was found that 25.0 cm3 of 0.05 M ClO4-(aq)
  required 50.0 cm3 of 0.20 M acidified
  titanium(III) solution to reach the end point.
• Deduce the oxidation state of the product
  containing chlorine and Write a balanced ionic
  equation.
• n(Ti3)/n(ClO4-) (50x0.2)/(25x0.05) 8
• Change in oxidation state of Cl in ClO4- 8, Cl-
  is the product
• ClO4- (aq) 8Ti3(aq) Cl- (aq) 8Ti4
  (aq)
• ClO4- (aq) 8Ti3(aq) 8H(aq) Cl- (aq)
  8Ti4 (aq)
• ClO4- (aq) 8Ti3(aq) 8H Cl- (aq) 8Ti4
  (aq) 4H2O(l)

3
Balancing Redox Reactions (2)

• 0.126 g of ethanedioic acid crystal, H2C2O4.nH2O
  required 20 cm3 of acidified manganate(VII) of
  concentration 0.02 M for complete reaction at
  60oC. Calculate the value of n.
• MnO4- H2C2O4 Mn2 2CO2
  
  7 3 2
  4
• 2MnO4- 5H2C2O4 2Mn2 10CO2
• 2MnO4- 5H2C2O4 6H 2Mn2 10CO2
• 2MnO4- 5H2C2O4 6H 2Mn2 10CO2
  8H2O
• 2MnO4- 5H2C2O4 nH2O 6H 2Mn210CO2(85n)
  H2O
• 0.02 x 0.02 x 5/2 n(H2C2O4.nH2O) 0.001
  0.126/FM
• FM 126 or n 2
• MnO2(s) PbO2(s) H(aq) MnO4-(aq)Pb2(aq)
  H2O(l)
• 2MnO2 3PbO2(s) 4H(aq) 2MnO4- 3Pb2
  2H2O(l)
• H2O2(aq) Cr(OH)4-(aq) OH-(aq)
  CrO42-(aq) H2O(l)
• 3H2O2 2Cr(OH)4- 2OH- 2CrO42- 8H2O

4
Electrochemical Cells

• Metal Ion/Metal Systems (Half Cells)
• When a metal electrode is dipped into a solution
  containing ions of the same metal, M(s) -
  ne Mn(aq), oxidation
• Mn(aq) ne
  M(s), reduction occur
  to different extents, depending on the nature of
  the particular metal ion/metal system under
  consideration.

----
----
-- --
- - - -
oxidation
reduction

Eventually an equilibrium is set up between the
two reactions Mn(aq) n e
M(s)
5
Metal-Metal Ion Systems

• The overall charge on the electrode depends on
  which of the 2 opposing processes occurring more
  readily and thus on the final position of
  equilibrium. This equilibrium position is
  dependent on many factors, including the nature
  of the metal ion/metal system, the concentration
  of ions the temperature
• If the equilibrium lies to the right, reduction
  predominates the electrode acquires a positive
  charge.
• If the equilibrium lies to the left, oxidation
  predominates and the electrode acquires a
  negative charge.
• In either case there is a separation of charge
  and thus a potential difference between the
  electrode and the ions in the solution. The metal
  ion/metal system is a half cell.

6
Metal Ion/Metal Systems

• For different metals the equilibrium position of
  Mn(aq) n e- M(s) will differ and this
  affects the extent of charge separation or the
  potential set up in the respective half cell.
• If 2 of these different half cells are connected
  externally by a wire, the difference in
  potentials between the 2 half cells is sufficient
  to push electrons through the wire from one half
  cell to another. However, such a current would
  only flow for an instant then cease, because of
  charge build-ups in the two systems. If electrons
  flow from the right to the left, the left system
  becomes negatively charged, the right system
  becomes positively charged.
• This results in a charge separation, stopping
  further e- flow

7
Electrochemical Process
e-
e-
Ions
Ions
Anode
Cathode
A salt bridge (a U-tube filled with an
electrolyte) or a porous disk can be used to
connect these solutions. These devices allow ion
flow without extensive mixing of the solutions,
the circuit is then complete. Electrons flow
through the wire from reducing agent to oxidizing
agent ions flow from one compartment to the
other via the salt bridge to keep the net charge
zero
An electrochemical process involves electron
transfer at the interface between the electrode
the solution. Ion species receive electrons from
the cathode, atoms in anode lose electrons.
8
Electrochemical Process
The Daniell cell
e-
e-
-

Salt bridge KNO3
Ions
Ions
Anode (Zinc)
Cathode (Copper)

The reduction half-cell and the oxidation
half-cell are forced to react by the passage of
electrons in the external wire and by the ion
flow in the salt bridge
The salt bridge allows ions to move between the 2
half-cells. These cations anions replace those
which are discharged at the electrodes
By physically separating the oxidation
reduction half-cells the electron transfer is
forced to occur through an external wire The
electric current generated can produce useful
work.
9
Electromotive Force (e.m.f.)

• Each half cell has its own electrode potential
  (absolute) that cannot be measured. When the two
  half-cells are joined by a wire to form an
  electrochemical cell, the difference in
  potentials between them can be measured.
• As current is taken from the cell the cell
  reaction proceeds the concentration of oxidized
  species (Zn2) in the half-cell containing the
  negative terminal (anode) increases. On the other
  hand, the concentration of the oxidized species
  in the half-cell holding the positive terminal
  (cathode) decreases.
• Thus, the potential difference between the 2
  electrodes drops When eventually the redox
  reaction reaches equilibrium, the p.d. between
  the 2 electrodes is 0 the battery becomes flat

10
E.M.F. Measurement

• The maximum potential difference measured is
  called the electromotive force or e.m.f. of the
  cell. It occurs only when no current is taken
  from the cell. The cell e.m.f. is a measure of
  the relative tendencies of the electrode systems
  involved to liberate electrons by forming ions in
  solutions.
• The key to finding the maximum potential is to
  carry out the measurement under conditions of
  zero current so that no electrode reactions occur
  and no energy is wasted.
• A potentiometer (variable voltage device) can be
  inserted in opposition to the cell potential and
  the voltage is adjusted until no current flows in
  the cell circuit. Under such conditions the cell
  potential is equal in magnitude opposite in
  sign to the voltage setting of the potentiometer
  and is the maximum cell potential, since no
  energy is wasted in heating the wire.

11
Cell Diagram (IUPAC)

• An easy way to represent an electrochemical cell
  and its e.m.f. is by means of a cell diagram. The
  IUPAC conventions used for the Daniell cell are
• Zn(s) Zn2(aq) Cu2(aq) Cu(s)
  E 1.10 V
• The solid vertical line between the symbols
  represents the phase boundary where there is a
  change of phase present.
• 2 vertical broken lines is used to represent the
  salt bridge that separates 2 solutions. In the
  case of porous pot, one vertical broken line is
  used.
• Zn(s) Zn2(aq) KCl Cu2(aq) Cu(s)
  E 1.10 V
• By convention, the half-cell with the cathode()
  is placed on the right. E represents the cell
  e.m.f. in volts, the sign ( or -) of the e.m.f.
  value indicates the polarity of the right hand
  electrode.

-

Salt bridge
12
Standard Electrode

• With the cell diagram written with a positive
  e.m.f. the overall reaction can be predicted as
  proceeding from left to right in the cell
  diagram. Zn(s) Zn2(aq) Cu2(aq) Cu(s)
• When E is the standard cell e.m.f., the cell
  diagram can be Zn(s) Zn2(aq, 1M)
  Cu2(aq, 1M) Cu(s) Eocell 1.10 V
• As its impossible to measure the potential of a
  single electrode system, the standard hydrogen
  electrode has been selected as the reference
  electrode and is arbitrarily assigned a zero
  potential, and the potentials of all other
  systems are referred to this value.A convenient,
  relative scale of electrode potentials is set up
  for different systems so that their tendency to
  release electrons is compared with one another.

13
The Hydrogen Electrode

• Hydrogen gas does not conduct electricity, and
  the hydrogen half cell reaction is slow to reach
  equilibrium.
• To overcome the difficulty of using a hydrogen
  electrode, the hydrogen electrode has an
  electrode made of a piece of platinum coated with
  finely divided platinum black, which catalyzes
  the half reaction. Its porosity provides a
  surface on which H2(g) can be adsorbed. This
  electrode is dipped into 1 M HCl(aq). A slow
  stream of pure H2(g) at one atmosphere is passed
  over the platinized surface, so that the
  equilibrium

reduction H2(g)
2H(aq) 2e- oxidation
is set up fairly readily. The platinum offers a
good pathway for electrons to enter or leave the
electrode system.
14
The Hydrogen Electrode (2)

• The cell diagram for this hydrogen electrode is
  written as
• 2H(aq,1 M) H2(g, 1 atm) Pt(s) or 2H(aq,1 M)
  H2(g, 1 atm) Pt(s)
• When inert electrodes are present, the least
  oxidized form of the compound of such electrode
  system is written next to the inert electrode.

Salt bridge
H2(g), 1 atmosphere
H(aq) 1 M
Platinum coated with platinum black
The potential of an electrode system depends on
temperature, concentration and pressure. To used
as a standard the potential of the hydrogen
electrode under specified conditions is
arbitrarily taken as zero. Conditions
Ways to achieve these
in the laboratory H2(g) at 1 atm.
H(aq) 1 M pressure gauge, use of
standard acid and at 298 K
solution, thermostat/water bath
15
Relative Electrode Potential

• The electrode potentials of various systems can
  be measured against it by forming an
  electrochemical cell between such systems and the
  standard hydrogen electrode
• Pt(s) H2(g, 1 atm.) 2H(aq, 1 M) Mn(aq, 1
  M) M(s)
• The cell e.m.f. ER.H.S.- EL.H.S. Electrode of
  half cell- S.H.E
• The measured cell e.m.f. is automatically
  equivalent to the electrode potential of the Mn
  M system. The relative scale of electrode
  potentials of different systems is useful for
  comparing their tendency to release electrons.

Being less reactive than hydrogen, the reduction
potential of Cu2(aq, 1M) Cu(s) is positive
(0.337 V). With a negative reduction potential,
the Zn2(aq,1M) Zn(s) system acts as the anode
when connected to the standard hydrogen electrode
16
Electrochemical Series

• Half cell systems other than metal ion/metal
• By convention, the reduced form of the ion is put
  nearest to the inert electrode, and separate it
  from the oxidized form by a comma.
• Pt Br2(aq), 2Br -(aq) Fe3(aq), Fe2(aq)
  Pt
• It sometimes happen that the oxidized and
  reduced forms of an electrode system contain more
  than one chemical species (ion or molecule) which
  participate in the cell reaction. For example,
  MnO4-(aq) 8H(aq) 5e Mn2(aq)
  4H2O(l) These ions
  molecules must be included in the oxidized
  reduced forms of the half cell diagram
  
  MnO4-(aq) 8H(aq), Mn2(aq) 4H2O(l), Pt
• In the electrochemical series, electrode systems
  with the largest negative potential are at the
  top of the list, and they have the greatest
  tendency to exist as cations in solution, i.e.
  they are the strongest reductants. (Li e-
  Li, Eo -3.05V)

17
Using Standard Reduction Potentials

• A cell will always run spontaneously in the
  direction that produces a positive e.m.f.

Cu2(aq) 2 e- Cu(s) Zn2(aq) 2 e-
Zn(s)
If we reverse the half reaction involving zinc, a
positive cell e.m.f. Cu2(aq) Zn(s)
Cu(s) Zn2(aq) Eo Ecathode - Eanode
0.337 - (-0.763) 1.10 V
The Daniell cell can be represented by the cell
diagram below Zn(s) Zn2(aq, 1 M) Cu2(aq, 1
M) Cu(s) Eocell 1.10 V
The value of electrode potential is not changed
when a half-reaction is multiplied by an integer.
Since a standard reduction potential is
an intensive property (doesnt depend on how many
times the reaction occurs), the potential is not
multiplied by the integer required to balance
the cell reaction.
18
Predicting Feasibility of Redox Reactions

• The reaction A(s) Bn(aq) B(s) An(aq)
  will be feasible when the e.m.f. of the cell A(s)
  An(aq) Bn(aq) B(s) is ve
• The Ecell has no indication of the reaction rate
  (kinetic factor)
• S2O82-(aq) 2 e- 2SO42-(aq)
  Eo 2.01 V
• Cr2O72-(aq)14H(aq) 6e- 2Cr3(aq)7H2O(l),
  Eo1.33V
• Eocell Ecathode - Eanode 2.01 - (1.33)
  0.68 V
• A reaction has not occurred despite positive Eo
  value.
• Ag(aq) e- Ag(s)
  Eo 0.779 V
  AgCl(s) e- Ag(s) Cl-(aq)
  Eo 0.220 V O2(g)
  2H2O(l) 4e- 4OH-(aq) Eo 0.401 V
• Ag(s) AgCl(s) Cl-(aq), O2(g)2H2O(l)4e-,
  OH-(aq) Pt
• Eocell 0.401 -(0.220) 0.181 V Silver will
  be oxidized spontaneously in the presence of
  oxygen, H2O Cl- (silver cannot be oxidized in
  the presence of O2 H2O only)

19
Primary cells

• The anode of the Leclanche cell consists of a
  zinc container that is in contact with manganese
  dioxide and an electrolyte, which consists of
  NH4Cl ZnCl2 in water. Starch is added to
  thicken the solution to a paste-like constituency
  so that it is less likely to leak. A graphite rod
  serves as the cathode, which is immersed in the
  electrolyte in the center of the cell.
• Anode Zn(s)
  Zn2(aq) 2 e-
• Cathode 2NH4 (aq)2MnO2(s) 2e- Mn2O3(s)
  2NH3(aq) H2O(l)
• Zn(s) 2NH4(aq) 2MnO2(s) Zn2(aq)
  2NH3(aq) Mn2O3(s)H2O(l)
• Zn(s) Zn2(aq) MnO2(s),2NH4(aq),Mn2O3(s),2N
  H3(g) C(s)
• NH3(aq) will accumulate at the cathode whereas
  Zn2(aq) will accumulate at the anode. The
  equilibrium of the overall reaction will shift to
  the left, leading to a drop in electrode
  potential.

20
Alkaline Cells

• The mercury battery can be represented by the
  cell diagram Zn(Hg)(s) ZnO(s) KOH(aq)
  HgO(s) Hg(l) steel
• HgO(s) H2O(l) 2 e- Hg(l) 2OH-(aq)
  Eo 0.098 V
• ZnO(s) H2O(l) 2e - Zn(Hg)(s) 2OH
  -(aq) Eo -1.216 V
• Zn(Hg)(s) HgO(s) ZnO(s) Hg(l)
• Eocell Ecathode - Eanode (0.098) 1.216
  1.314 V

insulation
Steel cathode
Anode (zinc container)
Solution of HgO/Hg in basic medium of KOH and ZnO
21
Secondary Cells

• In the charging process, a non-equilibrium
  mixture of reactants is formed an external
  source of electricity. When the cell is in use it
  produces electricity as the reaction approaches
  equilibrium again
• In the lead acid accumulator, lead serves as the
  anode lead coated with lead(IV) oxide serves as
  the cathode. During discharge Anode
  Pb(s) HSO4-(aq) PbSO4(s) H(aq) 2 e-
• Cathode PbO2(s) 3H(aq)HSO4-(aq) 2e-
  PbSO4(s)2H2O(l)
• Net Pb(s) PbO2(s) 2H(aq)HSO4-(aq)
  2PbSO4(s)2H2O(l)
• Pb(s) PbSO4(s) H2SO4(aq) PbO2(s) PbSO4(s)

Lead anode
Lead(IV) oxide as cathode
22
Secondary Alkaline Cell

• The nickel-cadmium battery has electrodes of
  solid cadmium solid nickel(IV) oxide, NiO2,
  coated on a conductor. When this battery
  discharges, Cd(OH)2(s) Ni(OH)2(s) are formed
  The electrolyte is a basic medium of potassium
  hydroxide.
• (-) Cd(s) 2OH-(aq) Cd(OH)2(s) 2 e-
  Eo0.76 V
• () NiO2(s)2H2O(l)2e- Ni(OH)2(s)2OH-(aq)
  Eo0.49V
• Cd(s) 2H2O(l) NiO2(s) Cd(OH)2(s)
  Ni(OH)2(s)
• Eocell 0.49 - (-0.76) 1.25 V
• Conductor Cd(OH)2(s) Cd(s) NiO2(s) Ni(OH)2(s)
  Conductor

The Eo values of the Cd(OH)2 Cd half cell
indicate that the half reaction must be reversed
so as to obtain a positive e.m.f.
23
Corrosion of Iron and its prevention
70 b were lost in the U.S. annually on corrosion

• 40 of the steel made in the US is used to
  replace steel lost by rusting
• Corrosion results in the formation of cracks
  crevices which weaken Fe
• Non-uniformities in steel cause areas where the
  iron is easily oxidized. In the anodic regions
  each iron atom gives up 2 electrons to form Fe2
  ion. The electrons released flow through the
  steel, as they do through the wire of an
  electrochemical cell, to a cathodic region where
  they react with O2
• O2(g) 2H2O(l) 4 e - 4OH- (aq)
  Eoanode 0.40 V
• Fe(s) Fe2(aq) 2e -
  Eo 0.44 V
• The Fe2(aq) ions formed in the anodic regions
  travel to the cathodic regions through the
  moisture on steel surface, just as ions travel
  through a salt bridge. In the cathodic regions
  Fe2 ions react with oxygen to form rust, which
  is hydrated iron(III) oxide of variable
  composition
• 4Fe2(aq) O2(g) (4 2n) H2O(l)
  2Fe2O3 .nH2O (s) 8H(aq)
• Iron(III) oxides lack the rigidity of the metal
  flake off. Water acts as a kind of salt bridge
  between anodic cathodic regions of this process.

24
Corrosion and its prevention

• In industrial areas, the acid rain further
  accelerate rusting.
• With the metal at the centre of water drop being
  the anodic site, whereas the metal at the
  periphery of the drop being the cathodic site.
  The cell diagram used to represent the process.
  Fe(s) Fe2(aq) OH (aq) O2(g) Fe(s)
  impurities
• Dissolved electrolyte (NaCl, NaHCO3) from
  atmosphere, a low pH and a high temperature, will
  accelerate rusting.
• Iron copper must be electrically insulated from
  each other
• Paint coatings are porous cannot effectively
  exclude water O2. They slow down ion movement,
  slowing down rusting

Water droplet
rust
O2
Fe2
anodic area
Cathodic area
Iron dissolves forming a pie
e-
Cathode reaction O2(g) 2H2O(l)
4e- 4OH-(aq)
Anode reaction Fe(s) Fe2(aq) 2 e-