Electrochemistry
Electrochemistry
Electrochemistry
ELECTROCHEMISTRY
ELECTROCHEMICAL CELLS
An electrochemical cell consists of two electrodes (metallic conductors) in contact with an electrolyte
(an ionic conductor).
An electrode and its electrolyte comprise an Electrode Compartment.
(ii) Galvanic Cells which produce electricity as a result of a spontaneous cell reaction
ELECTROLYSIS
The decomposition of electrolyte solution by passage of electric current, resulting into deposition of
metals or liberation of gases at electrodes is known as electrolysis.
ELECTROLYTIC CELL
This cell converts electrical energy into chemical energy.
The entire assembly except that of the external battery is
known as the electrolytic cell
ELECTRODES
The metal strip at which positive current enters is called anode; anode is positively charged in electrolytic
cell. On the other hand, the electrode at which current leaves is called cathode. Cathodes are negatively
charged.
GALVANIC CELL
This cell converts chemical energy into electrical energy.
Galvanic cell is made up of two half cells i.e., anodic and cathodic. The cell reaction is of redox kind.
Oxidation takes place at anode and reduction at cathode. It is also known as voltaic cell. It may be
represented as shown in Fig. Zinc rod immersed in ZnSO4 behaves as anode and copper rod immersed
in CuSO4 behaves as cathode.
Oxidation takes place at anode:
Zn Zn2+ + 2e– (loss of electron : oxidation)
Reduction takes place at cathode:
Cu2+ + 2e– Cu (gain of electron ; reduction)
Over all process:
Zn(s) + Cu2+ Cu(s) + Zn2+
In galvanic cell like Daniell cell; electrons flow from anode (zinc rod) to the cathode (copper rod)
through external circuit; zinc dissolves as Zn2+ ; Cu2+ ion in the cathode cell picks up two electron and
become deposited at cathode.
SALT BRIDGE
Two electrolyte solutions in galvanic cells are seperated using salt bridge as represented in the Fig. salt
bridge is a device to minimize or eliminate the liquid junction potential. Saturated solution of salt like
KCI, KNO3, NH4Cl and NH4NO3 etc. in agar-agar gel is used in salt bridge. Salt bridge contains high
concentration of ions viz. K+ and NO3 at the junction with electrolyte solution. Thus, salt bridge carries
whole of the current across the boundary ; more over the K+and NO3 ions have same speed. Hence,
salt bridge with uniform and same mobility of cations and anions minimize the liquid junction potential &
completes the electrical circuit & permits the ions to migrate.
Representation of a cell (IUPAC conventions ): Let us illustrate the convention taking the example of Daniel
cell.
(i) Anodic half cell is written on left and cathodic half cell on right hand side.
Zn(s) | ZnSO4 (sol) || CuSO4 (sol) | Cu(s)
(ii) Two half cells are separated by double vertical lines: Double vertical lines indicate salt bridge or any type
of porous partition.
(iii) EMF (electromotive force) may be written on the right hand side of the cell.
(iv) Single vertical lines indicate the phase separation between electrode and electrolyte solution.
Zn | Zn2+ || Cu2+ | Cu
(Illustration of Phase boundary)
(v) Inert electrodes are reprsented in the bracket
Zn | ZnSO4 || H+ | H2, Pt
G = G° + RT ln Q ..(1)
where G and G° are free energy and standard free energy change; ‘Q’ is reaction quotient.
Let n, Faraday charge is taken out from a cell of e.m.f. (E) then electrical work done by the cell
may be calculated as,
Work done = Charge x Potential = nFE
From thermodynamics we know that decrease in Gibbs free energy of a system is a measure of
reversible or maximum obtainable work by the system if there is no work due to volume expansion
G = nFE and G° = nFE°
Thus from Eq. (i),we get nFE = -nFE° + RT lnQ
0.0591
At 25°C, above equation may be written as E E0 log Q
n
Where ‘n’ represents number of moles of electrons involved in process.
E, E° are e.m.f. and standard e.m.f. of the cell respectively.
In general , for a redox cell reaction involving the transference of n electrons
aA + bB cC + dD, the EMF can be calculated as:
0.0591 [C]c [D]d
ECell = E°Cell – log
n [A ]a [B]b
Prediction and feasibility of spontaniety of a cell reaction.
Work done by the cell = nFE;
It is equivalent to decrease in free energy G = –nFE
Under standard state G0 = –nFE0 (i)
(i) From thermodynamics we know, G = negative for spontaneous process. Thus from eq.(i) it is
clear that the EMF should be +ve for a cell process to be feasible or spontaneous.
(ii) When G = positive, E = negative and the cell process will be non spontaneous.
(iii) When G = 0 , E = 0 and the cell will attain the equilibrium.
Reactions G E
Spontaneous (–) (+)
Non- spontaneous (+) (–)
Equilibrium 0 0
Standard free energy change of a cell may be calculated by electrode potential data.
Substituting the value of E0 (i.e., standard reduction potentialof cathode- standard reduction potential of
anode) in eq. (i) we may get G0.
Let us see whether the cell (Daniell) is feasible or not: i.e. whether Zinc will displace copper or not.
Zn | (s) | ZnSO4 (sol) || CuSO4 (sol) | Cu(s)
E 0Zn 2 / Zn 0.76volt ; E 0Cu 2 / Cu 0.34volt
0.0591 nE 0
0 K eq anti log
0E logK eq or
n 0.0591
Heat of Reaction inside the cell: Let n Faraday charge flows out of a cell of e.m.f. E, then
G = nFE (i)
Gibbs Helmholtz equation (from thermodynamics ) may be given as,
G
G = H + T T (ii)
p
G
G H T
T p
From Eqs. (i) and (ii), we have
G G
TST or S
T p T p
E
or SnF
T p
E
where T is called temperature coefficient of cell e.m.f.
p
E E
0.0591
0
log
pH 2 1/ 2
H /H H /H 1 H
2 2
H (c)e 1 / 2H 2 ( p2 )
1 / 2H 2 ( p1) 1 / 2H 2 ( p2 )
1/ 2
2.303RT p 2
E log
F p1
2.303RT p 2 0.059 p1
or E log
, At 25 C ,
0 E log
2F p1 2F p
2
For spontanity of such cell reaction, p1>p2
Electrolyte concentration cells:
Zn(s) | ZnSO4 (C1) || ZnSO4 (C2) | Zn(s)
In such cells, concentration gradient arise in electrolyte solutions. Cell process may be givewn as,
Zn s Zn 2 C1 2e (Anodic process)
Zn 2 (C 2 ) 2e Zn (s)
(Over all process)
Zn 2 (C 2 ) Zn 2 (C1 )
The m vs C plot of strong electrolyte being linear it can be extrapolated to zero concentration.
Thus, m values of the solution of the test electrolyte are determined at various concentrations the
concentrations should be as low as good.
m values are then plotted against C when a straight line is obtained. This is the extrapolated to zero
concentration. The point where the straight line intersects m axis is 0m of the strong electrolyte.
However, the plot in the case weak electrolyte being non linear, shooting up suddenly at some low
concentration and assuming the shape of a straight line parallel to m axis. Hence extrapolation in this
case is not possible. Thus, 0 of a weak electrolyte cannot be determined experimentally. It can, however,
be done with the help of Kohlrausch's law to be discussed later.
Kohlrausch's Law of Independent Migration of Ions
Kohlrausch determined 0 values of pairs of some strong electrolytes containing same cation say KF
and KCl, NaF and NaCl etc., and found that the difference in 0 values in each case remains the same:
0m (KCl) – 0m (KF) = 0m (NaCl) – 0m (NaF)
He also detemined 0 values of pairs of strong electrolytes containing same anion say KF and NaF, KCl
and NaCl etc.and found that the difference in 0 values in each case remains the same.
0m (KF) – 0m (NaF) = 0m (KCl) – 0m (NaCl)
This experimental data led him to formulate the following law called Kohlrausch's law of independent
migration of ions.
At infinite dilution when dissociation is complete, every ion makes some definite contribution towards
molar conductance of the electrolyte irrespective of the nature of the other ion which with it is associated
and that the molar conductance at infinite dilution for any electrolyte is given by the sum of the contribution
of the two ions. Thus,
0m = 0 0
Where 0 is the contribution of the cation and 0 is the contribution of the anion towards the
molar conductance at infinite dilution. These contributions are called molar ionic conductances at
infinite dilution. Thus, 0 is the molar ionic conductance of cation and 0 is the molar ionic
conductnace of anion, at infinite dilution. The above equation is, however, correct only for binary
electrolyte like NaCl, MgSO4 etc.
Application of Kohlrausch's law :
(1) Determination of 0m of a weak electrolyte:
In order to calculate 0m of a weak electrolyte say CH3COOH, we determine experimentally 0m values
of the following three strong electrolytes:
(a) A strong electrolyte containing same cation as in the test electrolyte, say HCl
(b) A strong electrolyte containing same anion as in the test electrolyte, say CH3COONa
(c) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl.
0m of CH3COOH is then given as:
0m (CH3COOH) = 0m (HCl) + 0m (CH3COONa) – 0m (NaCl)
Proof:
0
0m (HCl) = H Cl ......................(i)
0
0m (CH3COONa) = CH3COO Na ......................(ii)
0 0
0m (NaCl) = Na Cl ......................(iii)
Adding equation (i) and equation (ii) and subtracting (iii) from them:
0 0 0
0m (HCl) + ( CH 3COONa ) – ( NaCl) (H ) (CH3COO0 ) 0( CH3COOH)
0
ELECTROLYTIC CELL :
Q.12 3A current was passed through an aqueous solution of an unknown salt of Pd for 1Hr. 2.977g of Pd+n
was deposited at cathode. Find n.
Q.13 50mL of 0.1M CuSO4 solution is electrolyzed with a current of 0.965 A for a period of
200 sec. The reactions at electrodes are:
Cathode : Cu2+ + 2e Cu(s) Anode : 2H2O O2 + 4H+ + 4e.
Assuming no change in volume during electrolysis, calculate the molar concentration of Cu2+, H+ and
SO42at the end of electrolysis.
Q.14 A metal is known to form fluoride MF2. When 10A of electricity is passed through a molten salt for 330
sec., 1.95g of metal is deposited. Find the atomic weight of M. What will be the quantity of electricity
required to deposit the same mass of Cu from CuSO4?
Q.15 10g fairly concentrated solution of CuSO4 is electrolyzed using 0.01F of electricity. Calculate:
(a)The weight of resulting solution (b)Equivalents of acid or alkali in the solution.
Q.16 An electric current is passed through electrolytic cells in series one containing Ag(NO3)(aq.) and other
H2SO4(aq). What volume of O2 measured at 250C and 750mm Hg pressure would be liberated from
H2SO4 if
(a) one mole of Ag+ is deposited from AgNO3 solution
(b) 8 x 1022 ions of Ag+ are deposited from AgNO3 solution.
Q.17 Cadmium amalgam is prepared by electrolysis of a solution of CdCl2 using a mercury cathode. How
long should a current of 5A be passed in order to prepare 12% CdHg amalgam on a cathode of 2gm
Hg (Cd=112.4)
Q.18 After electrolysis of NaCl solution with inert electrodes for a certain period of time. 600 mL of the
solution was left. Which was found to be 1N in NaOH. During the same time, 31.75 g of Cu was
deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage yield of
NaOH obtained.
Q.19 Three electrolytic cells A, B, C containing solution of ZnSO4, AgNO3 and CuSO4, respectively are
connected in series. A steady current of 2 ampere was passed through them until 1.08 g of silver deposited
at the cathode of cell B. How long did the current flow? What mass of copper and of zinc were deposited?
Q.20 Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A
constant current of 2 mA was passed for 16 minutes. It was found that after electrolysis the concentration
of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate
in the original solution.
Q.21 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 ampere for 20
mintue. What mass of Ni is deposited at the cathode?
Q.22 A current of 3.7A is passed for 6hrs. between Ni electrodes in 0.5L of 2M solution of Ni(NO3)2.
What will be the molarity of solution at the end of electrolysis?
GALVANIC CELL :
Representation of Cell diagrams, complete and half cell reactions :
Q.23 Make complete cell diagrams of the following cell reactions :
(a) Cd2+ (aq) + Zn (s) Zn2+ (aq) + Cd (s)
(b) +
2Ag (aq) + H2 (g) 2H+ (aq) + 2Ag (s)
(c) Hg2Cl2 (s) + Cu (s) Cu2+ (aq) + 2Cl– (aq) + 2Hg (l)
(d) Cr2O 72 (aq.) + 14H+ (aq) + 6Fe2+ (aq) 6Fe3+ (aq) + 2Cr3+ (aq) + 7H2O (l)
Q.27 Determine the standard reduction potential for the half reaction :
Cl2 + 2e– 2Cl–
Given Pt2+ + 2Cl– Pt + Cl2, E oCell = – 0.15 V
Pt2+ + 2e– Pt E° = 1.20 V
o
Q.28 What is E Cell if
2Cr + 3H2O + 3OCl– 2Cr3+ + 3Cl– + 6OH–
3+
2Cr + 3e – Cr,, E° = – 0.74 V
OCl + H2O + 2e–
– Cl– + 2OH–, E° = 0.94 V
o
G°, E Cell and Keq :
Q.29 Is 1.0 M H+ solution under H2SO4 at 1.0 atm capable of oxidising silver metal in the presence of 1.0 M
Ag+ ion?
o
Eo = 0.80 V, E H |H = 0.0 V
Ag |Ag 2 ( Pt )
o o o
Q.31 If E Fe2 | Fe = – 0.44 V, E Fe3 | Fe 2 = 0.77 V. Calculate E Fe3 | Fe .
o o o
Q.32 If E Cu |Cu = 0.52 V, E Cu 2 |Cu = 0.34 V, what is E Cell of the cell reaction
Cu + Cu2+ 2Cu+?
is cell reaction spontaneous?
Q.33 Calculate the EMF of a Daniel cell when the concentration of ZnSO4 and CuSO4 are 0.001 M and
0.1M respectively. The standard potential of the cell is 1.1V.
Q.34 Calculate the equilibrium constant for the reaction Fe + CuSO4 FeSO4 + Cu at 250C.
Given E0 (Fe/Fe2+) = 0.44V, E0 (Cu/Cu2+) = 0.337V.
Q.35 For a cell Mg(s) | Mg2+(aq) || Ag+ (aq) | Ag, Calculate the equilibrium constant at 250C. Also find the
maximum work that can be obtained by operating the cell.
E0 (Mg2+/Mg) = 2.37V, E0 (Ag+/Ag) = 0.8 V.
Q.36 The standard reduction potential at 250C for the reduction of water
2H2O + 2e H2 + 2OH is 0.8277 volt. Calculate the equilibrium constant for the reaction
2H2O l H3O + OH at 250C.
+
Q.37 At 250C the value of K for the equilibrium Fe3+ + Ag Fe2+ + Ag+ is 0.531mol/litre. The standard
electrode potential for Ag + + e – Ag is 0.799V. What is the standard potential for
Fe3+ + e– Fe2+ ?
Q.38 The EMF of the cell M | Mn+ (0.02M) || H+ (1M) | H2(g) (1 atm), Pt at 250C is 0.81V. Calculate the
valency of the metal if the standard oxidation of the metal is 0.76V.
Q.39 Equinormal Solutions of two weak acids, HA (pKa = 3) and HB (pKa = 5) are each placed in contact
with standard hydrogen electrode at 250C. When a cell is constructed by interconnecting them through
a salt bridge, find the emf of the cell.
Q.40 In two vessels each containing 500ml water, 0.5m mol of aniline (Kb= 109) and 25mmol of HCl are
added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of
cell made by connecting them appropriately.
Q.41 Calculate E0 and E for the cell Sn | Sn2+ (1M) || Pb2+ | Pb(103M), E0 (Sn2+| Sn) = 0.14V,
E0 (Pb2+| Pb) = 0.13V. Is cell representation is correct?
Q.42 At what concentration of Cu2+ in a solution of CuSO4 will the electrode potential be zero at 250C?
Given : E0 (Cu | Cu2+) = 0.34 V.
Q.43 A zinc electrode is placed in a 0.1M solution at 250C. Assuming that the salt is 20% dissociated at this
dilutions calculate the electrode potential. E0 (Zn2+| Zn) = 0.76V.
Q.44 From the standard potentials shown in the following diagram, calculate the potentials E1 and E 2 .
Q.45 For the reaction, 4Al(s) + 3O2 (g) + 6H2 O + 40H– 4 [Al(OH)4– ] ; Ecell = 2.73 V. If
G f (OH ) = –157 kJ mol–1 and G f ( H 2 O) = –237.2 kJ mol–1, determine G f [Al (OH)4–].
Concentration cells :
Q.50 EMF of the cell Zn | ZnSO4 (a1= 0.2) || ZnSO4(a2) | Zn is 0.0088V at 250C. Calculate the value of a2.
CONDUCTANCE
Conductivities and cell constant:
Q.51 The resistance of a conductivity cell filled with 0.01N solution of NaCl is 210 ohm a t 1 8 o C .
Calculate the equivalent conductivity of the solution. The cell constant of the conductivity cell is
0.88 cm1.
Q.52 The molar conductivity of 0.1 M CH3COOH solution is 4.6 S cm2 mole1 . What is the specific
conductivity and resistivity of the solution ?
Q.53 The conductivity of pure water in a conductivity cell with electrodes of cross sectional area 4 cm2
and 2 cm apart is 8 x 107 S cm1.
(i) What is resistance of conductivity cell ?
(ii) What current would flow through the cell under an applied potential difference of 1 volt?
Q.54 Resistivity of 0.1M KCl solution is 213 ohm cm in a conductivity cell. Calculate the cell constant
if its resistance is 330 ohm.
Q.55 Resistance of a 0.1M KCl solution in a conductance cell is 300 ohm and specific conductance of
0.1M KCl is 1.29 x 10-2 ohm-1 cm-1. The resistance of 0.1M NaCl solution in the same cell is 380
ohm. Calculate the equivalent conductance of the 0.1M NaCl solution.
Q.56 For 0.01N KCl, the resistivity 709.22 mho cm. Calculate the conductivity and equivalent
conductance.
Q.57 A solution containing 2.08 g of anhydrous barium chloride is 400 CC of water has a specific
conductivity 0.0058 ohm–1cm–1. What are molar and equivalent conductivities of this solution.
Application of Kohlrausch's law:
Q.58 Equivalent conductance of 0.01 N Na2SO4 solution is 112.4 ohm–1 cm2 eq–1. The equivalent
conductance at infinite dilution is 129.9 ohm–1 cm2. What is the degree of dissociation in 0.01 N
Na2SO4 solution?
Q.59 Specific conductance of a saturated solution of AgBr is 8.486×10–7 ohm–1cm–1 at 250C. Specific
conductance of pure water at 250C is 0.75×10–6 ohm–1 cm–2. m for KBr , AgNO3 and KNO3
are 137.4 , 133 , 131 ( S cm2 mol–1) respectively. Calculate the solubility of AgBr in gm/litre.
Q.60 Saturated solution of AgCl at 250C has specific conductance of 1.12×10–6 ohm–1 cm–1. The
Ag+ and Cl– are 54.3 and 65.5 ohm–1 cm2 / equi. respectively. Calculate the solubility product
of AgCl at 250C.
Q.62 The value of m for HCl, NaCl and CH3CO2Na are 426.1, 126.5 and 91 S cm2 mol–1 respectively..
Calculate the value of m for acetic acid. If the equivalent conductivity of the given acetic acid is 48.15
at 25° C, calculate its degree of dissociation.
Q.63 Calculate the specific conductance of a 0.1 M aqueous solution of NaCl at room temperature,
given that the mobilities of Na + and Cl– ions at this temperature are 4.26×10 –8 and
6.80×10–8 m2 v–1 s–1, respectively.
Q.64 For the strong electroytes NaOH, NaCl and BaCl2 the molar ionic conductivities at infinite dilution
are 248.1×10–4, 126.5 ×10–4 and 280.0 ×10-4 mho cm2 mol–1 respectively. Calculate the molar
conductivity of Ba(OH)2 at infinite dilution.
Q.65 At 25°C, (H+) = 3.4982 ×10 –2 S m2 mol–1 and (OH–) = 1.98 ×10–2 S m2mol–1.
Given: Sp. conductnace = 5.1 ×10–6 S m–1 for H2O, determine pH and Kw.
PROFICIENCY TEST
1. In highly alkaline medium, the anodic process during the electrolytic process is
4OH– O2 + 2H2O + 4e–.
2. Compounds of active metals (Zn, Na, Mg) are reducible by H2 whereas those of noble metals (Cu, Ag,
Au) are not reducible.
EIt
3. The mass of a substance deposited on the cathode or anode during electrolysis is given by w = .
F
4. Faraday’s second law of electrolysis is related to the equivalent, mass of the electrolyte.
5. Equivalent conductance at infinite dilution of salt AB is equal to the sum of equivalent conductances of
ions, A+ and B– at infinite dilution.
6. The standard reduction potential of Cl– | AgCl | Ag half-cell is related to that of Ag+ | Ag half-cell through
the expression E E RT In K (AgCl).
Ag |Ag Cl |AgCl |Ag
SP
F
8. A half-cell reaction is A (x+n) + ne– Ax+. It is possible to determine the value of n from the
measurements of cell potential.
9. In a galvanic cell, the half-cell with higher reduction potential acts as a reducing agent.
10. In an electrode concentration cell, the cel reaction Zn(c1) Zn(c2) will be spontaneous if c1 > c2.
12. All chemical reactions used in galvanic cells are redox reactions.
13. The amount of the product formed by the passage of 1 coulomb of electricity through electrolyte is called
electrochemical equivalent of the substance.
14. The redox reaction involved in galvanic cell is a non- spontaneous process.
16. The specific conductance of a 0.1 N KCl solution at 23°C is 0.012 ohm–1 cm–1. The resistance of the
cell containing the solution at the same temperature was found to be 55 ohms. The cell constant is ____.
18. The electrical conductivity of a solution of acetic acid will _______ if a solution of sodium hydroxide is
added.
21. A cell in which two electrodes of the same metal are dipped in solutions of metal ion of different
concentrations in called___________.
23. During discharge of lead storage battery, the overall reaction is___________.
25. Temperature coefficient and change in enthalpy are related by the expression__________.
26. In salt bridge, the electrolyte used should be _________.
27. In electrochemical cell, the electrical neutrality in two half cells is maintained by _________.
30. Coulomb refers to _______ of electricity while ampere refers to ________ at which it flows.
32. During electrolysis of aqueous solution of CuSO4 using Pt electrodes the product at anode is ______.
33. The quantity of electricity required for complete reduction of 0.5 mole MnO 4 to Mn2+ is ______C.
Q.1 Same quantity of electricity is being used to liberate iodine (at anode) and a metal x (at cathode). The
mass of x deposited is 0.617g and the iodine is completely reduced by 46.3 cc of 0.124M sodium
thiosulphate. Find the equivalent mass of x.
Q.2 The standard reduction potential values, E0(Bi3+|Bi) and E0(Cu2+|Cu) are 0.226V and 0.344V
respectively. A mixture of salts of bismuth and copper at unit concentration each is electrolysed at 250C.
to what value can [Cu2+] be brought down before bismuth starts to deposit, in electrolysis.
Q.3 In a fuel cell, H2 & O2 react to produce electricity. In the process, H2 gas is oxidized at the anode & O2
at the cathode . If 67.2 litre of H2 at STP react in 15 minutes, what is the average current produced ? If
the entire current is used for electrode deposition of Cu from Cu (II) solution, how many grams of Cu
will be deposited?
Anode : H2 + 2OH 2H2O + 2 e– Cathode : O2 + 2 H2O + 4e 4 OH–
Q.4 One of the methods of preparation of per disulphuric acid, H2S2O8, involve electrolytic oxidation of
H 2 SO 4 at anode (2H 2 SO 4 H 2 S 2 O 8 + 2H + + 2e ) with oxygen and hydrogen as
byproducts. In such an electrolysis, 9.722 L of H2 and 2.35 L of O2 were generated at STP. What is
the weight of H2S2O8 formed?
Q.5 During the discharge of a lead storage battery the density of sulphuric acid fell from 1.294 to
1.139 g.ml1. H2SO4 of density 1.294 g mL1 is 39% and that of density 1.39 g mL1 is 20% by
weight. The battery holds 3.5L of acid and the volume practically remains constant during the discharge.
Calculate the number of ampere hours for which the battery must have been used. The discharging
reactions are:
Pb + SO42 PbSO4 + 2e (anode)
+
PbO2 + 4H + SO4 + 2e 2 PbSO4 + 2H2O (cathode)
Q.6 The emf of the cells obtained by combining Zn and Cu electrode of a Daniel cell with N calomel electrode
in two different arrangements are 1.083V and 0.018V respectively at 250C. If the standard reduction
potential of N calomel electrode is 0.28V and that of Zn is 0.76 V, find the emf of Daniel cell.
Q.7 Given t he standard reduction pot entials Tl+ + e TI, E 0 = 0.34V and
TI3+ + 2e TI+, E0 = 1.25V. Examine the spontaneity of the reaction, 3TI+ 2TI + TI3+. Also
0
find E for this disproportionation.
Q.8 The emf of the cell Ag|AgI|KI(0.05M) || AgNO3(0.05M) |Ag is 0.788V. Calculate the solubility product
of AgI.
Q.9 The cell Pt, H2(1 atm) | H+(pH=x) || Normal calomel Electrode has an EMF of 0.67V at 250C. Calculate
the pH of the solution. The oxidation potential of the calomel electrode on hydrogen scale is 0.28 V.
Q.10 Estimate the cell potential of a Daniel cell having 1 M Zn++ & originally having 1 M Cu++ after sufficient
NH3 has been added to the cathode compartment to make NH3 concentration 2 M.
Kf for [Cu(NH3)4]2+ = 1 x 1012, E0 for the reaction,
Zn + Cu2+ Zn2+ + Cu is 1.1 V..
Q.11 Consider the cell Ag|AgBr(s)|Br ||AgCl(s), Ag | Cl at 25º C . The solubility product constants of AgBr
& AgCl are respectively 5 x 1013 & 1 x 1010 . For what ratio of the concentrations of Br & Cl ions
would the emf of the cell be zero ?
Q.12 The pKsp of Agl is 16.07 . If the Eº value for Ag+Ag is 0.7991 V . Find the Eº for the half cell reaction
AgI (s) + e Ag + I.
Q.13 Voltage of the cell Pt, H2 (1 atm)|HOCN (1.3 x 103 M)||Ag+ (0.8 M)|Ag(s) is 0.982 V . Calculate the
Ka for HOCN . Neglect [H+] because of oxidation of H2(g) .
Ag+ + e Ag(s) = 0.8 V..
Q.14 The normal potential of Zn referred to SHE is 0.76V and that of Cu is 0.34V at 250C. When excess of
Zn is added to CuSO4, Zn displaces Cu2+ till equilibrium is reached. What is the ratio of Zn2+ to Cu2+
ions at equilibrium?
Q.15 Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally
contained 0.1M MnO4 and 0.8M H+ and which was treated with 90% of the Fe2+ necessary to reduce
all the MnO4 to Mn+2.
MnO4 + 8H+ + 5e Mn2+ + 4H2O, E0 = 1.51V
Q.16 Kd for complete dissociation of [Ag(NH3)2]+ into Ag+ and 2NH3 is 6 x 108. Calculate E0 for the
following half reaction; Ag(NH3)2+ + e Ag + 2NH3
+
Ag + e Ag, 0
E = 0.799 V
Q.17 The overall formation constant for the reaction of 6 mol of CN with cobalt (II) is
1 x 1019. The standard reduction potential for the reaction
[Co(CN)6]3 + e Co(CN)64 is 0.83 V. Calculate the formation constant of [Co(CN)6]3
Given Co3+ + e Co2+ ; E0 = 1.82 V
Q.18 Calculate the emf of the cell
Pt, H2(1.0 atm) | CH3COOH (0.1M) || NH3(aq, 0.01M) | H2 (1.0 atm),
Pt Ka(CH3COOH) = 1.8 x 105, Kb (NH3) = 1.8 x 105.
Q.19 A current of 3 amp was passed for 2 hour through a solution of CuSO4 ,3 g of Cu2+ ions were deposited
as Cu at cathode. Calculate percentage current efficiency of the process.
Q.20 The Edison storage cell is represented as Fe(s) | FeO(s) | KOH(aq) | Ni2O3 (s) | Ni(s) The halfcell
reaction are
Ni2O3(s) + H2O(i) + 2e 2NiO(s) + 2OH E0 = + 0.40V
FeO(s) + H2O(l) + 2e Fe(s) + 2OH E0 = 0.87V
(i) What is the cell reaction?
(ii) What is the cell e.m.f.? How does it depend on the concentration of KOH?
(iii) What is the maximum amount of electrical energy that can be obtained from one mole of Ni2O3?
Q.21 For the galvanic cell : Ag|AgCl(s)| KCl (0.2M) || K Br (0.001 M)| AgBr(s) | Ag,
Calculate the EMF generated and assign correct polarity to each electrode for a spontaneous process
after taking into account the cell reaction at 250C.
[K sp( AgCl ) 2.8x1010 ;K sp (AgBr ) 3.3x10 13 ]
Q.22 An aqueous solution of NaCl on electrolysis gives H2(g), Cl2(g) and NaOH according to the reaction:
2Cl(aq) + 2H2O 2OH(aq) + H2(g) + Cl2(g)
A direct current of 25 amperes with a current efficiency of 62% is passed through 20 liters of NaCl
solution (20% by weight). Write down the reactions taking place at the anode and the cathode. How
long will it take to produce 1Kg of Cl2? What will be the molarity of the solution with respect to hydroxide
ion? (Assume no loss due to evaporation).
Q.23 An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolyzed until all the copper is deposited.
The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml and the
current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis.
Q.24 In the refining of silver by electrolytic method what will be the weight of 100 gm Ag anode if
5 ampere current is passed for 2 hours? Purity of silver is 95% by weight.
Q.25 Hydrogen peroxide can be prepared by successive reactions:
2NH4HSO4 H2 + (NH4)2S2O8
(NH4)2S2O8 + 2H2O 2NH4HSO4 + H2O2
The first reaction is an electrolytic reaction the second is steam distillation. What amount of current
would have to be used in first reaction to produce enough intermediate to yield 100 gm pure H2O2 per
hour? Assume 50% anode current efficiency.
Q.26 Dal lake has water 8.2×1012 litre approximately. A power reactor produces electricity at the rate of
1.5×106coulomb per second at an appropriate voltage.How many years would it take to electrolyse the
lake?
Q.27 Calculate the potential at 25°C for the cell
Cd | Cd2+ (2.00 M) || Pb2 (0.0010 M) | Pb
Given E°cell = 0.277 V.
Q.28 Calculate E° for the following reactions at 298 K,
Ag ( NH 3 ) 2 + e– Ag + 2NH3
Ag (CN ) 2 + e– Ag + 2CN–
Given: E 0.7991 V, K Ins [Ag ( NH 3 ) 2 ] = 6.02 × 10–8 and K Ins [Ag(CN ) 2 ] = 1.995 ×10–19
Ag |Ag
Q.29 Determine the degree of hydrolysis and hydrolysis constant of aniline hydrochloride in M/32 solution of
salt at 298 K from the following cell data at 298 K.
Pt | H2 (1 atm) | H+(1M) || M/32 C6H5NH3Cl | H2 (1 atm) | Pt ; Ecell= – 0.188 V.
Q.30 The emf of the cell, Pt | H2 (1 atm), | H+ (0.1 M, 30 ml) || Ag+ (0.8 M) | Ag is 0.9 V. Calculate the emf
when 40 ml of 0.05 M NaOH is added.
Q.31 Given, E° = –0.268 V for the Cl– | PbCl2 | Pb couple and – 0.126 V for the Pb2+ | Pb couple, determine
Ksp for PbCl2 at 25°C?
Q.32 The equivalent conductance of 0.10 N solution of MgCl2 is 97.1 mho cm2 equi–1 at 250C. a cell
with electrode that are 1.5 cm2 in surface area and 0.5 cm apart is filled with 0.1 N MgCl2 solution.
How much current will flow when potential difference between the electrodes is 5 volt.
Q.33 A dilute aqueous solution of KCl was placed between two electrodes 10 cm apart, across which a
potential of 6 volt was applied. How far would the K+ ion move in 2 hours at 250C? Ionic
conductance of K+ ion at infinite dilution at 250C is 73.52 ohm–1 cm2 mole–1?
Q.34 When a solution of specific conductance 1.342 ohm–1 metre–1 was placed in a conductivity cell with
parallel electrodes, the resistance was found to be 170.5 ohm. Area of electrodes is 1.86×10–4 m2.
Calculate separation of electrodes.
Q.35 The specific conductance at 250C of a saturated solution of SrSO4 is 1.482×10–4 ohm–1 cm–1while
that of water used is 1.5×10–6 mho cm–1. Determine at 250C the solubility in gm per litre of SrSO4
in water. Molar ionic conductance of Sr2+ and SO42– ions at infinite dilution are 59.46 and
79.8 ohm–1 cm2 mole–1 respectively. [ Sr = 87.6 , S = 32 , O = 16 ]
ZENITH
Q.1 A lead storage cell is discharged which causes the H2SO4 electrolyte to change from a concentration of
34.6 % by weight (density 1.261g ml–1 at 25°C) to 27 % by weight. The original volume of electrolyte
is one litre. Calculate the total charge released at anode of the battery. Note that the water is produced
by the cell reaction as H2SO4 is used up. Over all reaction is
Pb(s) + PbO2(s) + 2H2SO4(l) 2PbSO4(s) + 2H2O(l)
Q.2 Assume that impure copper contains only iron, silver and a gold as impurities. After passage of 140 A, for
482.5s of the mass of the anode decreased by 22.260g and the cathode increased in mass by 22.011 g.
Estimate the % iron and % copper originally present.
Q.3 100ml CuSO4(aq) was electrolyzed using inert electrodes by passing 0.965 A till the pH of the resulting
solution was 1. The solution after electrolysis was neutralized, treated with excess KI and titrated with
0.04M Na2S2O3. Volume of Na2S2O3 required was 35 ml. Assuming no volume change during
electrolysis, calculate:
(a) duration of electrolysis if current efficiency is 80% (b) initial concentration (M) of CuSO4.
Q.5 Calculate the equlibrium concentrations of all ions in an ideal solution prepared by mixing 25.00 mL of
0.100M Tl+ with 25.00mL of 0.200M Co3+.
E0 ( Tl+ /Tl3+ )= –1.25 V ; E0 (Co3+/Co2+ ) = 1.84 V
Q.12 At 25°C, Hf (H2O,l) = –56700 cal / mol and energy of ionization of H2O (l) = 19050 cal/mol. What
will be the reversible EMF at 25°C of the cell,
Pt | H2(g) (1 atm) | H+ || OH– | O2(g) (1 atm) | Pt, if at 26°C the emf increas by 0.001158 V.
Q.13 Calculate the cell potential of a cell having reaction: Ag2S + 2e– 2Ag + S2– in a solution buffered at
pH = 3 and which is also saturated with 0.1 M H2S.
For H2S : K1 = 10–8 and K2 = 1.1 × 10–13, Ksp(Ag2S) = 2 × 10–49, EAg / Ag 0.8.
Q.14 Calculate the solubility and solubility product of Co2 [Fe(CN)6] in water at 250C from the following
data:
Conductivity of a saturated solution of Co2[Fe(CN)6] is 2.06 × 10–6 –1 cm–1 and that of water used
4.1 × 10–7–1 cm–1 . The ionic molar conductivities of Co2+ and Fe(CN)64– are 86.0 –1 cm2 mol–1
and 444.0 –1 cm–1mol–1.
Q.15 A sample of water from a large swimming pool has a resistance of 9200 at 25°C when placed in a
certain conductance cell. When filled with 0.02 M KCl solution, the cell has a resistance of 85 at
25°C. 500 gm of NaCl were dissolved in the pool, which was throughly stirred. A sample of this solution
gave a resistance of 7600 . Calculate the volume of water in the pool.
Given : Molar conductance of NaCl at that concentration is 126.5 –1 cm2 mol–1 and molar conductivity
of KCl at 0.02 M is 138 –1 cm2 mol–1.
SUHANA SAFAR
OBJECTIVE
Q.1 A dilute aqueous solution at Na2SO4 is electrolysed using platinum electrodes.The products at the anode
and cathode are
(A) O2, H2 (B) S2O82 , Na (C) O2, Na (D) S 2 O 82 , H2
Q.2 The standard reduction potentials of Cu2+/ Cu and Cu2+ / Cu+ are 0.337 and 0.153 V respectively. The
standard electrode potential of Cu+ / Cu half cell is
(A) 0.184 V (B) 0.827 V (C) 0.521 V (D) 0.490 V
Q.4 The standard reduction potential values of the three metallic cations X, Y, Z are 0.52, –3.03, and
–1.18 V respectively. The order of reducing power of the corresponding metals is
(A) Y > Z > X (B) X > Y > Z (C) Z > Y > X (D) Z > X > Y
Q.5 A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y– and 1 M Z– at 25°C. If the
reduction potential of Z > Y > X, then
(A) Y will oxidise X and not Z (B) Y will oxidise Z and X
(C) Y will oxidise both X and Z (D) Y will reduce both X and Z.
Q.6 For the electrochemical cell, M | M+ || X– | X, E° (M+/M) = 0.44 V and E° (X/X–) = 0.33V. From this
data , one can deduce that
(A) M + X M+ + X– is the spontaneous reaction
(B) M+ + X– M + X is the spontaneous reaction
(C) Ecell= 0.77 V
(D) Ecell= –0.77 V
Q.8 The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is
(A) LiCl > NaCl > KCl (B) KCl > NaCl > LiCl
(C) NaCl > KCl > LiCl (D) LiCl > KCl > NaCl
Q.9 Saturated solution of KNO3 is used to make salt bridge because
(A) velocity of K+ is greater than that of NO3
Q.10 Standard electrode potential data are useful for understanding the suitablilty of an oxidant in a redox
titration. Some half cell reactions and their standard potentials are given below:
MnO 4 (aq) + 8H+(aq) + 5e– Mn2+ (aq) + 4H2O (l); E° = 1.51 V
Cr2O 72 (aq) + 14 H+ (aq) + 6e– 2Cr3+ (aq) +7H2O (l); E° = 1.38 V
Fe3+ (aq) + e– Fe2+ (aq); E° = 0.77 V
Cl2 (g) + 2e – 2Cl– (aq); E° = 1.40 V
Identify the only incorrect statement regarding quantitative estimation of aqueous Fe(NO3)2
(A) MnO 4 can be used in aqueous HCl
Q.12 Zn | Zn2+ (a = 0.1M) || Fe2+ (a = 0.01M)|Fe. The emf of the above cell is 0.2905 V. Equilibrium
constant for the cell reaction is
(A) 100.32/0.0591 (B) 100.32/0.0295
(C) 100.26/0.0295 (D) e0.32/0.295
SUBJECTIVE
Q.14 The standard reduction potential for Cu2+ / Cu is 0.34 V. Calculate the reduction potential at
pH = 14 for the above couple. Ksp of Cu(OH)2 is 1 x 1019.
Q.15 Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2
as per the reaction, Mn2+aq + 2H2O MnO2(s) + 2H+aq + H2(g)
Passing a current of 27A for 24 hours gives one kg of MnO2. What is the value of current efficiency?
Write the reaction taking place at the cathode and at the anode.
Q.16 How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing
silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 Amperes? What is the area of
the tray if the thickness of the silver plating is 0.00254 cm? Density of silver is 10.5 g/cm3.
(b) 6.593 × 10–2 g of metallic Zn (amu = 65.39) was added to 100 ml of saturated solution of AgCl.
Calculate log10 Zn
2
, given that
Ag 2
Q.1 (a) 6.02 × 1022 electrons lost, (b) 1.89 × 1022 electrons gained, (c) (b) 1.80 × 1023 electrons gained
Q.2 (a) 0.75 F, (b) 0.69 F, (c)1.1 F Q.3 (i) 54 gm, (ii) 16.35 gm
Q.4 5
1.023 × 10 sec Q.5 1.12 mol, 12.535 litre
Q.6 0.112 litre Q.7 0.24 gms
Q.8 Rs. 0.75x Q.9 (i) 2.1554 gm ; (ii) 1336. 15 sec
Q.10 115800C, 347.4 kJ Q.11 t = 193 sec
Q.12 n=4 Q.13 Cu2+ = 0.08M, H+ = 0.04M, SO24 = 0.1M
Q.14 A = 114, Q = 5926.8C Q.15 Final weight = 9.6g,0.01Eq of acid
Q.16 (a) V(O2)=6.2L, (b)V(O2) = 0.824L Q.17 t = 93.65 sec.
Q.18 60 % Q.19 (i) 482.5 sec (ii) 0.3175 gm (iii) 0.327 gm
Q.20 7.958 ×10 M –5 Q.21 1.825 g Q.22 2M
Q.23 2+ 2+ + + 2+ –
(a) Zn | Zn | | Cd | Cd, (b) Pt, H2 | H | | Ag | Ag , (c) Cu | Cu | | Cl | Hg2Cl2 | Hg
(d) Pt | Fe2+, Fe3+ | | Cr2O 72 , Cr3+ | Pt
Q.24 (a) 2Ag + Cu2+ 2Ag+ + Cu, (b) MnO 4 + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O
(c) 2Cl– + 2Ag+ 2Ag + Cl2, (d) H2 + Cd2+ Cd + 2H+
Q.25 Anode Cathode
(a) Zn | Zn2+ H+, H2 | Pt
(b) Pt | Sn2+, Sn4+ Fe3+, Fe2+ | Pt
(c) Pt | Fe2+, Fe3+ MnO 4 , Mn2+ | Pt
(d) Pb | Pb2+ Br2, Br– | Pt ]
Q.26 1.61 V Q.27 1.35 V Q.28 1.68 V
Q.29 – 0.80 V, NO Q.30 0.53 V, disproportionation Q.31 – 0.0367 V
o
Q.32 E Cell = – 0.36 V, not spontaneous Q.33 E =1.159V
Q.34 Kc = 2.18 x 10 26 107 0
Q.35 Kc = 1.864 x 10 , G = – 611.8 kJ
Q.36 Kw 10 14 Q.37 E0 = 0.7826 V Q.38 n = 2
Q.39 E = 0.059 Q.40 E = 0.395 V
Q.41 E0cell = +0.01V, Ecell = 0.0785V, correct representation is Pb|Pb2+ (103M)||Sn2+(1M)|Sn
Q.42 [Cu2+] = 2.97 x 1012M for E =0 Q.43 E = 0.81eV
Q.44 0.52 V, 0.61 V 3
Q.45 –1.30 ×10 kJ mol –1
1. T 2. F 3. T 4. T 5. T
6. F 7. T 8. T 9. F 10. T
11. T 12. T 13. T 14. F 15. F
16. 0.66 cm–1 17. O2 18. increase 19. higher
20. O2 & H2 21. Electrolyte concentration cell
22. Cr2O 72 (aq.), Cr3+(aq.), H+ | Pt
23. Pb(s) + PbO2(s) + 2H2SO4 2PbSO4(s) + 2H2O (l)
24. Hg2Cl2(s) + 2e– 2Hg(l) + 2Cl–(aq.)
dE
25. H = nF T E 26. inert, i.e., should not interfere with net cell reaction
dT
27. Salt bridge or porous partition 28. zero 29. E°red cathode
30. Amount, rate 31. reduction process 32. Oxygen
33. 2.5 × 96500 C
MIDDLE GAME
ZENITH
Q.1 1.21 × 105 coulomb Q.2 Cu = 98.88%, Fe = 0.85% Q.3 1250 s, 0.064 M
Q.4 K = 10268 Q.5 Tl+ = 10–8; Co3+ = 2 × 10–8 Q.6 1.66V
Q.7 –1.188V Q.8 10–2 Q.9 5.24 x 1016
Q.10 1.143V Q.11 (a) –0.124 V, (b) 7.1, (c) calomel electrode
Q.12 0.4414 V Q.13 – 0.167 V Q.14 KSP = 7.682 × 10–17
Q.15 2 × 105 dm3
SUHANA SAFAR
OBJECTIVE
SUBJECTIVE
Q.14 E0 = 0.22N
Q.15 = 94.8%; Cathode : 2H+ + 2e H2, Anode : Mn2+ Mn4+ + 2e
Q.16 WAg = 272.2g, area =1.02 x104 cm2 Q.17 Kc = 7.6 x 1012 Q.18 KC = 6.26 x 107