Transcript Chapter 19 Redox Equilibria

Chapter 19
Redox Equilibria
19.1 Redox Reactions (reduction-oxidation)
• is an equilibrium of the competition for ebetween the 2 species
• reducing agent - undergo oxidation
oxidizing agent - undergo reduction
Oxidation
Addition of oxygen
Reduction
Removal of oxygen
Removal of hydrogen
Addition of hydrogen
Loss of electron(s)
Gain of electron(s)
Increase in oxidation number
Decrease in oxidation number
19.2
Oxidation States
Element
Free elements
For simple ions
All alkali metal ions
Oxidation
number
0
equal to its
charge
+1
All alkaline earth metal ions
+2
Al in all compounds
+3
Hydrogen in most compound
except hydride ion H- (-1)
+1
Oxygen is most compound
except peroxide (-1) & superoxide (-0.5)
-2
Fluorine in all compound
-1
More electronegative elements in covalent
compounds
-ve
Sum of oxidation number of all atoms in
compound
0
Sum of oxidation number of all atoms in an ion
Equal to the
charge of ions
Ex. 19.1
• a) KMnO4 0 =(+1) + x + 4(-2)
oxidation number of Mn is +7
x=+7
• b) POCl3 0 = x + (-2) + 3(-1)
oxidation number of P is +5
x=+5
• c) K2Cr2O7 0 = 2(+1) + 2x + 7(-2) x=+6
oxidation number of Cr is +6
• d) CuSO4
0 = +2 + x + 4(-2)
oxidation number of S is +6
x=+6
Ex. 19.1
• e) CaH2 0 = +2 + 2x
x=-1
oxidation number of H is -1
• f) Na2S2O3 0 = 2(+1)+2x +3(-2)
oxidation number of S is +2
x=+2
• g) Na2S2O8 0 = 2(+1)+2x +8(-2) x=+7
oxidation number of S is +7
• h) Na2S4O6 0 = 2(+1)+4x +6(-2 ) x=+2.5
oxidation number of S is +2.5
Thiosulphate ion S2O32-2
-2
-2
-2
Average oxidation number of S is +2
Peroxodisulphate ion S2O82+6
+6
oxidation number of S is +6
2-
Tetrathionate ion S4O6
Average oxidation number of S is +4
Ex. 19.2
• a) Mg(s) + H2O(l) → MgO(s) + H2(g)
0
+1
+2
0
reducing agent oxidizing agent
• b) Cr2O72-(aq)+2OH-(aq)→2CrO42-(aq)+H2O(l)
+6
+1
+6
+1
Not a redox reaction - only conversion under diff. pH
• c) 2CuCl(aq) → Cu(s) + CuCl2(aq)
+1
0
+2
CuCl acts as the reducing agent and the oxidizing
agent at the same time.
• 2CuCl(aq) → Cu(s) + CuCl2(aq)
+1
0
+2
• Cu+ (aq) + e- → Cu(s)
reduction
Cu+ (aq) → Cu2+(aq) + e- oxidation
___________________________________
• Cu+ (aq)+ Cu+ (aq) → Cu(s)+ Cu2+(aq)
• Disproportionation is a chemical change in
which a particular chemical species is
simultaneously oxidized and reduced.
19.3 Balancing Redox Equations
• a) SO32-+H2O→ SO42-+2H++2e• b) VO2++ H2O → VO3-+ 2H+
(not redox)
• c) MnO4- +4H+ + 3e-→ MnO2 + 2H2O
MnO4- +4H2O + 3e-→ MnO2+2H2O+4OHMnO4- + 2H2O + 3e-→ MnO2+ 4OH-
• d) MnO4- + 8H+ + 5e-→ Mn2+ + 4H2O
• e) BrO3-+6H+
+
6e-→
Br-
• f) N2H4 → N2+ 4H+ + 4e• g) Cl2+ 6H2O → 2ClO3- + 12H+ + 4e• h) NO3- + 4H+ +3e- → NO + 2H2O
+3H2O
Ex. 19.4
• a) Br2 +2I-
→ 2Br- + I2
• b) 3S2O82- + 2Cr3++7H2O → 6SO42- +
Cr2O72- + 14H+
• c) IO3- + 6Fe2+ + 6H+ → I- + 6Fe3+
+3H2O
19.4 Electrochemical Cells
• When a metal is dipped into a solution
containing ions of the same metal, a metal/metal
ion system is set up.
• Mn+(aq) + ne- → M(s)
M(s) → Mn+(aq) + ne-
ne-
ne-
M
M
Mn+
Mn+
• Mn+(aq) + ne- <======> M(s)
• The equilibrium position depends on
- nature of the M / Mn+ system,
-
concentration of ions
temperature
• Equilibrium lies on the right
 oxidation ? Reduction? predominates
• Charge on electrode?
 positive ? Negative?
ne-
ne-
M
M
Mn+
Mn+
• a separation of charge
 a potential difference between the
electrode and the ions in solution
• metal/metal ion system is called half cell
V
Metal Y
Y+
solution
Metal X
e-
e-
Y
X
Y+
X+
solution
X+
• connect 2 different half cells externally  e-s can
flow from one electrode to another
• Current flow through the external conducting wire
• after the charge is built up between the two system,
the current stop???? Solution???
Solutions:
• Using porous partition or salt bridge
• Functions:
- complete the circuit by allowing ions flow
- without extensive mixing of solutions of the 2
half-cell
• combination of the two half cell systems is
called electrochemical cell
• Also called galvanic cell
• is a device which converts chemical energy
into electrical energy
• Each half-cell has a tendency to accept
electrons from the other
• the half-cell with a stronger ability to gain
electrons will win the competition
• Each half-cell system has its own electrode
potential that cannot be measured.
• Only potential difference between 2 halfcells can be measured
• The maximum potential difference which
the cell can produce, called the
electromotive force (e.m.f.)
Daniel cell:
e-
e-
• Zn has a higher tendency to lose e- than Cu
• e-s are then pumped out from the Zn electrode
through the external circuit to the cell at Cu
electrode
Daniel cell:
• Oxidation (Anode / negative terminal) :
• Zn(s)  Zn2+(aq) + 2e• Reduction (Cathode / positive terminal):
• Cu2+(aq) +2e-  Cu(s)
• Overall Redox equation:
• Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Ex 19.5
a.
An electrochemical cell consists of Ag electrode with
AgNO3 as electrolyte and Cu electrode with CuSO4 as
electrolyte. The salt bridge is made of agar gel with
dissolved NaCl. However, no current flow through the
external wire.
Draw a diagram to show this
electrochemical cell and explain why there is no current?
Ag+ ion from the silver nitrate solution may
react with Cl- ion in the salt bridge to form
silver chloride which is insoluble in water.
Thus, ions cannot flow between the two system
and thus the current stops.
b. Explain why salt bridge of NaCl or
Na2SO3 should not be used when the
electrolyte is acidified KMnO4.
It is because chloride ion and sulphite ion
(sulphate(IV)) ion may be oxidized by
acidified KMnO4.

Cell Diagrams
• A representation of an electrochemical cell using
the IUPAC conventions
• For example, cell diagram of Daniel cell:
Zn(s)∣Zn2+(aq) Cu2+(aq)∣Cu(s) E =+1.1V
• solid vertical line (∣)= phase boundary

• a pair of vertical broken lines ( )= salt bridge
• single vertical broken line ( ) =
porous partition
• If the salt bridge is made of KCl,
Zn(s)∣Zn2+(aq) KCl Cu2+(aq)∣Cu(s)
• E represents the e.m.f. of the cell in volts
E=+1.1V
Zn(s)∣Zn2+(aq) Cu2+(aq)∣Cu(s)
E=+1.1V
Meaning???
L.H.S. :
Anode (negative electrode):
Zn(s)  Zn2+(aq) + 2eR.H.S.:
Cathode (positive electrode):
Cu2+(aq) +2e-  Cu(s)
• By convention, the electrode on LHS is
considered as the anode while the one on RHS is
considered to be the cathode
• sign of the e.m.f. indicates the polarity of the
right-hand electrode
• For a +ve e.m.f., the reaction proceeds from left
to right
Ex.
Cu(s)∣Cu2+(aq) Zn2+(aq)∣Zn(s)
19.6
E = ?
• E = -1.1V
• that means the reaction proceeds from
RHS to LHS
• or Cu + Zn2+  Cu2+ + Zn
is not spontaneous.
But Cu2+ + Zn  Cu + Zn2+
is spontaneous.
(a)
Ex. 19.7
• Sn(s)∣Sn2+(aq) Cu2+(aq)∣Cu(s) E=+0.50V
• Oxidation (anode):
Sn(s) Sn2+(aq) + 2e• Reduction (cathode)
Cu2+(aq) + 2e-  Cu(s)
• Overall equation:
Sn(s) + Cu2+(aq)  Sn2+(aq) + Cu(s)
(b)
• Cu(s)∣Cu2+(aq) Mg2+(aq)∣Mg(s) E=-2.70V
• Oxidation (anode):
Mg(s) Mg2+(aq) + 2e• Reduction (cathode)
Cu2+(aq) + 2e-  Cu(s)
• Overall equation:
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
(c)
• Ni(s)∣Ni2+(aq) Zn2+(aq)∣Zn(s) E=-0.70V
• Oxidation (anode):
Zn(s) Zn2+(aq) + 2e• Reduction (cathode)
Ni2+(aq) + 2e-  Ni(s)
• Overall equation:
Zn(s) + Ni2+(aq)  Zn2+(aq) + Ni(s)
(d)
• Ni(s)∣Ni2+(aq) Ag+(aq)∣Ag(s) E=+0.95V
• Oxidation (anode):
Ni (s) Ni2+(aq) + 2e• Reduction (cathode)
Ag+(aq) + e- Ag (s)
• Overall equation:
Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s)
Types of Half-cell
• 1.
Metal in contact with an aqueous solution of its metal ion
• Zn2+(aq)+2e- <====> Zn(s)
• Cell diagram :
Zn(s)∣Zn2+(aq)
Zn2+(aq)∣Zn(s)
2. Metal in contact with its insoluble salt and an
aqueous solution of the anion
• AgCl(s) + e- <====> Ag(s) +Cl-(aq)
• Cell diagram :
Pt(s)∣[Ag(s) + Cl-(aq)], AgCl(s)
AgCl(s), [Ag(s) + Cl-(aq)]∣Pt(s)
• PbSO4(s) + 2e- <====> Pb(s) +SO42-(aq)
• Cell diagram :
Pt(s)∣[Pb(s) + SO42-(aq,1M)], PbSO4(s)
PbSO4(s), [Pb(s) + SO42-(aq,1M)]∣Pt(s)
3. An inert electrode (e.g. Pt) and a gas in contact with an
aqueous solution from which the gas can be generated
• 2H+(aq) + 2e- <=====> H2(g)
• Cell diagram :
Pt(s)∣H2(g)∣2H+(aq) or Pt(s) [H2(g)]∣2H+(aq)
2H+(aq)∣H2(g)∣Pt(s) or
2H+(aq)∣[H2(g)] Pt(s)
• O2(g) +4e- +2H2O(l) <=====> 4OH-(aq)
• Cell diagram :
Pt(s)∣4OH-(aq), [O2(g) + 2H2O(l)]
[O2(g) + 2H2O(l)], 4OH-(aq)∣Pt(s)
4. An inert electrode and a solution containing both
the oxidized and reduced forms of the species
• Fe3+(aq) + e- ===== Fe2+(aq)
• Cell diagram :
Pt(s)∣Fe2+(aq), Fe3+(aq)
Fe3+(aq), Fe2+(aq)∣Pt(s)
• I2(aq) + 2e- <====> 2I-(aq)
• Cell diagram :
Pt(s)∣2I-(aq), I2(aq)
I2 (aq), 2I-(aq)∣Pt(s)
• MnO4-(aq)+8H+(aq)+5e- <==>Mn2+(aq)+4H2O(l)
• Cell diagram :
Pt(s)∣[Mn2+(aq)+4H2O(l)],[MnO4-(aq) +8H+(aq)]
[MnO4-(aq) +8H+(aq)], [Mn2+(aq)+4H2O(l)] ∣ Pt(s)
• The most reduced form is written next to the
inert electrode (i.e.Pt)
19.5 Standard Electrode and Relative Electrode
Potentials from E.m.f Measurements
• Standard Electrode
• Electrode potential of a half cell cannot be
measured.
• standard hydrogen electrode (s.h.e.)
- selected as the reference electrode
- its electrode potential is assigned as 0V at
25oC, 1atm and 1.0M H+.
• With the reference, the relative scale of
electrode potential is established for different
systems in order to compare their tendency to
release e-s
Standard hydrogen electrode:
A. Inlet for 1atm
H2 at 25oC
B. Platinum coated with
platinum black
C. 1M H+
D. outlet for H2
• has an electrode made of a piece of platinum
coated with finely divided platinum black
• function of this special electrode:
–catalyses the half-cell reaction :
H2(g) <=> 2H+(aq) + 2e–provides a surface on which the hydrogen can be
adsorbed
–provides an electrical connection to the voltmeter
• with the concentration of HCl(aq) of 1 mol dm-3
• a slow stream of pure hydrogen gas at 1atm is
bubbled over the platinized surface
• cell diagram
• Pt(s)∣H2(g)∣2H+(aq)
or Pt(s) [H2(g)]∣2H+(aq)
• Half -cell reaction:
• Reduction:
2H+(aq) + 2e- ===> H2(g)
• Oxidation:
H2(g) ===> 2H+(aq) + 2e-
• As the potential of an electrode system depends
on
,
and
. The
electrode potential of the reference electrode
can only be assigned as 0V under a specified
conditions:
• H2 gas at 1atm
• [H+] = 1 mol dm-3
• under temperature of 298K
• When the hydrogen electrode is used under the
above conditions, the hydrogen electrode is
known as a standard hydrogen electrode (s.h.e.)
• s.h.e. is always assumed to be the anode of the
system. (the left hand side in the cell diagram)
• Apart from hydrogen electrode, a calomel
electrode can also be used:
• Cell diagram :
Pt(s)∣[2Hg(l) + 2Cl-(aq,1M)], Hg2Cl2(s)
• Equation :
Hg2Cl2(s) + 2e-  2Hg(l) + 2Cl-(aq)
Relative Electrode Potential from e.m.f.
Measurements
• e.m.f. for the cell is a measure of the relative
tendencies of half cells to release or gain e-s
• e.m.f. of a cell depends on temperature,
concentration and also pressure
• all the measurement should be measured under
the standard conditions:
temp = 298K (25oC)
concentration of all solution = 1.0M
pressure of all gases = 1.0 atm
• standard electrode potential (or standard
reduction potential / Eo)
is the electrode potential (e.m.f.) of a half cell
measured when connected to a standard
hydrogen electrode under the standard
conditions
• Cell diagram :
Pt(s)∣H2(g, 1atm)∣2H+(aq,1M) Mn+(aq,1M)∣M(s) Eo=?
• Assumed Anode? Assumed Cathode?
• If s.h.e. undergoes oxidation, sign of Eo?
• For the standard reduction potential of the half
equation:
Cu2+(aq) + 2e-  Cu(s)
Eo = +0.34V
Meaning??
• Pt(s)∣H2(g, 1atm)∣2H+(aq,1M) Cu2+(aq,1M)∣Cu(s)
Eo = +0.34V
• Anode:
H2(g)  2H+(aq) + 2e• Cathode:
Cu2+(aq) + 2e-  Cu(s)
• Overall reaction:
H2(g) + Cu2+(aq)  2H+(aq) + Cu(s)
• for the following half equation:
Ni2+(aq) + 2e-  Ni(s)
Eo = -0.25V
• cell diagram:
•
•
•
•
Pt(s)∣H2(g, 1atm)∣2H+(aq,1M) Ni2+(aq,1M)∣Ni(s)
Eo = -0.25V
The value of Eo is
, the polarity of the nickel
electrode is
. That means nickel is the
and
undergoes
.
Anode:
Ni(s)  Ni2+(aq) + 2eCathode
2H+(aq) + 2e-  H2(g)
Overall Reaction:
Ni(s) + 2H+(aq)  Ni2+(aq) + H2(g)
• e.m.f. of this cell, Eocell = ER.H.S. - EL.H.S.
• The electrode system with the greatest negative
electrode potential is the strongest reducing
agent.
• By comparing the value of the standard
reduction potentials of different electodes, the
electrochemical series can be obtained.
19.6 Uses of Standard Reduction Potentials
• A cell will always run spontaneously in the direction that
produces a positive e.m.f.
• For an electrochemical cell:
Cu(s)∣Cu2+(aq,1M) Ag+(aq, 1M)∣Ag(s) Eocell=+0.462V
• 2Ag+(aq) + Cu(s) → Cu2+(aq) +2Ag(s) Eocell = +0.462V
• From data book,
Cu2+(aq) + 2e- → Cu (s)
Eo =+0.337V
Ag+(aq) + e- → Ag(s)
Eo = +0.799V
• Anode: Cu(s) → Cu2+(aq) + 2eEo =-0.337V
Cathode: 2Ag+(aq) + 2e- → 2Ag(s)
Eo = +0.799V
----------------------------------------------------------------------• 2Ag+(aq)+Cu(s) →Cu2+(aq)+2Ag(s) Eocell = +0.462V
• The value of electrode potential is not changed when a
half-reaction is multiplied by an integer because a
standard reduction potential is an intensive property
• The e.m.f. are usually shown for the reduction half
reaction. For an oxidation half reaction, the sign of the
e.m.f. must be reversed
• P.10
Ex. 19.9 (a)
Ni(s)∣Ni2+(aq) [NO3-(aq)+3H+(aq)],[HNO2(aq)+H2O(l)]∣Pt(s)
•
•
•
•
Ni2+(aq) → Ni(s)
Eo = -0.25V
NO3-(aq) → HNO2(aq) Eo = +0.94V
Eocell = Eocathode - Eoanode
Eocell = (+0.94V) - (-0.25V) = +1.19V
• Ni(s)+NO3-(aq)+3H+(aq)→Ni2+(aq)+HNO2(aq)+H2O(l)
(b)
Pt(s)∣2H2SO3(aq),[4H+(aq) + S2O62-(aq)] Cr3+(aq)∣Cr(s)
• Given: S2O62-(aq) → H2SO3(aq) Eo = +0.57V
Cr3+(aq) → Cr(s)
Eo = -0.74V
• Eocell = Eocathode - Eoanode
• Eocell = (-0.74V) - (+0.57V) = -1.31V
• S2O62-(aq)+4H+(aq)+ 2e-→ 2H2SO3(aq)
x3
• Cr(s) → Cr3+(aq) + 3ex2
----------------------------------------------------------------• 3S2O62-(aq)+12H+(aq)+ 2Cr(s) → 6H2SO3(aq) + 2Cr3+(aq)
Prediction of the Feasibility of Redox Reactions
• more -ve Eo ====> more reducing the system
more +ve Eo ====> more oxidizing the system
• By comparing Eo values, an electrochemical series is
formed.
• sign on the e.m.f. of cell ===> to predict the cell reaction
• Simiarly, use to predicting whether the reaction is
feasible or not.
• overall redox reaction with Eocell = Eocathode - Eoanode > 0 is
thermodynamically (energetically) feasible. The redox
reaction will run spontaneously
Example 19.1
Pb(s) + Mn2+(aq) → Pb2+(aq) + Mn(s) feasible?
Cell diagram:
Pb(s)∣Pb2+(aq) Mn2+(aq)∣Mn(s)
Half-equations:
Pb2+(aq) +2e-  Pb(s)
Eo = -0.13V
Mn2+(aq) + 2e-  Mn(s) Eo = -1.18V
•ΔEo = EoR - EoL = (-1.18V) - (-0.13V) = -1.05V
• The negative sign shows that the given reaction will
not proceed spontaneously or say the reaction is
not feasible.
Ex. 19.10
• a) Br2(aq) + 2I-(aq) → 2Br-(aq) + I2(aq)
• Eocell =(+1.07V) - (+0.54V) = +0.53V
As the emf is positive, the reaction will proceed
spontaneously. (the reaction is feasible)
• b) 2MnO4-(aq) + 16H+(aq) + 10Br-(aq) →
8H2O(l) + 2Mn2+(aq) + 5Br2(aq)
• Eocell =(+1.51V) - (+1.07V) = +0.44V
As the emf is positive, the reaction will proceed
spontaneously. (the reaction is feasible)
Limitations of Predictions Made Using Eo Values
• Is it true that all the redox reactions with Eo > 0 will
take place???
• A cell e.m.f. only gives information about the
feasibility of a redox reaction from an energetic
point of view.
It cannot tell how fast a feasible reaction is likely to
proceed. Therefore, some feasible redox reactions
may not appear to take place just because they are
too slow.
• The cell e.m.f. may be not taken under standard
conditions.
• A a rule of thumb, if the cell e.m.f. > 0.4V, it is
quite safe to say that the prediction is valid.
• The electrochemical series deals only with
aqueous solutions of ions, and reducing
tendencies may be quite different in other
solvents or for reactions involving gases and
solids
Testing Predictions about the Feasibility of
Redox Reactions
• By simple tests or simple observation
• For example, to test whether the reaction between
Br2 and KI is feasible
• 3ml aqueous solutions (about 0.1M) of both Br2 and KI can
be mixed in test tubes
• 1,1,1-trichloroethane is added and the lower layer became
purple
• Deductions : I2 is present
• The reaction has taken place
purple
orange
19.7 Secondary Cell and Fuel Cell
• a device that stores chemical energy for later
releases as electricity.
(chemical energy ==> electrical energy)
• Basically, there are 3 types of cells and batteries:
1. Primary Cells
2. Secondary Cells
3. Fuel Cells
• 1. Primary Cells
- galvanic cells
- cannot be recharged after reaching the eqm
• 2. Secondary Cells
- rechargeable
- a non-equilibrium mixture of reactants is
produced by an external source of electricity in
the charging process
• 3. Fuel Cells
- a type of primary cell
- the reactants are continuously applied from
outside while the cell is in use
Lead Acid Accumulator
• most common secondary cell
• used as motor car storage battery
• Anode :
Lead
Cathode :
Lead(IV) coated with lead dioxide
Electrolyte : sulphuric acid
• a number of cathode alternate with several anode plates ==>
increase current
eV
PbO2
Pb
H2SO4(aq)
• Anode : Pb(s) + SO42-(aq) → PbSO4(s) + 2eCathode:PbO2(s)+4H+(aq)+SO42-(aq)+2e-→ PbSO4(s)+2H2O(l)
-------------------------------------------------------------------------------Cell reaction :Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)
• Alternatives,
Anode : Pb(s) + HSO4-(aq) → PbSO4(s) +H+(aq) + 2eCat:PbO2(s)+3H+(aq)+HSO4-(aq)+2e-→ PbSO4(s)+2H2O(l)
-----------------------------------------------------------------------------Cell rxn :Pb(s)+PbO2(s)+2HSO4-(aq)+2H+(aq)→2PbSO4(s)+2H2O(l)
• Cell
diagram:
Pb(s)∣[Pb(s)+SO42-(aq)],PbSO4(s)∣H2SO4(aq)∣
[PbO2(s)+SO42-(aq)+4H+(aq)],[PbSO4(s)+2H2O(l)]∣Pb(s)
• Alternatives,
Pb(s)∣PbSO4(s)∣H2SO4(aq)∣PbO2(s) ∣PbSO4(s)∣Pb(s)
After discharging for some time,
V
PbSO4
• No current
• Reasons:
- both electrode coated with PbSO4
- density of liquid drops (conc. H2SO4 drops)
• When charging:
Cathode :PbSO4(s) + 2e- → Pb(s) + SO42-(aq)
Anode :PbSO4(s) + 2H2O(l) →
PbO2(s) + 4H+(aq) + |SO42-(aq) + 2e----------------------------------------------------------------------• Overall :2PbSO4(s) + 2H2O(l) →Pb(s)+PbO2(s)+2H2SO4(aq)
• Pb(s) + PbO2(s) + 2H2SO4(aq) <===> 2PbSO4(s) + 2H2O(l)
discharge
charge
Hydrogen-Oxygen Fuel Cell
• is a primary cell
• converts the chemical energy of continuous supply of
reactants (fuels) into electrical energy.
Anode
1. Stream of H2
(fuel)
e-
KOH(aq)
as
Cathode
5. e-s flow through an
external circuit from
anode to cathode.
3. Stream of O2
(oxidizing agent)
electrolyte
2. H2 diffuses through the
porous anode (e.g. Ni). When
is comes into contact with the
electrolyte, KOH(aq), adsorbed
H2 is oxidized:
2OH-(aq) + H2(g) → 2H2O(l) + 2e-
4. O2 diffuses through the
porous cathode (e.g. Ni).
Adsorbed O2 is reduced:
O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
Overall:2H2(g) + O2(g) → 2H2O(l)
Function of porous nickel electrodes
• Acts as electrical conductors connecting to
external circuit
• acts as catalyst for the reaction
• allows adsorption of reactants(fuels) and
oxidizing agent for reaction
Fuel cells differ from other electrochemical cells
•
•
•
•
do not store reactants and electrolytes
electrodes do not undergo permanent change
reactants are gases
product and operating procedures are non-polluting
(only water is produced)
• high efficiency of energy production
(continuous supply of fuel)
• can supply energy at a constant rate
Disadvantages:
High cost
(limited in spacecrafts)
19.8 Corrosion of Iron and its Prevention
• Rusting - corrosion of iron
• an electrochemical process
• the slow deterioration of iron to hydrated
iron(III) oxide
• this oxide is permeable to both air
and water can’t protect iron from
further corrosion (like Al2O3)
• Anode:
Fe(s) → Fe2+(aq) + 2eCathode: O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
• Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
2Fe(OH)2(s) + 1/2O2(g) + (x-2)H2O(l) → Fe2O3 •xH2O(s)
• Overall :
4Fe2+(aq) + O2(g) + (4+2n)H2O(l) → 2Fe2O3n•H2O(s) + 8H+(aq)
Factors that speed up rusting
• a low pH - presence of acid, or even CO2, NO2
and SO2 in rain water
• Presence of electrolytes (i.e. NaCl)
- increases the conductivity
• High temperature
- increase the rate of chemical reaction
• in contact with less reactive metal such as Cu
and Sn
• Sharply pointed regions
- serves as anode
Prevention of Rusting
• 1. Protective Coating
A protective layer is coated on the surface of iron to
exclude oxygen and water
a)
Painting
b) Oiling and greasing
c) Galvanizing
(zinc-plating)
d) Tin-plating
e) Electroplating
2. Sacrificial protection
3. Alloying
4. Cathodic protection
iron is treated as the cathode
and connected to the negative
terminal of a battery
Ex. 19.11
• a) i) (+1.23)-(-0.45) = +1.68V
ii) (+0.40)-(-0.45) = +0.85V
• b) As Eocell in (i) is more positive, iron lose
electrons more readily to form Fe(II) ions. Thus,
Iron would rust more rapidly under acidic
conditions.
• c) Iron would rust more rapidly in sea water than
in pure water because the process of corrosion can
be speeded up by electrolyte like NaCl, which
increases the conductivity of the solution.
End