Chapter 6.3 Redox Equilibrium
Transcript of Chapter 6.3 Redox Equilibrium
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6.3 Redox Equilibria
A) Redox Reactions
(a) Redox reactions in terms of electron transfer
Oxidation loss of electron(s)
Reduction gain of electron(s)
e.g.(1) loss of 2 e-(oxidation)
Cu + Cl2 CuCl2
gain of e-(reduction)
(2) loss of 2 e-(oxidation)
Zn + Cu2+ Zn2+ + Cu
gain of 2 e-(reduction)
(b) Oxidation State (Number)
Rules for Assigning Oxidation State
1. Oxidation state of an atom in an element = 0
e.g. Na OS = 0 H2 OS = 0 O2 OS = 0
2. The more electronegativeatom in a compound has the
ve oxidation state
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e.g. NH3 OS of N = -3 NF3 OS of N = +3
3. Sum of oxidation state of a neutral compound = 0
e.g. NaCl (+1) + (-1) = 0 H2O (+1 x 2) + (-2) = 0
4. Sum of oxidation state of a simple / complex ion = the
charge on the ion.
e.g. SO42- (-6) + 4(-2) = -2 NH4+ (-3) + 4(+1) = +1
5. Some elements always have the same oxidation state in their
compounds
(i) Group I metals: +1
(ii) Group II metals: +2
(iii) Al: +3
(iv) H: +1 in covalent compounds with non-metals
-1 with group I & II metals (e.g. NaH, MgH2)
(v) F: -1
(vi) O: -2 except
1. +2 in OF2
2. 1 in peroxides (e.g. Na2O2, H2O2)
3. 1/2 in superoxides (e.g. KO2)
e.g. POCl3 P + (-2) + 3(-1) = 0 P = +5
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HPO3- (+1) + P + 3 (-2) = -2 P = +3
6. Assigning oxidation state to elements in covalent compounds
Oxidation state = formal charge on the atom when the
shared electrons are assigned to the more electronegative
one of the bonded atoms.
e.g. CH4, EN of C > H
OS of C = -4
OS of H = +1
N2H4 EN of N > H
OS of N = -2
OS of H = +1
S2O82
S
O
O
O
O S
O
O
O
O
-1
-1
-2
-2
-2
-2
-2
-2
+6
+6
C
H
H H
H
-1 -1 +1
+1
+1
+1
-1-1
N
H
N
H
H H
+1 +1
-1
-1+1 -1 +1
-10 0
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S2O32- 2(S) + 3(-2) = -2 S = +2
+2 is only the average OS of S
or
Ex 6.3-1
Try to assign the OS of the underlined elements.
OH ClO S4O62-
CO32 ClO3
NH4+ ClO4
CrO42 MnO4
Cr2O72 MnO42
(c)Redox reactions in terms of oxidation number
Oxidation increase in oxidation state
Reduction decrease in oxidation state
S
S
OO
-
O-
0
+4
-2
-2 -2
S
S
OO
-
O-
-2
-2 -2
+6
-2 more electropositive
owing to its direct
bonding to more
electronegative oxygen
atom
SO S
S
S
O
O-
OO
-
O
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e.g. 0 +1 oxidation
2Na(s) + Cl2(g)2NaCl(s)
0 1 reduction
Disproportionation a chemical reaction in which a particular
chemical species is simultaneously oxidized and reduced
e.g. reduction
3ClO- 2Cl- + ClO3-
+1 -1 +5
oxidation
Oxidizing agent- substance which oxidize others but itself being
reduced
Reducing agent substance which reduce others but itself being
oxidized
Oxidizing agent and reducing agent are relative only. A
compound may be an oxidizing agent in one reaction but it may
be a reducing agent in another reaction.
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iii) In an acidic solution, balance Hby adding the appropriate
no. of H+to the side deficient in H.
In an alkaline solution, balance H by adding the
appropriate no. H2Oto the side deficient in H and then add
an equal no.of OH-on the opposite side.
iv) Then balance the charge by adding e(s) to the side deficient
in negative charge.
3. Multiply the two balanced half-equations by appropriate no. so
that the no. of egained in one half equation is equal to that lost
in the other.
4. Add the 2 half-equations and eliminate the e. Collect like terms
and cancel any duplications on both sides if necessary.
Example
(1) MnO4-+ Fe2+ Mn2++ Fe3+
MnO4- Mn2+
MnO4- Mn2+ + 4H2O
MnO4- + 8H+ Mn2+ + 4H2O
MnO4- + 8H+ + 5e Mn2+ + 4H2O(1)
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Fe2+ Fe3+
Fe2+ Fe3+ + e ..(2)
(1) + (2) x 5
MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+
(2) Cr2O72-+ SO2 Cr3++ SO42-
Cr2O72- Cr3+
Cr2O72- 2Cr3+
Cr2O72- 2Cr3+ + 7H2O
Cr2O72- + 14 H+ 2Cr3+ + 7H2O
Cr2O72- + 14 H+ + 6 e 2Cr3+ + 7H2O..(1)
SO2+ 2H2O SO42-
SO2+ 2H2O SO42- + 4H+
SO2+ 2H2O SO42- + 4H++ 2e..(2)
(1)+ (2) x 3
Cr2O72- + 14 H+ + 3SO2+ 6H2O2Cr3+ + 7H2O + 3SO42-+ 12H+
Cr2O72- + 2H+ + 3SO2 2Cr3+ + H2O + 3SO42-
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(3) MnO4-+ H2O2MnO2+ O2 (basic)
MnO4- MnO2
MnO4- MnO2+ 2H2O
MnO4-+ 4H2O MnO2+ 2H2O + 4OH-
MnO4-+ 2H2O + 3e MnO2+ 4OH-.(1)
H2O2 O2
H2O2 + 2OH- O2+ 2H2O
H2O2 + 2OH- O2+ 2H2O + 2e (2)
(1)x 2 + (2) x 3
2MnO4-+ 4H2O + 3H2O2 + 6OH-2MnO2+ 8OH-+ 3O2+ 6H2O
2MnO4-+ 3H2O2 2MnO2+ 2OH-+ 3O2+ 2H2O
(4) SnO22- Sn + SnO32- (basic)
SnO22- Sn + 2H2O
SnO22-+ 4H2O Sn + 2H2O + 4OH-
SnO22-+ 2H2O + 2e Sn + 4OH-(1)
SnO22- SnO32-
SnO22-+ H2O SnO32-
SnO22-+ H2O + 2OH- SnO32-+ 2H2O
SnO22-+ 2OH- SnO32-+ H2O + 2e (2)
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(1)+ (2)
2SnO22- + 2H2O + 2OH- Sn + 4OH-+ SnO32-+ H2O
2SnO22- + H2O Sn + 2OH-+ SnO32-
Ex 6.3-3
Balance the following equations
i) Cu + NO3Cu2++ NO
ii) MnO4+ IMn2++ I2
iii) SO32+ Br2SO42+ Br(acidic)
iv) Mn3+MnO2+ Mn2+(acidic)
v) Sn2++ H2O2Sn4++ H2O (acidic)
vi) MnO2+ PbO2MnO4+ Pb2+(acidic)
vii) Cl2Cl+ ClO3(basic)
viii)Cl2Cl+ ClO(basic)
ix) H2O2+ Cr(OH)4-CrO42-+ H2O (basic)
B) Electrochemical Cells
(a) Introduction
Redox reaction can be split into two half reversible
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Mn+(aq) + ne M(s)
At equilibrium: rate of oxidation = rate of reduction
An electrode potential is set up at the interface of the
solid and solution phases single electrode
potential
The position of equilibrium depends on
1. nature of metal / metal ion system,
2. concentration of metal ions,
3. temperature.
Single electrode potential cannot be measured absolutely
c) Electrochemical Cells Made up of Metal / Metal Ions Systems
Made up of 2 different metal / metal ion half cells
Different metal / metal ion half cells will have different
electrode potentials
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If we connect an electrical wire between the two different metal
electrodes.
The potential difference will drive electrons through the
conducting wire from one cell to another (but only
instantaneously).
Current will stop after a while as charges would be built up in
the two half cells which prevent further electrons movement.
The problem can be solved by adding a salt bridge or a
porous partition. It enables electrical contact while
preventing the mixing of solutions which leads to the direct
redox reaction.
Salt bridge contains a strong electrolyte (e.g. saturated KCl,
KNO3 or NH4NO3) held in a jelly like matrix in a inverted
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Utube or just a piece of filter paper soaked with strong
electrolyte
Porous partition contains tiny passage that allows hindered
flow of ions.
e.g. Daniell cell (composed of Cu and Zn half cell)
Zn electrode: Zn(s)Zn2+(aq) + 2e anode (-ve pole)
Cu electrode: Cu2+(aq) + 2eCu(s) cathode (+ve pole)
Electrons flow from Zn electrode through external circuit
to Cu electrode.
Overall reaction: Zn(s) + Cu2+
(aq) Zn2+(aq) + Cu(s)
d) Electromotive Force (e.m.f.) of Electrochemical Cell
Absolute potentialof a half cell CANNOTbe measured.
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Only thepotential differencebetween 2 different half cells can be
measured.
The potential difference of 2 half cells can be measured by the
following circuit.
Potential difference (measured by the voltmeter) increases with
the increasing resistance (decreasing current).
NO currentflowmaximum potential differencee.m.f. of the
cell
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e) E.m.f. Measurement of Electrochemical Cell
A voltmeter cannot measure the e.m.f. of an electrochemical cell
accurately because current is taken by voltmeter.
E.m.f. can be measured by:
(1) Use a voltmeter of high resistance (faster method)
e.g. a transistor voltmeter, impedance voltmeter or
a digital multimeter.
(2) A potentiometer
When no current flows (i.e. no deflection on
galvanometer),
e.m.f. of the cell under measurement =
known e.m.f. of the standard cellACdistance
ADdistance
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(f) Cell Diagrams (IUPAC Convention)
Commonly used notations
1. Solid line () boundaries between electrodes and
solution (phase boundaries).
2. Dotted line (M)porous partition
(MM
)salt bridge3. E or Ecell e.m.f. of a cell in volt.
4. Sign (+ or) polarity of theRIGHTHANDelectrode.
e.g. Daniell cell
Anode Cathode
Zn(s)Zn2+(aq)MCu2+(aq)Cu(s) E= + 1.1 V
Zn(s)Zn2+(aq)Cu2+(aq)Cu(s)
Reaction represented:
Zn(s)+ Cu2+(aq)Zn2+(aq)+ Cu(s)
If the cell diagram is represented as:
Cu(s)Cu2+(aq)MZn2+(aq)Zn(s) E=1.1 V
Cu(s)Cu2+(aq)Zn2+(aq)Zn(s)
Reaction represented:
Zn2+(aq)+ Cu(s)Zn(s)+ Cu2+(aq)
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C) Standard Electrode Potential and Electrochemical Series
a) Standard Hydrogen Electrode
Since only potential difference between 2 electrodes
can be measured standard electrodemust be
chosen so as to compare electrode potential of
different system.Standard hydrogen electrodeis one of the standard
electrodes and its electrode potential is arbitrarily
assigned as 0.
At equilibrium: 2H+(aq) +2e H2(g)
Cell diagram: Pt(s)H2(g, 1 atm)2H+(aq, 1.0 M)MM
Platinum electrode is used as an inert electrode by which
electrons can leave or enter the electrode system. Platinum
black acts as a catalyst, being porous, it retains a comparatively
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large quantity of hydrogen.
b) Standard Electrode Potential E
Electrode potential depends on [ion] and temp. Thus we have to
standardize the potential measurement conditions.
Eis a value of an electrode potential relative to the standard
hydrogen electrode under the conditions:
1.[ion] = 1.0 M
2.temp. = 298 K
3.pressure of gas = 1 atm.
c)StandardElectrode Potential from e.m.f. Measurements
Standard electrode potential of a metal / metal ion system can be
measured by using standard hydrogen electrode under standard
conditions.
Pt(s)H2(g, 1 atm)2H+(aq, 1.0 M)MMMn+(aq, 1.0M)M(s)
E= Ecell = E
R.H.S.EL.H.S.= ER.H.S.
where ER.H.S..= the electrode to be measured
EL.H.S= the standard Hydrogen electrode = 0
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Electrode potential to be measured is always written on R.H.S.
and thus it always refers to the Reduct ion Potential.
Mn+(aq)+ ne M(s)
The more ve the E, the higher the tendency to release eand
vice versa.
Oxidized form Reduced form E/ VoltsK+(aq)+ e
K(s) 2.92Ca2+(aq)+ 2e
Ca(s) 2.87Na+(aq)+ e
Na(s) 2.71Mg2+(aq)+ 2e
Mg(s) 2.37Al3+(aq)+ 3e
Al(s) 1.66Zn2+(aq)+ 2e
Zn(s) 0.76Fe2+(aq)+ 2e Fe(s) 0.44Pb2+(aq)+ 2e
Pb(s) 0.13
d) Redox Equilibria Extended to Other Systems
Reduction potential of a redox system other than metal / metal
ion system can also be measured. Pt is used as an inert
electrode.
e.g. Br2(aq)+ 2e 2Br(aq) Non-metal / non-metal ion
MMBr2(aq), 2Br(aq)Pt(s)
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Fe3+(aq) + e Fe2+(aq) Ion / ion
MMFe3+(aq), Fe2+(aq)Pt(s)
MnO4(aq)+ 8H+(aq)+ 5e
Mn2+(aq)+ 4H2O(l)
MM[MnO4(aq)+ 8H+(aq)], [Mn2+(aq)+ 4H2O(l)Pt(s)
2IO3-(aq) + 12H+(aq) + 10 e I2(aq) + 6H2O(l)
MM[2IO3-
(aq, 1M)+ 12H
+
(aq, 1M)], [I2(aq, 1M)+ 6H2O(l)]Pt(s)
Rules for constructing cell diagram of system other than metal
/ metal ion.
1. The reduced form of ions is put nearest to the inert electrode.
2. Reduced form is separated oxidized form by comma
3. For oxidized & reduced forms containing more than one
chemical species, they must be included in the oxidized &
reduced forms of the half cell diagram by square brackets.
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e) Electrochemical Series
E.C.S. is formed when the electrode potentials of various
redox systems are tabulated together.
The more ve electrode potential are at the TOP
Greater tendency to lose e
Stronger Reducing Agent
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D) Uses of Standard Reduction Potentials
(a) To construct a working electrochemical cell and calculate its
e.m.f.
Consider the following standard reduction potential at 298 K.
Cu2+(aq) + 2e Cu(s) E
= +0.337V
Ag+(aq)+ e Ag(s) E
= +0.799V
Zn2+(aq)+ 2e Zn(s) E=0.763V
As to construct a working cell, i.e. a cell which runs
spontaneously, the e.m.f. of the cell MUST BE +ve .
Ecell = Eright - Eleft
Case 1:
Cell diagram:
Cu(s)Cu2+(aq 1.0 M)MMAg+(aq 1.0 M)Ag(s) E= +0.462V
Ecell= -0.799 (0.337) = +0.462 V
2Ag+(aq)+ Cu(s) 2Ag(s)+ Cu2+(aq) E= +0.462V
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Case 2:
Cell diagram:
Zn(s)Zn2+(aq 1.0 M)MMCu2+(aq 1.0 M)Cu(s) E= +1.1V
Ecell= (+0.337) (- 0.763) = + 1.1 V
Cu2+(aq)+ Zn(s) Cu(s)+ Zn2+(aq) E= +1.100V
Ex 6.3-4
When a standard Ni2+(aq)Ni(s)half cell is connected to a standard Cu2+(aq)Cu(s)
half cell, the cell voltage is 0.59 V. The copper electrode is the cathode.
Determine the standard reduction potential for Ni2+(aq)Ni(s)half cell. (Given:
Eof Cu2+(aq)Cu(s)= +0.34 V)
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Ex 6.3-5
A cell is based on the reaction of
5Fe(s)+ 2MnO4(aq)+ 16H+(aq)5Fe2+(aq)+ 2Mn2+(aq)+ 8H2O(l)
Given that: Fe2+(aq)Fe E=0.44 V
MnO4(aq)Mn2+(aq) E= +1.51 V
a) Write balance equations for the reactions occurring at the cathode
and the anode.
b) State the direction in which electrons move when the electrodes
are connected externally.
c) Calculating the e.m.f. of the cell
d) Write the cell diagram.
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b) To Predict the Feasibility of Redox Reactions
E = +ve energetically feasible
= 0 in equilibr ium
= -ve not energetically feasible
Evalues predicts whether a reaction is energetically feasible or
not, it does not tell the rate at which the reaction takes place.
Eis an intensive property, i.e. independent of the amount of
the no. of etransferred.
In using Eto predict feasibility of a reaction, care must be taken
that prediction apply only to standard conditionsonly.
In general, ifEcell> 0.4 V, we can predict with confidence.
Ex 6.3-6
Given that: Cu2+(aq)+ 2e Cu(s) E
= +0.34 V
Ag+(aq)+ e Ag(s) E= +0.80 V
i) Write an ionic equation for the spontaneous reaction
ii) Calculate Ecell
iii) Write the cell diagram for this reaction.
iv) Explain why the e.m.f. measured by a potentiometer differs from that
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calculated in (iii).
Ex 6.3-7
Given that: Ag+(aq) + e Ag(s) E
= + 0.80 V
AgCl(s)+ e Ag(s)+ Cl
(aq) E
= + 0.22 V
i) Write the equation for the spontaneous reaction.
ii) Calculate the cell e.m.f.
iii) Write the cell diagram for the spontaneous reaction.
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Ex 6.3-8
Given that: Zn2+(aq) + 2e Zn(s) E
=0.76 V
Hg2Cl2(aq) + 2e 2Hg(l)+ 2Cl
(aq) E
= +0.28 V
i) Write an ionic equation for the spontaneous reaction.
ii) Calculate Ecell.
iii) Write the cell diagram for this reaction.
Ex 6.3-9
(a) Given that
Br2(aq) + 2e 2Br-(aq) E= +1.09 V
O2(g) + 2H2O(l) + 4e 4OH-(aq) E= +0.40 V
Predict whether Br2or O2is a stronger oxidizing agent.
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(b) Given that:
MnO42(aq) + 4H+(aq) +2e
MnO2(s)+ 2H2O(l) E= +2.26 V
MnO4(aq) + e
MnO42(aq) E
= +0.56 V
Show that MnO42(aq)disproportionate in acidic medium.
Ex 6.3-10
Given that: S4O62(aq), 2S2O32(aq)Pt(s) E= +0.10 V
Br2(aq), 2Br(aq)Pt(s) E= +1.10 V
I2(aq), 2I(aq)Pt(s) E= +0.55 V
i) Write the cell diagram for the cell with highest e.m.f. and the equation for the
chemical change that takes place.
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ii) Which is the positive electrode?
iii) An aqueous solution of I2oxidizes S2O32(aq)to S4O62
(aq), but an aqueous
Br2(aq)oxidizes it further to SO32(aq).
a) What does this tell about the relative oxidizing power of Br2and I2?
b) What conclusion can you draw about the Eof the of the following half
cell?
[2SO32(aq)+ 6H+(aq)], [2S2O32(aq)+ 3 H2O(l)]Pt(s)
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E) Effect of Concentration Change on Electrode Potential
e.g. Cu(s) Cu2+(aq,1M)MAg+(aq)Ag(s)
Measure Ecellfor different concentration of Ag+(aq)
Plot Ecellvs log [Ag+(aq)]
Cell reaction Cu(s) + 2Ag+(aq) Cu
2+(aq) + 2Ag(s)
(a) E
cell= EAg - E
Cu= (+0.8) (+ 0.34) = + 0.46 V
Ag+(aq) + e Ag(s) E= + 0.8 V
By Le Chatelier Principle
[Ag+] increase in tendency to form Ag+
equilibrium shifts to LHS
electrode potential of Ag becomesless +ve
Ecellbecomes less +ve
+0.46 Vlog [Ag+] = -8.8
Log [Ag+]
Ecell
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(b) From the graph Ecellas [Ag+]
consistent with the prediction
And Ecelllog [Ag+] EAglog [Ag
+]
(c) When Ecell= 0 log [Ag+(aq)] = -8.8
Ecell= 0 equilibrium is established
[Ag+]eqn= 1.6 x 10-9M
29-2
2
c)10x(1.6
1
]Ag[
][CuK ==
+
+
Kc= 4 x 1017dm3mol-1
(d) log [Ag+] < -8.8 Ecell= -ve
(e) Nernst Equation
E = E
+[ion]ln
Fz
RT
R - gas constant = 8.31 JK-1 T temp
F Faraday constant z charge of metal ion
E = E+ [ion]log
F
RT2.310
z
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E = E+ [ion]log
z
0.0610
Ex 6.3-11
Al3+
(aq) + 3e Al(s) E= -1.66 V
Cu2+(aq) + 2e Cu(s) E= +0.337 V
Calculate Ecellif [Cu
2+
] = 0.1 M & [Al
3+
] = 0.01M
F) Practical Cell for Electricity Production through Chemical
Reactions
There are basically 3 types of cells and batteries
i) Primary cells
ii) Secondary cells
iii) Fuel cells
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i) Primary Cells
CANNOT be recharged.
Most common, convenient and cheapest to use
Produce environmental problems with their disposal after use
e.g.Leclanche (Zinc Carbon) cell
Anode: Zn container
Cathode: Graphite
Electrolyte: NH4Cl and ZnCl2 +
MnO2(in paste form)
Anode: Zn(s)Zn2+(aq)+ 2e
Cathode:
2NH4+(aq)+2MnO2(s)+2eMn2O3(s)+2NH3(aq)+H2O(l)
Overall: Zn(s) + 2NH4+(aq)+ 2MnO2(s)
Zn2+(aq)+ Mn2O3(s)+ 2NH3(aq)+ H2O(l)
Cell diagram: Zn(s)Zn2+(aq)MMnO2(s)Mn2O3(s)C(s) E
= +1.50 V
When a current is drawn for sometimes
NH3(g)& Zn2+accumulates
Equilibrium shifts to LHS
Ecelldrops
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When the cell is stored for some time without use
Zn + NH4+Zn2++ NH3+ H2occurs
[NH4+] decreases & eqn shifts to LHS
Ecelldrops
e.g. Silver oxide battery
Silver oxide cells look like buttons.
The () electrode (anode) is zinc powder.
The (+) electrode (cathode) is silver oxide.
These materials are tightly compacted powders separated by
a moist paste of silver oxide containing some potassium
hydroxide. Moistened paper serves as the salt bridge.
The cell produces 1.5V.
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Anode: Zn(s)+ 2OHZnO(s)+ H2O(l)+ 2e
Cathode: Ag2O(s)+ H2O(l)+2e2Ag(s)+ 2OH(aq)
Overall: Zn(s)+ Ag2O(s)ZnO(s)+ 2Ag(s)
The silver oxide cell is smalland lasts for a long time. It
also gives a steady current. However, it is more
expensivethan other types of dry cells.
Cell diagram:
Zn(s)ZnO(s)KOH(aq)Ag2O(s)Ag(s)
ii) Secondary Cells
Must be charged up first before it can be used.
Usually rechargeable
e.g. Lead acid accumulator
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Cathode: Pb coated with PbO2
Anode: Pb metal
WhenDISCHARGING
Anode: Pb(s)+ SO42(aq)PbSO4(s)+ 2e
Cathode: PbO2(s)+4H+(aq)+SO42(aq)+2e
PbSO4(s)+2H2O(l)
discharge
Cell rxn: Pb(s)+PbO2(s)+4H+(aq)+2SO42(aq) 2PbSO4(s)+2H2O(l)
recharge
Cell diagram:
Pb(s) PbSO4(s) H2SO4(aq) PbO2PbSO4(s) Pb(s)
E= -0.35 V E
= +1.69 V
E
cell= +1.69 (-0.35) = + 2.05 V
When the battery is recharged by connecting to an external source,
electrons are forced in the opposite direction
Reverse the cell reaction
It is mainly used in motor cars. Usually 6 cells are joined in
series to give a total of 12 V
Life time: 3-5 years
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Ex 6.3-12
The rechargeable nickel-cadmium battery has electrodes
of solid cadmium and solid nickel (IV) oxide, NiO2,
coated on a conductor. When this battery discharges,
solid Cd(OH)2and solid Ni(OH)2are formed. The
electrolyte is a basic medium of KOH. Given
Cd(OH)2 Cd E= -0.76 V
NiO2 Ni(OH)2 E= +0.49 V
(a) Write balanced equations for the reactions occurring
at cathode and anode when the battery discharges.
(b) Write the overall cell reaction and find Ecell.
(c) What is the direction of electrons flow when the
battery discharges?
(d) Write the cell diagram by IUPAC convention.
(e) Why is the battery re-chargeable and how, in principle,
this is carried out?
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iii) Fuel Cells
Differences between fuel cells and electrochemical
cells:
Fuel cell is a primary cell which efficiently converts the chemical
energy into electrical energy without losing to heat, mechanical
linkages
Environmentally friendly
Continuous supply of electrical energy
Open system
Flexible: micro to mega Watts
e.g. Hydrogen-oxygen fuel cell
Fuel (H2) and oxidizing agent (O2) are supplied in a continuous
flow into the anode and cathode compartments respectively.
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Porous Ni electrodes act as the electrical conductor and also a
catalyst for the reactions.
Anode: 2H2(g)+ 4OH(aq)4H2O(l)+ 4e
Cathode: O2(g)+ 2H2O(l)+ 4e4OH(aq)
Overall reaction: 2H2(g)+ O2(g)2H2O(l)
Cell diagram:
Ni(s)H2(g)H2O(l)KOH(aq)O2(g)2OH(aq)Ni(s)
It is used as the energy source in spacecraft and the crew can
drink the water produced during the reaction.
It cannot be used as a source of portable power because of heavy
weight (500 lb)
G) Corrosion of Iron and Its Prevention
(a) Chemistry of Rusting
(i) Electrochemical Process Involving in Rusting
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Around the edge of water droplet higher [O2]
act as CATHODE
O2(g)+ 2H2O(l)+ 4e4OH(aq) E= +0.40 V
(turn phenophthalein pink)
At the centre of water droplet lower [O2]
act as ANODE
Fe(s)Fe2+(aq)+ 2e E=0.44 V
(turn K3Fe(CN)6blue)
O2(g) + 2H2O(l) + 2Fe(s)4OH-(aq) + 2Fe2+(aq)E= +0.84V
E
cellis +ve spontaneous reaction
Fe2+(aq)formed in the anodic regions travel to the cathodic regions
where it reacts with the OHproduced in the cathodic regions.
Fe2+(aq)+ 2OH(aq)Fe(OH)2(s)
Fe(OH)2(s)further react with O2(g)in air
4Fe(OH)2(s)+ O2(g)+ 2 H2O(l)4Fe(OH)3(s)
Fe(OH)3(s)Fe2O3nH2O(s) (rust)
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ii) Essential Conditions for Rusting
1. Moisture (water)
2.Air (O2)
iii) Factors Increasing the Rusting Rate
1. Low pH because it favours FeFe2++ 2e
2. High temperature
3. Present of dissolved electrolytes e.g. NaCl because
conductivity
b) Prevention of Rusting
i) Application of a protective layer (coating)
Nonmoving objects: Painting, varnishing or coating with
plastic
Moving objects: Greasing or oiling
ii) Cover with a thin layer of another metal which is resistant to
corrosion
1. Tinplating: By immersing Fe in molten Sn (used to make
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Zn protects Fe from rusting by sacrificial protection. Galvanized
iron is not used for canning food because Zn2+is poisonous.
3. Chromiumplating, Nickelplating: By electrolysis
iii) Alloying
Stainless steel: An alloy of Fe with Cr (min. 12%), Ni and
Mn. Cr forms a layer of impermeable and
insoluble oxide layer on surface of alloy
grains.
iv) Sacrificial protection
Use more reactive metals, e.g. Mg as sacrificial anode which
supply eto the iron (cathode) to prevent it from oxidation, e.g.
attach large Mg blocks to the steel of the hull of ship, offshore
oil drilling platform or underground oil tank.
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v) Cathodic protection
Iron is treated as the cathode and connected to the ve
terminal of a battery. This can inhibit the formation of Fe2+.
iv) Socioeconomic Implications of Corrosion and Prevention
Corrosion of iron can cost much to the society
i) Money spent on:
1. Replacement of the corroded articles
2. Prevention of corrosion
3. Indirect cost, e.g. those for the maintenance of machines,
those due to lost production when machines fail down or
where they are shut down for maintenance. (In 1980 in US,
70 billions US dollars were lost annually because of
corrosion)
ii) Wastage of natural resources.
It is estimated that 1 tonne of steel is converted into rust very
90 sec in Britain and about 40 % of the steel made in US is
used to replace steel lost by rusting.
iii) Inconvenience to human beings and even loss of life.
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Corrosion results in the formation of cracks and crevices
which weaken the strength of metals
concrete in buildings may fall off
aged vehicles may cause more road accidents
Disadvantages of corrosions are outweighed by the relative
low cost, abundance and ease of extraction
The society should assume a greater responsibility for
corrosion prevention in order to have a greater improvement
of the situation.
HKAL Past Paper Questions
86 IIA 2(b) 87 IIA 2(b)88 IA 1(c) 89 IIA 1(b)90 IA 2(b), IIA 1(b) 91 IA 2(a), IIA 2(b)92 IIA 1(c) 93 IA 1(d), IIA 1(c), 3(a)94 IA 1(d)(e), IIA 3(a)(i) 95 IIA 3(c)96 IIB 6(d)(e) 97 IA 2(c), IIA 4(a)(b)98 IIA 4(a) 99 IIA 1(b)00 IIA 3(d) 01 IA 4(a)(b), IIA 4(a)02 IA 1(b), IIA 3(b)