Electrochemistry 47

Previous Years' CBSE Board Questions

3.1 Electrochemical Cells State:

(i) Which one of its electrodes is negatively
SAL (2 marks) charged.
(ii) The reaction taking place at each of its
1Define electrochemical cell. What happens if
external potential applied becomes greater than
electrode.
(ii) The carriers of current within this cell.
Ecell of electrochemical cell? (2/5, AI 2016)
(Delhi 2008)
3.2 Galvanic Cells SAII (3 marks)
VSA (1 mark) 8. A cell is prepared by dipping copper rod in 1 M

copper sulphate solution and zinc rod in

1 M
2. Represent the galvanic cell in which the
zinc sulphate solution. The standard reduction
reaction
potential of copper and zinc are 0.34 V and
Zn +Cuag)Znag) + Cu takes place.
-0.76 V respectively.
(1/3, Delhi 2013C)
i) What will be the cell reaction?
What is the necessity to use a salt bridge in a (ii) What will be the standard electromotive
Galvanic cell? (Delhi 2011C) force of the cell?
(ii) Which electrode will be positive?
SAI (2 marks)
(Delhi 2011C)
On the basis of standard electrode potential
values stated for acid solutions, predict whether 9. Depict the galvanic cell in which the reaction
Zn t 2Agag) Zna)+2Ago takes place.
Ti species may be used to oxidise Fe" to Fe
Further show :
Reactions:
T +eTi"; +0.01
(i) Which of the electrode is negatively

Fe+eFe*; +0.77 (AI 2007) charged?

(ii) The carriers of the current in the cel.
Two half-reactions of an electrochemical cell
(i) Individual reaction at each electrode.
are given below:
(Delhi 2010C)
MnO4taa) +8H(ag) + 5e> Mn(ag) + 4H,O,E
1.51V
=+
3.3 Nernst Equation
ag)Sn (ag) t 2e , E + 0.15 V
=

Sn
Construct the redox equation from the standard S A (2 marks)
potential of the cell and predict if the reaction 10 Calculate A,G° for the reaction:
is reactant favoured or product favoured. Mg) + Cuau Mgaug) + Cug)
(Delhi 2011, AI 2010, 2009) 96500 C mol
( Given Feell +2.71 V, 1F
= =

Given that the standard electrode potential (E°) (2/3,AI 2014)

of metals are constant (K) for the given cell
K/K = -2.93 V, Ag'/Ag = 0.80 V, Equilibrium
Cu"/Cu =0.34 V,
Mg"/Mg = -2.37 V, Cr"/Cr =-0.74 V,
reaction is 10. Calculate E cell
A+ Ba) A ag) t+ B (
(2/3,Foreign 2014)
Fe/Fe -0.44 V. for
these metals in an increasing order of 12. The standard electrode potential (E)
Arrange AG° for the
their reducing power (AI 2010) Daniell cell is +1.1 V. Calculate the

Formulate the galvanic cell in which the reaction.

7.
following reaction takes place:
Zn+Cu(ag) Znag) +Cu (AI 2013)
Zn)t 2Agiaq) Zn {ug) t 2Ag)
(1 F= 96500 C mol )
 w t CBSE Chapterwise-Topicwise Chemistry
48
13. The standard electrode potential for Daniell cell 0.25 V and'IAU.60
is 1.1 V.Calculatethestandard Gibbs energy for
(3/5, Delhi 2015C)
the cell reaction
cell potential of the
(F 96.500 C mol ) (Delhi 201C) 22Calculate the standard
the following reaction
14 A zinc rod isdipped in 0.1 M solution of ZnSO, galvanic cell in which
The salt is 95% dissociated at this dilution at 298 takesplace:
Fetau)t Agla) Feay) t Ag
Cakulate the eetde potential constant of
Calculate the A,G" and equilibrium
(Delhi 2012C)
n-0.76 V] the reaction also,
Determine the value of equilibrium constant
e20.77 V
Ag'Ag =0.80 V;eFe
(K)and AG for the following rcactions (3/5, Delhi 2015C)
N+2Aging) Ni (a)+ 2Ag, E" =1.05 V the emf of following cell at 298
K:
F 96500 C mol') 23 Calculate
(Delhi 2011, Foreign 2011) Mg Mg (0.1 M)|| Cu"(0.01 M)|Cumol ]
Given: Eel 2.71 V. 1 F 96500 C
= + =

16 Calculate the emf for the given cell at

25° C: (3/5, Delhi 2014)

CriCr(0.1 M) Fe" (0.01 M)| Fe

|| Estimate the minimum potential difference
24.
GIven : needed to reduce Al,0, at 500°C. The Gibbs
= -0.74 V, E2 0.44V
energy change for the decomposition reaction

(2/5, Delhi 2009C) 2 4

Al,O,Al+0, is 960 kJ.
(F 96500C mol') (3/5,Delhi 2014C)
SAU (3 marks)
17. Calculate e.m.f. of the following cell at 298 K 25. Calculate the emfofthe following cell at 298 K:
FeFe(0.001 M)||H"(1 M)|H2( (1 bar), Pt
2Cr3Fe"(0.1M) -> 2Cr"(0.01M) +3Fe +0.44 V) (Delhi 2013)
Given: Ecr|C= -0.74 V,EGpe2"|fe) =-0.44 V (GivenEedl =

(Delhi 2016) 26. Calculate the emf of the following cell at 25°C
reaction at
AgalAg(10 M)||Cu" (10 M)|Cu
Calculate Eel for the following Given: Ecell = +0.46V and log 10" = n.
298 K (AI 2013)
2A 3Cu"(0.0IM) 2Al"(0.01M)+3Cu used in watches, the
Given E l l = 1.98 V (3/5,AI 2016) 27. In the button cell, widely
following reaction takes place.
19. Calculate emf of thefollowing cell at 25°C: ntAg.O + H,O- Znu+ 2AS
Fe Fe (0.001 M)||H (0.01 M)|H(I bar)|Pt, +20H
EFFe) -0.44 V, E(H'|H,) 0.00 V
= =

Determine E and AC° for the reaction.

(Delhi 2015) iven: agIAg= +0.80 V, EznZn = -076 Y
20. Calculate the emf of the following cell at 25°C (3/5, Delhi 2012)

ZnZn (0.01 M) ||H'(001M)| Hg (I bar)| Pl, 28. A voltaic cell is set up at 25°C with the following
-0.76 V,t =
0.00 V half cells:
-976 , Al/AI (0.001 M) and Ni/Ni" (0.50 M)
(oreign 2015) Write an equation for the reaction that occurs
21. For the cell reaction hen the cell generates an electrice current and

Ni Nia|AgulAg) determine the cell potential.

Calculate the equilibrium constant at 25°C.
Ni IN -

0.25 V and EAPAI =

- 1.66 V.
How much maximum work would be obtained (log8x 10 = - 5.09) (3/5, AI 2012, 2011,
by operation of this cell? 3/5, Foreign 2011, 3/5, Delhi 200
Electrochemistry 49
29. The cell in which the following reaction occurs: Calculate the cell voltage
2Fe) t 21(ag) 2Fe (ag) t l2 NIN =-0.25 V, EALA1.66 V
has Eel = 0.236 V at 298 K, Calculate the
standard Gibbs encrgy and the (3/5, Delhi 2009)
equilibrium
constant of the cell reaction. 36. Calculate the equilibrium constant for the
(Antilog of6.5 3.162x 10; of8.0 reaction
10x 10:of 8.5 =3.162 x 10") Feg +Cdan) Feag)+ Cd
Given:
(Delhi 2012C)
0. Calculate the potential for half-cell containing FcaICd-0.40 V,Epe21Ee0.44 V
0.10 M K,.Cr,O-ag), 0.20 M Cr" (a) and (3/5, Delhi 2009, 2009C, 3/5, AI 2009)
1.0 x 10
MHgThe halfcell reaction is 37. One half-cell in a voltaic cell is constructed
CrOa)+14H ()+ 6e> 2Cr(ag) t 7H,0% from a silver wire dipped in silver nitrate
and the standard electrode potential is given as solution of unknown concentration. Its other
P = 1.33 V. (3/5, AI 2011) half-cell consists of a zinc electrode dipping in
31. For the cell 1.0 M solution of Zn(NO,)2. A voltage of 1.48 V
is measured for this cel. Use this information
Zn Zn (2 M) || Cu" (0.5 M) | Cu
to calculate the concentration of silver nitrate
(a) Write equation for each half-reaction.
solution used.
(b) Calculate the cell potential at 25°C
Given: EZ2Z-0.76 V, EAe'lAe =+0.80 VV]
(3/5, Delhi 2009)
Ezn21Z=-0.76 V;E2C +0.34V
38. Calculate the standard cell potential of a
(Delhi 2011C) galvanic cell in which the following reaction
32. Calculate the equilibrium constant, K for the
takesplace:
reaction at 298 K, 2C+3Cdta)2Ca)+ 3Cd
Zn +Cua) Znag)+Cu Calculate A,G° and equilibrium constant, K of
Given: AG° = - 212.300 kJ mol the above reaction at 25°C.
Given
-0.76 V; C2Ca+0.34 V
Zn 1Zn -0.74 V, 2tFe -0.44 V
(AI 2011C) cICr
(F= 96,500 C mol)
33. A copper-silver cell is set up. The copper ion (Delhi 2008C)
concentration is 0.10 M. The concentration 39. (i) Formulate the electrochemical cell
of silver ion is not known. The cell potential representingthe reaction;
when measured was 0.422 V. Determine the 2Cr +3Fean) 2Crau) +3Fet
concentration of silver ions in the cell. (is) Calculate E cell
Given (ii) Calculate Eoei at 25°Cif
= +0.80 V, E21C =+0.34 V Cr]=0.1 M and [Fe*" =0.01 M
Ag/Ag Given:
(3/5, Delhi 2010, AI 2009)
E
Cr/Cr
= -0.74 V,E2Fe = -0.44 V
34. A voltaic cell is set up at 25° with the following
(3/5, AI 2008C)
halfcells:
Ag (0.001 M) | Ag and Cu" (0.10 M) | Cu LA (5 marks)
What would be the voltage of this cell?
40. Calculate e.m.fand AG for the following cell
(Eell=0.46V) (AI 2009)
Mg| Mg (0.001 M)||Cu" (0.0001 M)| Cu,
35. A voltaic cell is set up at 25°C with the following
F(M /Mg) = -2.37 V,cucu)
= +0.34 V
half-cells:
(AI 2015)
Al| AN (0.001 M) and Ni |Ni" (0.50 M)
 50 M.tGCBSE Chapterwise-Topicwise Chemish
41. Calculate the standard electrode
potential of SAI(2 marks)
INi electrode if emf of the cell
52. Define the term degree ofdissociation. Write a
NiNi (0.01 M)||Cu (0.IM)|Cu, is 0.059V. expression that relates the molar conductivity
a weak electrolyte to its degree of dissociation
Given E2 +0.34 V] (Delhi 2009C)
(2/5, Delhi 2015
42. Calculate the cell emf and A,G° for the cell
53. Define conductivity and molar conductivit
reaction at 25°C
for the solution of an electrolyte. Discuss theji
Zn Zn (0.1 M) || Cd* (0.01 M)| Cd
yariation with concentration. (2/5, AI 2015C)
Given
ZnZn-0.763 V, Ecd?*cH-0.403 V 54 State Kohlrausch law of independent migratior
Zn
of ions. Why does the conductivity of a solutio

IF 96,500 C mol, R =8.314J K molr

decrease with dilution? (A12014)
(AI 2009C) 55. Define the terms conductivity and mola
conductivity for the solution of an electrolyte,
3.4 Conductance of Electrolytic Comment on their variation with temperature
(Delhi 2014C)
Solutions 56. The resistance of0.01 M NaCl solution at 25°Cis
200 2. The cell constant of the conductivity cell
VSAA mark)
used is unity. Calculate the molar conductivity
43Define limiting molar conductivity. Why of the solution. (2/3, AI 2014C)
conductivity of an electrolyte solution
57. Define conductivity and molar conductivity
decreases with the decrease in concentration?
for the solution of an electrolyte. Discuss their
(1/2, Delhi 2015)
variation with change in temperature.
Sate Kohlrausch's law ofindependent migration (AI 2014C)
of ions. Write its one application. 58. The conductivity of 0.20 M-solution of KCd
(1/2, Foreign 2015) at 298 K is 0.025 S cm Calculate its molar
conductivity. (Delhi 2013, 2008, AI 2007)
45. Define the following term:
Molar conductivity (A,,) (1/5, Delhi 2015C) 59. The conductivity of 0.001 M acetic acid is
46. Define the following term:
4x 10 S/cm. Calculate the dissociation
constant of acetic acid, if molar conductivity at
Kohlrauschs law of independent migration of
infinite dilution for acetic acid is 390 S cm/mol
1ons. (1/5, Delhi 2015C)
(2/3, Delhi 2013C, 2012C)
47. Define the following term: Limiting molar 60. Express the relation among cell constant,
conductivity (1/5, Delhi 2014) resistance of the solution in the cell and
48. State and explain Kohlrausch's law. conductivity of the solution. How is molar
(1/3, Delhi 2013C) conductivity of a solution related to is
49. Express the relation between conductivity and conductivity? (AI 2012, 2010, 2/5,Delhi 2009)
molar conductivity ofa solution held in a cel1? 61. The molar conductivity of a 1.5 M solution o
(Delhi 2011)
an electrolyte is found to be 138.9 S cm mol
Calculate the conductivity of this solution.
50. Express the relation among the conductivity (AI
2012, 20100)
of solution in the cell, the cell constant and the
62. The resistance of a conductivity cell containing
resistance of solution in the cell. (Delhi 2011) 0.001 M KCl solution at 298 K is 1500 2. Whal
51. Express the relation between the conductivity is the cell constant if conductivity of 0.001
and the molar conductivity of a solution. KCl solution at 298 K is 0.146 x 10 S cm
(AI 2008) (Delhi 201OC, 2009C, 2008, 2007
 Electrochemistry 51

63. Define molar conductivity of a substance and 72. When a certain conductance cell was filled with
describe how for weak and strong electrolytes, 0.1 M KCI, it has a resistance of85 ohms at 25°C.
molar conductivity changes with concentration When the same cell was filled with an aqueous
of solute. How is such change explained? solution of 0.052 M unknown electrolyte,
(2/5, Delhi 2009) the resistance was 96 ohms. Calculate the
molar conductance of the electrolyte at this
64. Define the term molar conductivity. How is it
concentration.
related to conductivity of the related solution?
[Specific conductance of 0.1 M KC
(2/5, Delhi 2009) = 1.29 x
10 ohm cm
65. State Kohlrausch'slawofindependent migration (Al2012CJ
of ions. How can the degree of dissociation of 73. Calculate the degree of dissociation of acetic
acetic acid in a solution be calculated from its acid at 298 K, given that
Am(CH,CoOH) = 11.7 S cm 2 mol
molar conductivity data? (2/5, AI 2008C)
A°m(CH,COO) = 49.9 S cm mol
66. Explain with examples the terms weak and
AmH) = 349.1 S cm mol(Delhi 2011C)
strong electrolytes. (Delhi 2007)
74. The resistance of a conductivity cell when filled
SAI (3marks) with 0.05 M solution of an electrolyte X is
mol Lsolution 100 ohms at 40°C. The same conductivity cell
h e conductivity of 0.001 filled with 0.01 M solution of electrolyte Y has
of CH,COOH is 3.905 x 10S cm.
a resistance of 50 ohms. The
Calculate its molar conductivity and degree of conductivity of
0.05 M solution 1.0 x 10S
ofelectrolyte Xis cm
dissociation (a).
Calculate
Given (H) = 349.6 S cm mol and
i) Cell constant
A°(CH,CO0) = 40.9 S cm mol
(i) Conductivity of 0.01 M Ysolution
(3/5, AI 2016) (ii) Molar conductivity of 0.01 M Y solution
(3/5, AI 2008C)
68. The conductivity of 0-20 mol L solution of
KCI is 2.48 x 10S cm
Calculate its molar
LA |(5 marks)
conductivity and degree of dissociation (a).
75. (a) Define molar conductivity ofa solution and
Given (K") = 73.5 S cm' mol
explain how molar conductivity changes
and" (CI) = 76.5 S cm mol". (AI 2015) with change in concentration of solution
69. Resistance of a conductivity cell filled with for a weak and a strong electrolyte.
0.I molL KClsolution is 1002. Ifthe resistance (b) The resistance of a conductivity cell
ofthe same cell when filled with 0.02 mol L KCI containing 0.001 M KCl solution at 298 K
solution is 5202, calculate the conductivity and is 1500 2. What is the cell constant if the
molar conductivity of0.02 molL KCl solution. conductivity of 0.001 M KCl solution at
The conductivity of 0.1 mol L' KCI solution is 298 K is 0.146 x 10d S cm? (AI 2012)
1.29x 10 Q' cm. (3/5, AI 2014) 76. (a) State Kohlrausch's law of independent
migration of ions. Write an expression for
70Ahe value ofAmofAl;(SO,), is 858 S cm mol, the molar conductivity of acetic acid at
while so; is 160Scm mol calculatethe infinite dilution according to Kohlrausch's
limiting ionic conductivity of Al*. law.
(AI 2013C) Calculate Am for acetic acid.
(b)
71. The electrical resistance of a column of 0.05 M Given that
NaOH solution of diameter 1 cm and length A m(HCI) = 426 S cm mol
50 cm is 5.55 x 10' ohm. Calculate its resistivity. Am(NaCl) = 126 S cm' mol
conductivity and molar conductivity. A,mCH,COONa) = 91 S cm mol
(AI 2012) (Delhi 2010)
 MtG CBSE Chapterwise- Topicwise Chemisty
85. Predict the products of electrolysis in each of
77Conductivity of 0.00241 M acetic acid is
7.896 x 10 S cm', Caleulate its molar the following:
An aqueous solution of AgNO, with
conductivity if A"for acetic acid is (i)
platinum electrodes.
390.5 S cm mol . What is its dissociation
of H,SO, with
(Delhi, Al 2008) (ii) An aqueous solution
constant?
platinum electrodes. (2/5, Delhi 2014C)
3.5 Electrolytic Cells and Electrolysis 86,How much electricity
in terms of Faradays
is required to produce
20 g of calcium from
VSA (1 mark) molten CaCl,? (2/3, Delhi 2013C)
occur at cathode during the
78, Following reactions 87. Silver is uniformly electrodeposited
on a

electrolysis of aqueous silver chloride solution: metallic vessel of surtace area of 900 cm by
Agapte Agt E +0.80V current of 0.5 ampere for 2 hours,
passing a
Calculate the thickness of silver deposited
H H2(), E =0.00 V 10.5 g cm and
Given the density of silver is
atomic mass of Ag = 108 amu. (2/3, AI 2013C)
On the basis of their standard reduction
electrode potential (E°) values, which reaction How many coulombs are required to reduce
88.
is feasible at the cathode and why? I mole Cr,0 to Cr*? (2/3, Delhi 2012C)
(1/2, Delhi 2015)
89. How many moles of mercury will be produced
reduction
79. How much charge is required for the solution with
of 1 mol of Zn to Zn? (Delhi 2015) by electrolysing 1.0 M Hg(NO,)2
a current of 2.00 A for 3 hours? (2/5, AI 201H
80. Following reactions occur at cathode during
90. A of CuSO4 is electrolysed for
solution
the electrolysis of aqueous copper (II) chloride
10 minutes with a current of 1.5 amperes. What
solution :
E = +0.34 V is the mass of copper deposited at the cathode?
Cu (aa) +2e > Cui)
(AI 2009)
E° = 0.00 V
Hu)te ,2(g) 91. Explain why electrolysis of aqueous solution
On the basis of their standard
reduction of NaCl gives H at cathode and Cl, at anode.
electrode potential (E°') values, which reaction Write overall reaction.
is feasible at the cathode and why? Given
(1/2, Foreign 2015) -0.83 V,
Nat INa-2.71 V, Eu,o /H,
=

8State the Faraday's first law of electrolysis

(Delhi 2015C) = +1.23V
ECL,C+.36
Cl,/C
V,
Ht/H,/tH,O
82. How much charge is required for the reduction
(2/5, Delhi 2009C)
of I mole of Cus" to Cu? (Delhi 2007)
92. Consider the reaction:
S A (2 marks) Cr,0+14H + 6e >2Cr +8H,O
83. State Faraday's first law of electrolysis.
How What is the quantity of electricity in coulombs
2008)
much charge in terms of Faraday is required for needed to reduce I mol of Cr,0; ? (AI
the reduction of I mol of Cu to Cu. 93. Explain why electrolysis of an aqueous
solution
(2/5, Delhi 2014) anode.
of NaCi at cathode and Cl, at
gives H,
between
84. A solution of Ni(NO,), is electrolysed -2.71 V, = -0.83 V

platinum electrodes using

a current of FNa' INa =
0/H,
What mass of nickel
5.0 ampere for 20 minutes. = +1.36 V,
will be deposited at the cathode? a, /2C
= +1.23 V
(Given: At. mass Ni 58.7 g mol",
of =
E
IF=96500 Cmol') 2H0,/H,O (2/5, Delhi 2008C)
(Foreign 2014)
 55
Electrochemistry

DetailedSolutionsS
8. (i) The cell reactions are
1. The device which converts the chemical energy
liberated during the chemical reactionto Zn Zn(a) + 2e (Anode)
electrical energy is called electrochemical cell. Cua)+2e >Cu, (Cathode)
If external potential applied becomes greater than Net reaction:
Zn) + Cua) Znag) + Cu)
Ecll of electrochemical cell then the cell behaves
as an electrolytic cell and the direction of
flow of (ii) cell =E°right -Eeh =0.34 V-(-0.76 V) =1.10V
which
current is reversed. (ii) Copper electrode will be positive on
2. Representation df the galvanic cell for the given reduction takes place.
reaction is 9. The reaction is
Zn | Zna ll Cua) Cu Zn +2Ag(ag) Zn{an) + 2AS
Anode Salt Cathode Cell can be represented as
bridge
Zn | Znal| Aglay| Ag
3. The salt bridge allows the movement of
Flow of
ions from one solution to the other without mixing Electrons
Current
of the two solutions. Moreover, it helps to maintain Znelectrode Ag clectrod

of the solutions in the two (Anode) KCl salt bridge (Cathode)

the electrical neutrality
half cells.
Because standard electrode potential of
g
Ti /T: is less than that of Fe*/Fe*" so, it cannot Zn NO solution
soiution
Reduction
oxidise Fe" to Fe" Oxidation
Z n ) Z n a ) t 2e Agag)teAg()
5. At anode Sna) : Sn(a) +2e] x5 (i) The zinc electrode is negatively charged (anode)
At cathode: MnO4(ag) +8H (ag)+ 5e as it pushes the electrons into the
external circuit.
the cel.
Mnag)+4H,0] x 2 (1i) lons are the current carriers within
i ) The reactions occurring at two electrodes
are:
Net cell reaction:
2MnO(a) +5Sna)+ 16H(ag)
At zinc electrode (anode): Zng Zn{ay) 2e

2Mnaa)+5Sn(aa) +8H,0) At silver electrode (cathode): Agia) + e Ag)

1.36 V. 10. (a) Given: Eell =2.71 V
cel-E cathode E anode = 1.51 V
-0.15 V =

the reaction
Since, cell potential is positive therefore For the reaction,
2+ +
is product favoured. Mg+Cujay) Mg(aa) Cu
The reducing power increases with decreasing n 2,A,G° = ?
6.
value of electrode potential. Hence,
the order is
Using formula, A,G° nFEell =-

Ag< Cu< Fe < Cr <

Mg <K. A,G=-2 x 96500 Cmol'x2.71
VV
7. The cell reaction is
or A,G 523.03 kJ mol"
Zn +2Agag Zn(a)+2Ag
The cell is represented as
11. A+ Bia) Aag)+ B
Here, n =2
Zn Zniag)l| Agan) electrode
Ag will be negatively using formula,
(1 Anode i.e., zinc
0.059.
charged. Ecell - log K
(ii) At anode n

Zn (ag) 2e
Zn"ta) t (oxidation)
Atcathode: EP
E cell log 10
Aglag) t e Ag, (Reduction)
(111) lons are the carriers of current within the cell. Ecell =0.0295 V
 56 M.tGCBSE Chapterwise-Topicwise Chemist
12. Here n = 2, E°cell= 1.1 V
F 96500 C mol 1.98V Ecell 0.059 og 10
6
A,G = -nFEel
A,G = - 2 x 1.1 x 96500 - 212300 J mol
1.98 V E cell 0.0592
6
-212.3 k) mol
13. Refer to answer 12. cell 1.98 + 0.05911.99 V
6
14. The electrode reaction written as reduction
19. The cell reaction is
reaction is
Fe +2H(ag) Fe(an) t H2(
Zn2e Zn (n =2) 0.44 VV
Applying Nernst equation, we get, Ecell 0.00 (-0.44) =

= Ez* 1Zn
9.0510 1
2
FccllFcell
0.0591[Fe*]
2 HP
As 0.1 M Zn$O, solution is 95% dissociated, this 0.0591 0.001
means that in the solution, 0.44- log
2 (0.01)
95
Zn= x0.1 =0.095 M = 0.44 0.02955 =0.41045 V
100
Ezn1Zn -
0.76

= - 0.76 - 0.02955 (log 1000 - log 95)

-

0.0-oS n095
2
1 20.

Zn+2Han)
The cell reaction is

Zna) +H2(
0.765V
E cell = 0.00 -

(-0.76) =

= - 0.76 -0.0295 (3 -1.9777) = 0.79021 V

15. Nig + 2Aga) Niay) + 2Ag, E = 1.05 V Ecell=Ecell
0591Zn*]
log
2 IH*
Here, n =2
nE cell 0.0591 (0.001)
Using formula, log K
= 0.76
2
log
0.059 (0.01)
= 0.76 - 0.02955 = 0.730 V
2x1.05
or log K 35.5932
0.059 21. At anode : Ni Ni + 2e

K, =
antilog 35.5932 or K = 3.92 x 10* Atcathode [Ag +e : Ag] x 2
Again, AG° = -nFE°cell Cell reaction:Ni + 2Ag-> Ni** +2Ag
AG =-2 x 96500 x 1.05 = - 202650 J E cell= E cathodeE anode
AG = -202.65 kJ Agt/Ag-ENt/Ni 0.80 V (-0.25) = -

16. Ecell = Eright-E left-0.44- (-0.74) = 0.30 V Ecell = 1.05 V

Eell 0.30-
0.0591.
-10gg
6
(0.1) Ecell O.0log K n
(0.01)
X1.05 x2
= 0.30 0.0394 0.26 V
logK. cell
-
=

0.0591 0.0591
17. Refer to answer 16.
logk = 35.53
18. Given cell,
2Al+ 3Cu (0.01 M)-2Al"(0.01 M) +3Cu) antilog 35.53 3.38 x10
+0.80 V -0.77 V +0.03 V
Fcell 1.98 V, Fel ? = 22. Ecell = =

Using Nernst equation at 298 K A,G=-nFEcell -1 x 96500 x 0.03 =

= -2895 J mol= -2.895 k] mol

Ecell Ecell
0.0591
-log- JA AG -2.303 RT logK
ICu - 2895 = -2.303 x 8.314 x 298 x log K

0.059log 10 * or log K, = 0.5074

1.98 V E cell- 3.22
K= Antilog (0.5074)
=
6 [103
 Electrochemistry 51
23. The cell reaction can be represented as
(10
M g +C u ) Mgag) t Cu) cell0.46 V -0.0295 log
10
Given: Fell+2.71 V, T 298 K = 0.46 0.0295 log 100.46 -0.0295 (-5)
Accordingto the Nernst cquation:
= 0.46 + 0.0295x 5
=

0.6075 V

Egell=Eel
0.0591
log-
Mgiay) 27. Thecell reaction in button cell:
1
[Cu{ag) Zn+Ag.O1,+H,O Znn)*2Ag +20Hap)
0.0591 0.1 (i) Calculation of Ecll
= 2.71 = 2.6805 V
Reactions :
Anode:Zn Zn(a)t 2e
24. Al-O; (2A1 + 3 0 ) > 2Al+ O , n = 6e Cathode:
AgO+H,0t 2e 2Agi, + 20H(au) + 2e
AlO, - Al+O, n = x 6e = 4e n 2

AG = 960 x 1000 = 960000 Eell= cathode Eanode= EAg,O/AgE Zn2+1n

= +0.80 - (-0.76) V = 1.56 V
Now. AG =-nFEceli (ii) Calculation of A,G°
AG -960000
Ecel AG= -nFE cell
nF 4x965000 - 2 x 96500 C mol' x 1.56 V
Ecell=-2.487V 301080 CV mol
difference needed to reduce
Minimum potential 301080 J mol= - 301 kJ mol

AlO, is -2.487 V. 28. At anode : Al>Al(a) + 3e"] x2

25. Fe|Fe*"(0.001 M)||H'(1 M)|H2(1 bar)| Pt 2e>Ni(l x 3
At cathode: Nis +

Reactions
node: Fe, Feaa)+2e 2e
Cell reaction:2Al +3Ni(a)2Alay) + 3Ni)
the above cell reaction,
Applying Nernst equation to
Cathode: 2H(a) 2e t
0.0591 A
Cell reaction: Fe t+ 2H(aq) > Feaa) + H2( Ecell cell 2x3
n = 2.

Using Nernst equation at 298 K Now, Ecell =

Ey/NiEa/AI
= - 0.25 V-(-1.66) = 1.41 V
0.0591 Fe]x PH
log
log (10
Eell Ecell 0.0591
2 H* Fcell =1.41 V-
6
For the given cell,
(0.5)
- E pFe =1.41 V - .0591 1og (8 x 10)
EelE cathode anode 6
0-(-0.44) =+0.44 V =1.41 V- 0.0591
Given (Fe] 0.001 M; {H"] =1 M; P1,
=
=
1 bar
6
Putting in Nernst equation = 1.41V + 0.050 V = 1.46 V
0.0011
Ecell=9.44 0.0295 log 29. 2Fe" +2e" 2Fe2 and 21 >I, + 2e
2
Hence, for the given cell reaction, n =

= 0.44 0.0295 log 10 A,G°=-nFE cel=-2x96500x0.236=-45.55kJ

mol
= 0.44 [(0.0295) x (-3) A,G-2.303 RTlog K,
0.44+0.0885 = 0.53V
log K, =
A,
or
26. The cell may be represented as 2.303RT

Ag lAg (10 M)||Cu (10 M)|Cu, -45.55 kJ mol = 7.983

0.0591 AgT 2.303x8.314x10 kJ Kmolx 298K

Using formula Eel = E cell 10g
ICu K =
Antilog (7.983)
9.616 10 = x
 M.tCBSE Chapterwise-Topicwise Chemisth
nisty
58

30. For half cell reaction,

FelEell-
0.0591 oCu"]
Cr,Olag) +14H (ag) 6e 2Cr(a)7H,O 2 AgP
0.0591
log
C - 0.0591,
0.1
EellE cell or, 0.422 V 0.46 V
2 Ag'
Given, Ecell 1.33 V, n =6, |Cr"]
=

0.2 M
0.038 V = - 0.0295 log
0.1
ICr,O:1-0.1 M.|H'| 1 x 10 " M Ag"
0.059 (0.20) 0.1 -0.038
= 1.288
Ecell 1.3.3V-
6 (0.1) (10) or, log Ag' -0.0295

0.0591
= 1.33V- log (4 x1055 0.1
antilog 1.288 = 19.41
6 or,

IAg
D.059 1log 4+ log 10 0.1
= 1.33 V -
Ag' _19.41= =5.1519 x 10
0.0591 Ag'= 7.1 x 10 M
= 1.33 V log 4 + 55 log 10
6 34. Refer to answer 26.
0.0591 35. Refer to answer 28.
= 1.33 V [0.602 + 55]
6 36. FcellEathode Eanode=-0.40V-(-0.44V) =0.04V
= 1.33 V - 0.548 V= 0.782 V
nE
Using formula, log K, =F at 298 K
31. (a) Oxidation halfreaction: 0.0591
Zn Znjag) t 2e 2x0.04 V
Reduction halfreaction: or, K=antilog
0.0591 V
Cuag)+2e Cu or, K = antilog 1.356 = 22.38
(b) Eel=0.34 (-0.76) =1.10 V
37. The cell may be represented as

E 0.0591Zn
el2 og(Cu Zn Zna(1 M)||Ag(ag)lAg)
Ecell=Ecathode Fanode 0.80 V = -

(-0.76 V) =
1.56 V
0.0591 106 2 Using formula,
Eell = 1.10 -

2
Eell E cell 0.0591Zn
=

= 1.10 -

= 1.10 -
o.051o5
0.059
x0.6021
2

or,
2

1.48 = 1.56 -
0.059
[Agt

= 1.10
2
0.0177 1.0823 V
or, log[Ag]= - 1.354
2
-Ag
or, Ag']= antilog(- 1.354)
32. AG°= -RT In K, = 2.303 RT log Ke
or, [Ag']=4.426 x10 M
-

- 212300 2.303 x 8.314 x 298 x

log Ke
212300
38. T 273 + 25°C =
298 K and n
=6
cell Ecathode E anode = -0.40 - ( - 0 . 7 4 ) =0.34 V
or
log 2.303x 8.314x298 37.2074
A,G=-nFE cell=-6x96500 x 0.34 =-196860Jmol
K, =Antilog 37.2074 1.6x 10" Again A,G= - 2.303 RTlog K
33. The given cell may be represented as 196860 2.303 8.314 298
= x
logK
-
-

x x

Cug Cu" (0.10 M)|| Ag (C)| Ag log K= 34.5014

cell = Ecathode- Fanode =0.80 V - 0.34 V = 0.46V
K antilog 34.5014 3.172 x 10*
 Electrochemistry 59
39. (i) The cell can be represented as 42. Fcellcathodeanode
CrlCran) | Fejg)| Feto =-0.403-(-0.763)= - 0.403 +0.763 =0.360 V
(ii) Ecell = Ecathode- anode

= - 0.44-(-0.74) = -0.44 +0.74 0.30 VN

Ecellcell
2.303RT,
-1og
Znag)
(ii) Refer to answer 16.
40. Mg Mg (0.001 M)|| Cu*(0.0001 M) | Cu)
nF
ca
Reactions
Anode: Mg Mg(ag) t 2e
Ecell= E cel 0.059log
Znan)
Cathode: Cu ag)t 2c Cu
n
cda
Net cell reaction: Mg( + Cua) Mglag) + Cu) 2.303RT
n = 2
= 0.059
F
Using Nernst equation:
0.059 0.059
= 0.36-o o 1 0 . 3 6 - l o g 1 0
Eel=Ecel- 2.303RT[Mg
log 2 0.01
nF Cu = 0.36 0.0295 = 0.3305 = 0.33 V

For the given cell A , = - nFE ell

cellEcathodeEanodeEcu2/CuEMg2+/Mg = - 2 x 96500 x 0.36 = - 69480 J mol
= 0.34 V -(-2.37 V) =2.71 V
Given [Mg"] = 0.001 M, [Cu*] =0.0001 M 69.48 kJ mol

Putting in Nernst equation at 298 K 43. The limiting conductivity of an

molar
0.059 0.001 electrolyte is defined as its molar conductivity when
Eell 2.71 V-
2
log 0.0001 the concentration of the electrolyte in the solution
2.71 -0.03 2.68 V
Ecell 2.71 0.0295 log 10
-
=
approaches zero.

A,G=- nFEcell Conductivity of an electrolyte decreases with dilution

= - 2 x 96500 C mol'x2.68 because the number of current carrying particles
- 517,240 J mol = - 517.24 kJ mol
i.e., ions present per cm' of the solution becomes less
41. We have and less on dilution.

0.059, Na 44. Kohlrausch's law of independent migration

Ee= E cel n log2 of ions: It states that limiting molar conductivity of
be represented the sum of the
an electrolyte can as

0.0590.01 Here n = 2 individual contributions of the anion and cation of

0.059 E cell
0.1 the electrolyte.
0.059 oe If ANat and A°ar are limiting molar conductivities
0.059 E cell then
of the sodium and chloride ions respectively
chloride
0.059 = Ecell - 0.059
-log 10) the limiting molar conductivity for sodium
is given by
0.059
0.059 E cell
2
A (NaC)A Na* A°ar of
Kohlrausch's law helps in the calculation of degree
0.059
Ecell = 0.059
dissociation of weak electrolyte like acetic acid.
2
0.059 The degree of dissociation a is given by
. cell = 0.0295 = 0.03
2
Now Ecll E cathode anode
=

0.03 0.34
m is the
E anode is the molar conductivity
and A°m
anode 0.34 0.03 =0.31 V
where A
Hence, E N2/N= +0.31 V limiting molar conductivity.
60 MtG CBSE Chapterwise-Topicwise Chemisth
45. Molar Conductivity : Molar conductivity of Variation of conductivity and molar conductivit
a solution at a dilution V is the conductance of all with concentration: Conductivityalways decreas
the ions with decrease in concentration, for both weak and
produced frommole of theone electrolyte
strong electrolytes. Because the number of ions Pe
dissolved in V'cm ofthe solution when the electrodes
are one cm apart and the area of the electrodes is so unit volume that carry the current in a solution
large that the whole solution is contained between decreases on dilution,.
them.
Ap=V 400 CH,COOH
It units is S cm'mol

46. Refer to answer 44. 200

47. Refer to answer 43. KCI
0 0.2 0.4
48. Refer to answer 44.
Cl (mol/L)/2
KX 1000 Molar conductivity vs C for acetic acid (weak electrolyte) and
49. A i n CGS units potassium chloride (strong electrolyte) in aqueous solutions
M
KX10 Molar conductivity increases with decrease in
in SI units concentration.Because that total volume, V, of
M
where K is the conductivity, M is the molar solution containing one mole of electrolyte also
concentration and A,, is molar conductivity. increases. It has been found that decrease in K on
dilution of a solution is more than compensated by
50. R A increase in its volume.
54. Refer to answers 43 and 44.
where K is the conductivity R is resistance and /A is 55. Refer to answer 53.
the cell constant.
Variation of conductivity and molar conductivity
51. Refer to answer 49. with temperature : Both increase with increase in
52. The fraction of the total number of molecules temperature as degree of ionisation increases.
present in solution as ions is known as degree of
dissociation.
56. Conductivity (K) =
xG=x1
R 200
Molar conductivity () a°m 5x 10
=
=
2 cm
where m is the molar conductivity at infinite Molar conductivity (Am
dilution.
Kx10005x10 x1000 = 500 2 cm mol
53. The
reciprocal of resistivity is known as specific M 0.01
conductance or simply conductivity. lt is denoted by
K (kappa). 57. Refer to answer 55.
58. Here, K = 0.025 S cm', Molarity = 0.20 M
K= or K = Gx
a Molar conductivity A =KX1000

Hence, conductivity of a solution is defined as Molarity

the conductance of a conductor of I cm 0.025x1000
length =
125 S cnm' mol
and having I sq. cm as the area of cross section. 0.20
Alternatively, it may be defined as conductance of 59. C= 0.001 M, K = 4 x
10 S cm,
one centimetre cube of the solution of the
electrolyte. A390Scm/mol
Molar conductivity ofa solution at a dilution Vis the KX1000
conductance of all the ions
produced from 1 mole C
of the electrolyte dissolved in V cm' of the Substituting the values,
solution.
It is represented by
A 4x10 x 1000
AKV A = 40 S cm'/mol
0.001
Electrochemistry 61

40 64. Refer to answers 45 und 49.

0.10250
390 65. Refer lo answer 14.

6. Weak electrolytes: The electrolytes which ares

CHCOOH H,CO0 H
not completely dissocialed into ions in solution are
called weak electrolytes eg., CH,COOH, NH,OH,
ICH,COO H1 HCN, etc.

CH,COOH Strong electrolytes: The electrolytes which are

(l-a) 1 -
completely dissociated into ions in solution are called
0.0010.103) 1.061x10
=1.18 x 10
strong electrolytes. e.g., HCI, KCI, NaOH, NaCI, etc.
l-0.103) 0.897
KX1000
Acter to ü1sWCTS 49 and 50.
67. Using formula, A3 C
KX 1000 Given K = 3.905 x 10 S cm
61. =
M C= 0.001 mol L
138.9x1.5
0.20835 Scm 3.905x10x1000 39.05 S cm' mol
1000 1000
0.001
62 Here. conductivity (k) = 0.146 x 10 S cm
The degree of dissociation,
resistance (R) = 1500 Q
39.05
Cell constant = Conductivity = 0.1

Conductance 390.5
= Conductivity x Resistance A = 349.6 + 40.9 = 390.5 S cm'mol'

68. Given : Conductivity, K = 0.0248 S cm

= xR ''conductance =
resistance. Molarity, C =0.20 M =0.20 mol L
= 0.146x 10 x 1500 =
0.219 cm Using formula, A = 1000 XK
63. Strong electrolyte: The molar conductivity of C
strong electrolyte decreases slightly with the (1000cm L)x(0.0248 S cm)
increase in concentration. This increase is due to (0.20 mol L)
increase in attraction as a result of greater number =
124 S cm' mol
of ions per unit volume.With dilution the ions are
ar apart, interionic attractions become weaker and 124
= 0.82
conductance increases. A, 73.5+ 76.5
Weak electrolyte : When the concentration of
69. Resistance of 0.1 M KCI solution R = 100 2
weak electrolyte becomes very low, its degree of
10nisation rises sharply. There is sharp increase in Conductivity K =1.29 Sm
the number of ions in the solution. Hence the molar Cell constant G = Kx R= 1.29 x 100 = 129 m

conductivity of a weak electrolyte rises steeply at low Resistance of 0.02 M KCI solution. R = 520 2

Concent ration. 129

Conductivity, K = cellconstant m

R 520 2
=0.248 S m'
Strong
clectrolyte Concentration, C=0.02 mol I.

1000 x 0.02 mol m 20 mol m

Weak 0.248 S m
electrolyte Molar conductivity, ,
C 20 mol m
= 0.0124 S m* mol
Concentration
62 MtGCBSE Chapterwise-1opicwise Chemi
70. Am Al(SO,), =
2°,AP + 3A°,S0 74. For electrolyte X:
858 =2mA" +3 x 160 Molarity = 0.05 M
resistance = 100 ohms
858-480
Al*= = 189 S cm* mol conductivity = 1.0 x 104S cm
2
71. Given: Diameter = I em, length = 50 cnm For electrolyte Y:
Molarity =0.01 M
R 5.5 x 10' ohm, M = 0.05 M
resistance= 50 ohms
p=? K=? A =? conductivity = ?
Area of the column,
Cell constant, G
= conductivity (K) x resistance (R)
nr =3.14 x = 100 x 1 x 10 = 10 cm
Conductivity of solution Y is
Resistivity,
3.14 cm
2 102 = 0.02 x 10=2x 10Scm
P=R5.5x 10 ohmx
4x 50 cmm R 50
KX1000
86.35 ohm cm Molar conductance, A,= M
2x10x1000
Again, conductivity, K= 20 ohm
=
cm mol
0.01
75. (a) Refer to answers 45 and 63.
=

86.35
1.158 x104 ohm cm (b) Refer to answer 62.
andmolar conductivity, 10 76. (a) Refer to answer 44.
Am=K*M M A(CH,COOH) = ^Ht + ACH,CO0
(b) A, (HC) = AH* * Aar
= 1.158 x 10 ohm cmx 103
5x10-2 Ap(NaCI) =Nat +cr
=
231.6 ohm' cm mol Am(CH,COONa) ACH,coo+ ANa =

72. K = 1.29 x 10 ohm cm A(CH,COOH) = \cCH,CO0 tAH*

K xCell constant =H*+aar+H,CO0-+A Nat-Aar -ANat

R =
AHCI) + Am(CH,COONa) Am NaCl -

Cellconstant=KxR=1.29 Smx85=109.65 m = 426 +91 -

126 =
391 S cm mol
For second solution, 77. Conductivity of acetic acid,
K = x Cell constant = 6 O x 109.65 m K 7.896x 10° S cm
R 96 2 A mfor acetic acid = 390.5 S cm mol.
=
1.142 2 m KX1000
Kx1000 1.142 2mx1000 cm Molar conductivity, Am
Molarity
Am M 0.052 7.896x10x1000 =
32.76 S cm mol
0.00241
1.142 2cmx10x1000cm
m Degree of dissociation,
0.052 mol 32.76
8.4 x 102
= 219.61 S cm mol m 390.5
73. According to Kohlrauch's law, Dissociation constant of acetic acid,
A°CH,COOH = 2°CH,COO +1°H
K,
Co?
(0.00241)x(8.4x102) = 1.86 x 10
Degree of dissociation, o. = 1-a 1-0.084
78. The species that get reduced at cathode is the
11.7 S cm'mol having higher
11.7
3x10-2
one value of standard reduction a
(49.9+349.1) S cm mol 390 potential. Hence, the reaction that will occur
cathode is Ag(ag) te Ag(
 Electrochemistry 63
79. Znag) + 2e Zn OH > OH e
mo
4OH 2 1 , 0 + O, + 4e
(ii) At the cathode, the following reduction reaction
One mole of Zn requires 2 moles ot clectrons for
reduction i.e.
occurs to produce H, gas.
Q 2x F 2x 96500 19300 C
80. The species that get reduced at cathode is the
At the anode, the following processes are possible
one which have higher value of standard reduction
2H,0 Oty + 4Ha) t 4e; E= + 1.23 V ..)
potential. Hence. the reaction that will occur at
cathode is 250ag) S,0ktan) +2e : E 1.96 V
= + ...ii)
Cu 2c For dilute sulphuric acid, reaction (i) is preferred to
Cu produce O, gas but for concentrated sulphuric acid,
SI. Faraday's first law of electrolysis :
During (ii) occurs.
electrolvsis the amount of any substance deposited
or liberated at an electrode is 86. Reaction for production of Ca from molten
directly proportional
to the quantity of electricity passed through the CaCl
electrolyte i.. CaCl2 >Ca*+2C1
Ca +2e Ca
waQor Wxx t Q=Ixt 2F 40 g
w=ZxIxt Electricity required to produce 40 g = 2 F
where. Z is a constant of proportionality known
. Electricityrequiredto produce 20g=0.5 x2 F = 1 F
as eiectrochemical equivalent of the substance
87. Calculation of mass of Ag deposited
deposited.
82. The electrode reaction is Cu" +2e Cu Theelectrodereaction is Ag'+e Ag
The quantity of electricity passed = Current x Time
Quantity of charge required for reduction of
mole of Cu = 2F = 2 x 96500 = 193000 C
= 0.5 (amp)x 2 x 60 x 60 (sec) =
3600 C
From the electrode reaction, it is clear that 96500 C
83. Refer to answers 81 and 82. of electricity deposit Ag = 108 g
84. Given Current I = 5 A; t = 20 x 60 s, w =? 3600 C of electricity will deposit Ag
Q=Ixt=5 x 20 x 60 =6000 C 108
x 3600 = 4.03 g
Reaction for deposition of Ni, 96500
N 2e
Ni Calculation of thickness:
2 moi mol Let the thickness of deposit be x cm
296500C 58.7 g Mass=volume x density = Area x thickness x density
Thus, 2x 96500 C of electricity produces 58.7 g Ni [ volume = area x thickness]
6000 C of electricity would produce 4.03 g= 900 (cm) x x (cm) x 10.5 (g cm)
58.7x6000 4.03
- = 1.825g cm = 4.26 x 10" cm.
2x96500 900x10.5
85. i) At cathode : The following reduction 88. The given reaction is
reactions compete to take place at the cathode. Cr,O+14H' + 6e 2Cr + 7H,0
e A g a iE = 0.80 V one mole Cr^0; requires 6 mol of electrons for
ABa reduction. Hence, quantity of electricity required
H H E =
0.00 V = 6 mol x 96500 C mol= 5.79 x 10 coulomb

The reaction with a higher value of E takes place at 89. Quantity of electricity passed
the cathode. Therefore, the deposition of silver will Q= Ixt= 2.0 A x 3 x 60 x 60s = 21600 C

take place at the cathode. Hg+2e Hg

Since, Pt electrodes are inert, the anode is not 2F I mol

attacked by NO, ions. Therefore, OH or NO, ions 2 x 96500C electricity produces I mole Hg
21600
can be oxidized at the anode. But OH ions having 21600 C will produce =
a lower discharge potential get preference and 2x 96500
= 0.112 mole of Hg
decompose to liberate OD