2. Electrochemistry
Electrochemistry is the study of production of electricity
from energy released during spontaneous chemical
reactions and the use of electrical energy to bring about
non-spontaneous chemical transformations. Galvanic
cells or voltaic cell: It is a cell in which electricity is
produced by a spontaneous reaction.
E.g. Daniell cell
Electrolytic cell: It is a device used to carry out non-
spontaneous chemical reactions using electricity.
4. Fig 18-14 Pg 877
The copper/zinc electrochemical cell. The voltmeter measures the
difference in electrical potentials between the two electrodes.
5. Fig 18-6 Pg 870
If the zinc-and copper-containing portions of this redox system are
physically separated, electron transfer can occur only through an
external wire.
6. Galvanic cells
Cu 2+ + 2e– → Cu (s) (reduction half reaction)
Zn (s) → Zn 2+ + 2e– (oxidation half reaction)
Zn (s)+Cu 2+
(aq)→Zn 2+
(aq)+Cu (s)Net cell reaction
• In a galvanic cell, the half-cell in which oxidation takes place is
called anode and it has a negative potential The other half-cell
in which reduction takes place is called cathode and it has a
positive potential with respect to the solution.
• A potential difference develops between the electrode and the
electrolyte which is called electrode potential.
• When the concentrations of all the species involved in a
half-cell is unity then the electrode potential is known as
standard electrode potential.
• According to IUPAC convention, standard reduction
potentials are now called standard electrode potentials.
7. Cell Potential
• The potential difference between the two electrodes of a
galvanic cell is called the cell potential and is measured in
volts. The cell potential is the difference between the electrode
potentials (reduction potentials) of the cathode and anode. It
is also called electromotive force (emf) of the cell
The emf of the cell is positive and is given by the potential of
the half-cell on the right hand side minus the potential of the
half-cell on the left hand side Ecell = Eright – Eleft
the anode is kept on the left and the cathode on the right
In anode metal is represented first followed by metal ion
In cathode metal ion is represented first followed by metal
put a vertical line between metal and metal ion
Put a double vertical line between the two electrodes.
9. Standard hydrogen electrode(SHE)
Reduction reaction
2H+(1M) +2e- H2(1atm)
Standard reduction potential of SHE is
E0 =0.00V.It can be connected with
any other half cell and electrode
potential any electrode can be
calculated
10. Standard Electrode Potentials
19.3
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):Zn (s)Zn2+ (1 M) + 2e
Cathode (reduction): 2e-+2H+(1M) H2
(1 atm)
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)
E0 = Ecathode – Eanode
=(Eo H+/H2 - E°(Zn
2+
/Zn)
0.76 V = 0 - E°(Zn
2+
/Zn)
EZn /Zn = -0.76 V
11. Standard Electrode Potentials
19.3
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) 2H+ (1 M) + 2e-
Anode (oxidation):
Cathode (reduction):
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)
E0 = Ecathode - Eanode
cell
0 0
E0 = 0.34 V
cell
Ecell = ECu /Cu – EH /H
2+ +
2
0 0 0
0.34 = ECu /Cu - 0
0 2+
ECu /Cu = 0.34 V
2+
0
12. 19.3
• E0 is for the reaction as
writtenEcell = Eright – Eleft
• Half cell with lower standard
electrode potential is taken as
anode and the higher value is
taken as cathode
• The more positive E0 the
greater the tendency for the
substance to be reduced
• The halfcell reactions are reversible
• The sign of E0 changes when
the reaction is reversed
• Changing the stoichiometric
coefficients of a half-cell
reaction does not change the
value of E0
13. NERNST EQUATION
NERNST showed that for the electrode reaction,
Mn+
(aq) + ne- M(s)
E(M
n+
/M) = E°(M
n+
/M) - (2.303RT/nF) log (1/[Mn+])
R is gas constant (8.314JK–1 mol–1)
F is Faraday constant (96487 C mol–1)
n is number of electrons
T is temperature in Kelvin and
[Mn+] is the concentration of the species, Mn+.
14. Calculation of EMF for Daniell cell
E(Cu
2+
/Cu) = E°(Cu
2+
/Cu) - (2.303RT/2F) log (1/[Cu2+])
E(Zn
2+
/Zn) = E°(Zn
2+
/Zn) - (2.303RT/2F) log (1/[Zn2+])
Ecell= E(Cu
2+
/Cu)- E(Zn
2+
/Zn)
= E°(Cu
2+
/Cu) - (2.303RT/2F) log (1/[Cu2+])- {E°(Zn
2+
/Zn)
- (2.303RT/2F) log (1/[Zn2+])}
= E°(Cu
2+
/Cu)- E°(Zn
2+
/Zn) - (2.303RT/2F) log ([Zn2+]/[Cu2+])
= E0
cell - (2.303RT/2F) log ([Zn2+]/[Cu2+])
Ecell = E0
cell –(0.059/2)log ([Zn2+]/[Cu2+])
It can be seen that Ecell increases with increase in concentration
of Cu 2+ and decreases with increase in concentration of Zn2+
15. Equilibrium constant and Gibbs energy
Ecell = E0
cell - (2.303RT/2F) log ([Zn2+]/[Cu2+])
At equilibrium there is no change in concentration takes
place and the volt meter reading will be zero
0 = E0
cell - (2.303RT/2F) log ([Zn2+]/[Cu2+])
E0
cell = (2.303RT/2F) log ([Zn2+]/[Cu2+])
E0
cell = (0.059/2)log ([Zn2+]/[Cu2+])
E0
cell = (2.303RT/nF) log Kc
G0= -nFE0
cel
G0=- 2.303RT* log Kc
16. Conductance of Electrolytic Solutions:Conductance of electrolytic
solutions depend on,Nature of electrolyte added, Nature of solvent and
its viscosity, Concentration of electrolyte, Size of ion produced and its
solvation,Temperature. Conductance is measured in Siemen (S). It is
equal to mho (Ω-1) .
Measurement of the conductivity of ionic solutions
Two main problem we face in the measurement of resistance of ionic
solution are
• passing of DC changes the composition of the solution.
• solution is difficult to connect to a bridge
• These problems are solved by passing AC instead of DC and by using
specially designed devices called conductivity cells.
17. CONDUCTIVITY CELLS
It consists of two platinum electrodes coated
with platinum black. The area of cross
section is ‘A’ and are separated by a distance
‘l’.
R = ρ.l/A ρ is resistivity
= l/κ A κ is conductivity
The quantity l/A is called cell constant
denoted by the symbol, G*. R=G*/ κ
G* = R. κ
18. Variation of Conductivity and Molar
Conductivity with Concentration
Conductivity always decreases with
decrease in concentration both, for weak
and strong electrolytes.
This is due to the fact that the number of
ions per unit volume that carry the
current in a solution decreases on dilution.
19. Molar conductivity of a solution at a given
concentration is the conductance of the
volume V of solution containing one mole of
electrolyte
Molar conductivity increases with
decrease in concentration. This is
because the total volume, V, of solution
containing one mole of electrolyte also
increases. It has been found that decrease in
κ on dilution of a solution is more than
compensated by increase in its volume.
20. Molar conductivity versus c½
Strong electrolyte
Weak electrolyte
c½
Λm
When concentration of
solution approaches zero,
molar conductivity is called
limiting molar conductivity.
On dilution molar
conductivity of weak
electrolyte increases steeply
as the number of ions
increases and for strong
elecrtolytes it increases
slightly as the strong
electrolytes are already
dissociated completely and
only mobility of ions
increases
21. For a strong electrolyte, Λm
0 can be
determined by extrapolating the graph, as
Λm varies linearly. But for a weak electrolyte,
the graph obtained is a curve. So it can not
be extrapolated to zero concentration. So for
weak electrolytes, Kohlrausch law of
independent migration of ions is used to
determine Λm
0
Determination of Limiting molar conductivityΛm
0
22. Kohlrausch law of independent
migration of ions
The law states thatlimiting molar conductivity
of an electrolyte can be represented as the sum
of the individual contributions of the anion and
cation of the electrolyte. In general, if an
electrolyte on dissociation gives v+ cations and v–
anions then its limiting molar conductivity is given
λm° = ν+ λ0
+ + ν– λ0
–
Here,λ +
0 andλ−
0 are the limiting molar
conductivities of the cation and anion respectively.
23. i f λ°Na+ and λ°Cl
– are limiting molar
conductivity of the sodium and chloride ions
respectively, then the limiting molar
conductivity for sodium chloride is given by the
equation:
Em°(NaCl) = λ0
Na
+ + λ0
Cl
–
Applications of Kohlrausch law
Λm
0 for weak electrolytes is obtained by using
Kohlrausch law of independent migration of
ions
24. Solve the numerical
Λ o
mfor NaCl, HCl and NaAc are 126.4, 425.9 and
91.0 S cm2 mol–1 respectively. Calculate Λofor
HAc.
Λ o
m(HAc) = λο
mH+ + λο
mAc–
= λο
mH
++ λο
mCl– + λο
mAc–+ λο
mNa
+ - λο
mNa
+-λο
mCl–
= Λ o
m (HCl)+Λ o
m(NaAc)- Λ o
m(NaCl)
= (425.9 + 91.0 – 126.4 ) S cm2 mol –1
= 390.5 S cm2 mol –1 .
25. Answer the following and check your understanding
1) Represent the galvanic cell in which the following reactions
take place. Zn(s) + 2Ag+(aq)→Zn2+(aq) + Ag(s)
i)Write all notation ii) which one of its electrode is negative charged
iii) The reaction taking place at each electrode iv) The carrier of current.
Ans: The cell represents as
i)Zn(s)/Zn2+ (aq) // Ag+(aq)/Ag(s)
ii)Zn electrode is negatively charged
iii)At anode Zn(s) → Zn2+ + 2e-
At cathode Ag+(aq) + e- → Ag(s)
iv)Ions are carrier of current within the cell.
2) Write Nernst equation and calculate the emf of the following cell
Sn(s)/ Sn2+ (.04M) // H+(.02M)/H2(g),Pt(s), E0Sn2+ /Sn= -0.14V
Ans : E cell = E o
Cell – 0.0591 2* log[Sn2+]/[H+]2
=(Eo H+/H2 – EoSn2+/ Sn) – 0.0591/2 *log 0.04/(0.02)2
= +0.14V – 0.0591 V
= 0.0809V
3.What is the unit of molar conductivity?
Ans:ohm-1cm2 mol-1
26. Fig 18-25
Pg 898
Courtesy Stephen Frisch
A photograph and schematic diagram of a cell for the
electrolysis of water. The reaction generates hydrogen and
oxygen in a 2:1 ratio.
Electrolytic cells and electrolysis of Water
27. Faraday’s Laws of Electrolysis
First Law: The amount of chemical reaction which
occurs at any electrode during electrolysis by a current
is proportional to the quantity of electricity passed
through the electrolyte (solution or melt).
Second Law: The amounts of different substances
liberated by the same quantity of electricity passing
through the electrolytic solution are proportional to
their chemical equivalent weights (Atomic Mass of
Metal ÷ Number of electrons required to reduce the
cation).`
28. For example, in the reaction:
Ag +
(aq) + e– → Ag(s)
One mole of the electron is required for the
reduction of one mole of silver ions.
charge on one electron is equal to 1.6021×10–19C.
Therefore, the charge on one mole of electrons
is equal to: 6.02 × 1023 mol–1 × 1.6021 × 10–19
C = 96487 C mol–1
This quantity of electricity is called Faraday
and is represented by the symbol F.
29. Products of Electrolysis
if we use molten NaCl, the products of
electrolysis are sodium metal and Cl2 gas.
Here there is only one cation (Na+) which is
reduced at the cathode
Na+ + e– → Na
Here there is only one anion (Cl–) which is
oxidised at the anode
Cl–→ ½Cl2+e–
30. During the electrolysis of aqueous sodium chloride solution, the
products are NaOH, Cl2 and H2.
In this case besides Na+ and Cl– ions we also have H+ and OH–
ions along with the solvent molecules, H2O.
At the cathode there is competition between the following
reduction reactions: Na+
(aq) + e– → Na(s) E0
cell = – 2.71 V
H+
(aq) + e– → ½ H2(g) E0
cell = 0.00 V The reaction with
higher value of E is preferred and, therefore, the reaction at the
cathode during electrolysis is: H+
(aq) + e– → ½ H2 (g)
but H+
(aq) is produced by the dissociation of H2O, i.e.,
H2O (l ) → H+
(aq) + OH–
(aq) Therefore, the net reaction at the
cathode may be written as the sum of the above two and we have
H2O (l ) + e– → ½H2(g) + OH–
ELECTROLYSIS OF AQUEOUS SODIUM CHLORIDE
SOLUTION
31. At the anode the following oxidation reactions are possible:
Cl–
(aq) → ½ Cl2 (g) + e– E0
cell =1.36 V
2H2O (l )→ O2 (g) + 4H+
(aq) + 4e– E0
cell =1.23 V
The reaction at anode with lower value of E0 is preferred and
therefore, water should get oxidised in preference to Cl–
(aq).
However, on account of over potential of oxygen, the first
reaction is preferred The net reactions may be summarised
as: NaCl (aq) ⎯⎯H2⎯O⎯→ Na+
(aq) + Cl–
(aq)
Cathode: H2O(l ) + e– → ½ H2(g) + OH–
(aq)
Anode: Cl–
(aq) → ½ Cl2(g) + e–
Net reaction:
NaCl(aq) + H2O(l) → Na+
(aq) + OH–
(aq) + ½H2(g) + ½Cl2(g)
34. Primary cells
Electrochemical cells which cannot be
recharched again and the reaction occurs only
once are called primary cells.eg.Daniel cell
Secondary cells
Electrochemical cells which can be recharched
by passing electric current in the opposite
direction that can be used again ie the reaction
can be reversed.eg. Lead storage battery
35. LACLANCHE CELL: DRY CELL
NH4Cl
Dry cell consists of Zinc
container as anode.
Graphite rod surrounded
by MnO2 and carbon as
cathode. A paste of
NH4Cl,ZnCl2 and MnO2
as electrolyte.
At Anode- Zn(S) —› Zn2+
+ 2e-
At Cathode -
2MnO2(s)+2NH4
+ +2e-
—› Mn2O3 +2NH3
+H2O
37. Mercury cell
Anode – Zinc amalgam.
Cathode – A paste of HgO and carbon.
Electrolyte - A paste of HgO and ZnO
Anode – Zn(Hg) + 2OH- —› ZnO + H2O +
2e-
Cathode – HgO + H2O + 2e- —› Hg + 2OH-
Overall reaction – Zn + HgO —› ZnO + Hg
The cell potential is approximately 1.35 V and
remains constant during its life as the overall
39. Lead storage battery
Lead plates act as anode. Grid of lead plates
coated with lead dioxide act as cathode. A
solution of H2SO4 acts as electrolyte.
Anode - Pb(s) + SO4
2- —› PbSO4 + 2e-
Cathode – PbO2(s) + 4H++ SO4
2- + 2e- —› PbSO4
+2H2O
Cell reaction- Pb(s) + PbO2(s) + 2H2SO4 —› 2PbSO4
+2 H2O
On recharging cell reaction is reversed.
40. Ni-Cad Battery
Cadmium metal act as anode. Metal grid containing
Cd(IV) act as cathode. KOH solution act as
electrolyte.
Cd(S) + 2Ni(OH)3(s) —›CdO(S) + 2Ni(OH)2(s) + H2O
Cell reaction is reversed on recharging. Ecell is
approximately 1.4 V .
It has longer life than lead storage cell
41. H2 as a Fuel
Cars can use electricity generated by H2/O2 fuel
cells.H2 carried in tanks or generated from
hydrocarbons
42. Hydrogen oxygen Fuel Cells
Invented by
WilliamGrooves.
• It is an
electrochemical
device used to
convert free energy
of combustion
reaction of fuels such
as H2,CH4,CO etc
into electrical energy
directly
• It consists of porous
carbon electrodes.
• Hydrogen and
oxygen are bubbled
43. Reactions are
Anode – 2[H2(g) + 2OH-
(aq) —› 2 H2O(aq) +
2e-]
Cathode – O2 + 2 H2O + 4e- —› 4OH-
(aq)
Cell reaction -2 H2(g) + O2 —› 2 H2O(aq)
EMF is about 0.9V
Advantages
1) 70% efficient which is greater than any
other conventional source.
2) Cell runs continuously as long as
reactants are supplied.
3) No pollution.
4) Electrodes are not affected
44. CORROSION
It involves the slow destruction of a metal as a
result of its reaction with moisture and gases
present in atmosphere.
More reactive metals corrode more easily.
Corrosion of Iron is called rusting.
Mechanism of Rusting(Electrochemical
theory of rusting)
Impure surface of iron act as an electrochemical
cell. Pure Iron act as anode and impure iron as
cathode. Carbonic acid act as the electrolyte.
45. Water
Rust
Iron Dissolves-
At anode Fe Fe+2 +2e-
e-
Salt speeds up process of rusting by increasing
conductivity
2Fe(s) + O2 + 4H+ —› 2Fe2+ + 2 H2O
Cathode is further oxidised by atmospheric oxygen
to form rust. Fe2O3 .x H2O
Fe2+
At cathode O2 +4H+ + 4e- 2 H2O
46. Prevention of corrosion
• Barrier protection – By coating with a suitable material
–paint, grease etc
• Sacrificial protection – Coating with a more reactive
metal. The process of coating the surface of iron with
Zinc is called Galvanization. More reactive metal act as
anode.
• Alloying with metals that form oxide coats.
• Cathodic protection - Here metal to be protected is set as
cathode by attaching a more reactive metal such as
magnesium to it. Now the more reactive metal undergo
oxidation.
• Anti rust solutions – Alkaline phosphate or chromate
solutions are applied on iron surface to form a heat
resistant iron phosphate or chromate coating
48. ANSWER THE FOLLOWING QUESTION
1)What type of battery is the lead storage battery
Ans : Secondary Cell
2)What is fuel Cell?
Ans: The cell which converts combustion energy of fuel
into electricity.
3) Write the cathode reaction in fuel cell
Ans: Cathode reaction O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
4)What is galvanisation?
Ans: Giving a coating of zinc on iron material
49. Conductivity
Λm – molar
conductivity
Conductance of all ions produced by
dissolving 1mol of electrolyte in V ml
Λm = K
Cm
Cm = concentration in mol/l
unit = Scm2/mol Or Ω-1 cm2/mol
Mo
promoter
Equivalent Conductivity
Λeq
Conductance of all ions
produced from one gram
equivalent of electrolyte
dissolved in V ml of solution.
Λeq = K
Ceq
Ceq = concentration of solution
/l
unit = Ω-1 cm2 mol-1
Specific conductivity (K)
It is the reciprocal of sp.
Resistance
ρ = sp. resistance
l/A Cell constant
R = Resistance
Molar conductivity of a weak electrolyte like
NH4OH, NH3 etc increases with decrease in
concentration of solute or with dilution; due
to increase in degree of ionization of
electrolyte. No. of ions increases so
conductivity too.
Λm = Λo
m-AC1/2
C= concentration
Λo
m = molar conductivity
at ∞ dilution
Λm
C1/2
A = Constant
- For Acetic Acid
Molar conductivity
Of strong electrolyte increases with dilution,
ions get parted and interionic attraction
decreases and conductivity increase
Eg:- HCl, NaOH
Λm = Λm – AC1/2
Λm
C1/2
- For KCl
50. Batteries
Fuel Cells
Primary Batteries
Cannot recharge
Dry Cell/ Leclanche Cell
Zn Anode, graphite + MnO2 + C
→Cathode, NH4Cl + ZnCl2 paste
electrolyte
Anode: Zn→ Zn2+ + 2e-
Cathode: MnO2 + NH4+ + e-→MnO(OH) +
NH3
Mn reduced to +4 → +3 state
Voltage = 1.5V
NH3 forms [Zn(NH3)4]2+ complex.
Secondary Cell
Can be recharged
Combustion of fuels like H2 , CH4 etc.
H2-O2 fuel cell
Cathode- C +O2
Anode- C+H2
Electrolyte:- Conc. NaOH solution
Cathode :- O2 + 2H2O +4e- → 4OH-
Anode:- 2H2 + 4OH-→4H2O + 4e-
Overall:- 2H2 + O2→ 2H2O
Advantage: Pollution free, cheap,
70% efficient.
Mercury Cell
Anode –Zn/Hg
Cathode- C + HgO paste
Electrolyte = KOH + ZnO
Anode = Zn/Hg + 2OH- → ZnO +H2O +2e-
Cathode HgO + H2O + 2e-→ Hg + 2OH-
Overall Zn/Hg + HgO → ZnO + Hg
Voltage 1.35V.
Ni-Cd Cell
Long life but more expensive
Reaction: Cd + Ni(OH)3 → CdO
+ 2Ni(OH)2 + H2O
Lead Storage Battery
Anode- Pb, Cathode: Pb + PbO2
Electrolyte – 38% H2SO4
Anode: Pb + SO4
2- →PbSO4 + 2e-
Cathode: PbO2 + SO4
2- + 4H+ +2e- → PbSO4
+2H2O
Overall Pb + PbO2 + 2H2SO4 → 2PbSO4 +
2H2O
On recharging reaction get reversed.
51. Electrolysis
Aqueous NaCl
At Cathode H2O + e- → ½ H2 + OH-
At Anode Cl- → ½ Cl2 + e-
NaCl + H2O → Na+ + OH- + ½ H2
+ ½ Cl2
AgNO3 using Ag
electrodes
At Anode Ag → Ag+ + e-
At Cathode Ag+ + e- → Ag
Aq. AgNO3 using Pt
electrodes
At Cathode Ag+ + e- → Ag
At Anode OH- → OH + e-
4OH- → 2H2O + O2(g)
Mo
promoter
Aqueous CuCl2 using Pt
electrodes
At Cathode
Cu2+ + 2e- → Cu
At Anode
2Cl- → Cl2 + 2e-
H2SO4 using Pt electrodes
At Cathode 2H+ + 2e- → H2
At Anode 4OH- → 2H2O + O2
Molten NaCl
At Anode Cl- → ½ Cl2 + 2e-
At Cathode Na+ + e- → Na(s)
52. Corrosion
Rusting of
Fe
Anode:- Fe
Reaction:- 2Fe → 2Fe2+ + 2e-
Cathode:- Fe covered with H2CO3
Reaction:- O2 + 4H+ + 4e- → 2H2O
Overall:- 2Fe + O2 + 4H+ → 2Fe2+ + 2H2O
Electrolyte → H2O + CO2 (H2CO3)
Fe2+ oxidation Fe3+ by oxygen
Fe3+ + xH2O + O2 → Fe2O3 .xH2O
Rust (Hydrated ferric oxide)
Sea shores enhances rusting due to presence of NaCl
Prevention
1.painting / chemicals
2.covering with Zn, Sn etc.
3.sacrificial protection using Mg, Zn
etc.
Eating away of metal
Eg: rusting of Fe ,tarnishing of
Ag green coating on Cu etc.
53. LEARN NUMERICALS
1) How many moles of mercury will be produced by
electrolyzing 1.0 M Hg (NO3)2 solution with
a current of 2.00 ampere for three hours?
[ gram atomic mass of Hg = 200.6 g]
Ans: Current = 2A, Time = 3 h = 3(60)(60) S, W= Zit
Z for Hg in compound = E96500=200.6/2(96500)
W= 200.6(3)(60)(60)2(96500), W = 22.45g
Number of moles = 22.45200.6 = 0.112 mol
2) How much change is required for the reduction of 1
mole of Cu2+ to Cu?
Ans: Statement Cu2+ + 2e → Cu
charge required for the reduction of 1 mole Cu2+ = 2F
2F = 2(96500 C) = 193000C
54. 3) Determine the value of equilibrium constant and ∆Go for the
following reaction Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), (Eo = 1.05 V)
Ans: ∆Go = -nFEocell
= - 2*(96500) C *(1.05) V
= - 202650J mol-1 = -202.65 KJ mol-1,
-nFEocell = -2.303 RT log Kc
logKc = nFEocell 2.303RT,
= Eocell0.0591
= 2(1.050.0591) = 35.532g
Kc = antilog 354.532, Kc = 3.411 x 1035
4) Calculate the standard electrode potential of Ni2+// Ni electrode if emf
of the Ni2+/Ni(0.01)//cu2+(0.1) Cu (s) is 0.059v. Given Eocell= 0.34v
Ans: Ecell = 0.059V, Eo Cu = 0.34 V (Cu+2) = 0.1 M
Ecell = Eo
cell - 0.0591/n *log [Ni2+ (aq)]/ [Cu2+(aq)]
0.059=Eocell- 0.059/2*log 0.01/0.1
0.059=Eocell -0.0295 log 1/10
0.059=Eocell –0.0295 (log 1- log 10), =Eocell -0.0295(-1)
0.059 =Eo cell + 0.0295, Eo cell = 0.05g-0.0295 = 0.0295v
Eo cell= Eo c- Eo a, 0.0295 = 0.34 – Eo anode
Eo anode = 0.34-0.0295 = 0.3105v
55. HOME WORK QUESTION
1.The resistance of a conductivity cell containig0.001M KCl solution at
298K is 1500 Ω. What is the cell constant if the conductivity of
0.01M KCl solution at 298k is 0.146(10-3)Scm-1.
2)The conductivity of 0.001M CH3COOH is 4.95x10-5 Scm-1.
calculate its dissociation constant. Given limiting molar conductivity
is 390.5s cm2 mol-1.[Hints : α=0.126]
3) Calculate the potential of the following cell reaction at 298k
Sn4+(1.5M) + Zn(s)→Sn2+(0.5M) + Zn2+(2.0M)
(Eo
cell = 0.89V) [ Hints: 3.17V]
4) How many moles of electrons are required to
i) Reduce 1 mol of MnO4
- to Mn2+
ii) Produce 10.0 g of Al from molten Al2O3.
[Hints: i) 5 mol of electron ii) 1.11 mol Al]
5) Calculate the emf of the following cell at 298k
Zn/Zn2+ (10-4M)││ Cu2+ (10-2M)/Cu.
Given: Eo
Zn2+/zn =-0.76V Eo
Cu2+/cu = +0.34V.
[ Hints: emf = 1.1591V]
