Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

Chemistry Class 12

Created by: Mohit Rajpoot

Class 12 : Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

The document Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes is a part of the Class 12 Course Chemistry Class 12.
All you need of Class 12 at this link: Class 12

Solved Example

Ex.1 The reaction : Zn2+  (aq) + 2e- → Zn (s) has a electrode potential of - 0.76 V. This means-
(A) Zn cannot replace hydrogen from acids
(B) Zn is reducing agent
(C) Zn is oxidizing agent
(D) Zn2+  is a reducing agent
Sol. (B) Negative electrode potential shows that Zn2+  is difficult to be reduced and therefore, Zn acts as reducing agent.

Ex.2 Certain quantity of current is passed through 2V connected in series and containing XSO4(aq) and Y2SO4(aq) respectively. If the atomic masses of X and Y are in the ratio of 2 : 1 the ratio of the masses of Y liberated to that of X is :
(A) 1 : 1 
(B) 1 : 2 
(C) 2 : 1 
(D) 3 : 2
Sol. (A) X2+ + 2e- → X ; Y+ +  e-→ Y

2 mol e- produce X = 1 mol = 1 × M g

2 mol e- produce Y = 2 mol = 2 × M/2 = M g

Hence, ratio of the masses of Y:X is M:M or 1:1

Ex.3 The equivalent conductivities at infinite dilution of the cation and the anion of a salt A2B are 140 and 80 ohm-1 cm2 eq-1 respectively. The equivalent conductivity of the salt at infinite dilution is- 
(A) 160 ohm-1 cm2 eq-1 
(B) 220 ohm-1 cm2 eq-1
(C) 60 ohm-1 cm2 eq-1 
(D) 360 ohm-1 cm2 eq-1
Sol. (B)Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes (A2B) = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes (A ) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes (B2-)

= 140 + 80 = 220 ohm-1 cm2 eq-1

Ex.4 The specific conductance of a 0.20 mol L-1 solution of an electrolyte at 20ºC is 2.48 x 10-4 ohm-1 cm-1. The molar conductivity of the solution is -
(A) 1.24 ohm-1 cm2 mol-1  
(B) 4.96 ohm-1 cm2 mol-1
(C) 1.24 ohm-1 cm2 
(D) 4.96 ohm-1 cm2
Sol. (A) Lm = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

= 1.24 ohm-1 cm2 mol-1
Ex.5 When an electric current is passed through acidulated, water, 112 mL of hydrogen gas at N.T.P. collects at the cathode in 965 seconds. The current passed, in amperes, is-
(A) 1.0 
(B) 0.5 
(C) 0.1 
(D) 2.0
Sol. (A) 22,400 mL of hydrogen at STP(or NTP)=2g
Therefore,112 mL of hydrogen at
STP = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 10-2 g
Therefore,2H  2e- → H
2F 1 mol

= 2 × 96,500 C = 2 g, 2 g hydrogen is deposited by 2 × 96,500 C

Therefore,10-2 g hydrogen will be deposited by = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 965 C

Q = i × t = 965 = i × 965
= i = 1

Ex.6 The charge required to deposit 40.5 g of Al (atomic mass = 27.0 g) from the fused Al2 (SO4)3 is-
(A) 4.34 × 105
(B) 43.4 × 105
(C) 1.44 × 105 
(D) None of these
Sol. (A) Al3+ + 3e- → Al
3F 1 mol = 27.0 g
to deposite 27g required charge =3×96,500 C
Therefore,to deposite 40.4g required charge
= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 4.34 × 10C

Ex.7 The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 0.3 g of nickel was deposited in the first cell, the amount of chromium deposited is (At. wt. Ni= 59, Cr=52)
(A) 0.1 g 
(B) 0.176 g 
(C) 0.3 g 
(D) 0.6 g
Sol. (B)Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
For Ni2+  and Cr3+ , we have : Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes mCr = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.176 g

Ex.8 Electrolytic conduction differs from metallic conduction. In case of metallic conduction -
(A) The resistance increases with increasing temperature
(B) The resistance decreases with increasing temperature
(C) The flow of current does not generate heat
(D) The resistance is independent of the length of electrolytic conductor
Sol. (A) With increase in temperature vibration of Kernal (Cation) increases and therefore, conduction decreases and hence, resistance of the metallic conductor increases.

Ex.9 Three faraday of electricity is passed through molten solutions of AgNO3, NiSO4 and CrCl3 kept in three vessels using inert electrodes. The ratio in mol in which the metals Ag, Ni and Cr will be deposited is-
(A) 1 : 2 : 3 
(B) 3 : 2 : 1 
(C) 6 : 3 : 2 
(D) 2 : 3 : 6
Sol. (C)
(i) Ag (aq) + e- → Ag (s)

1 mol = 1F          1 mol

3F         =              3 mol
(ii)
Ni2+ (aq) + 2e- → Ni (s)

2 mol = 2F           1 mol

3 F =3/2 mol
(iii)
Cr3+ (aq) + 3e- → Cr(s)

3 mol = 3F           1 mol

The required ratio of moles of Ag, Ni and Cr is :

3 mol Ag : 3/2 mol Ni : 1 mol Cr

or 6 mol Ag : 3 mol Ni : 2 mol Cr. 

Ex.10 In the reaction :
4 Fe + 3O2 → 4 Fe3+ + 6O2-
which of the following statements is correct
(A) A redox reaction 
(B) O2 is reducing agent
(C) Fe3+  is an oxidizing agent 
(D) Fe is reduced to Fe3+
Sol. (A) 

In this reaction, Fe is oxidized to Fe3+  and O2 is reduced to O2_.

Ex.11 Calculate current strength in ampere required to deposit 10 g Zn in 2hrs. At wt. of Zn = 65.
Sol. Q w = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes; Therefore,i = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes i = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 4.12 ampere

Ex.12 How many hour are required for a current of 3.0 ampere to decompose 18 g water.
Sol. H2O → H2 O2Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Therefore,Eq. of H2O = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesEquivalent weight of H2O = 18/2 as two electrons are used for
1 mole H2O to decompose in H2 and O2.
Therefore,Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Therefore,t = 64333.3 sec = 1072.2 minute = 17.87 hr

Ex.13 Calculate the Avogadro's number using the charge on the electron 1.60 × 10-19 C and the fact that 96500 C deposits 107.9 g silver from its solution.
Sol. Q 96500 coulomb deposits 107.9 g Ag
Here eq. wt = Atomic weight
because Ag is monovalent. Thus 96500 coulomb charge means charge on N electrons
where N in Av. no.
Thus N × e = 96500
N = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
= 6.03 × 1023 electrons

Ex.14 Calculate the volume of gases liberated at anode and cathode at NTP from the electrolysis of Na2SO4(aq.) solution by a current of 2 ampere passed
for 10 minute.
Sol. At cathode : 2H2O + 2e → H2 + 2OH-
At anode : 2H2O → 4H+ + 4e- + O2
Therefore,At anode Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 8 Therefore,Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.0995 g
At NTP : Volume of O2 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.0696 litre
Similarly at cathode Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.0124 g
At NTP : Volume of H2 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.139 litre

Ex.15 Fused Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 ampere for 20 minute. What mass of Ni is deposited at the cathode ?
Sol. Eq. of Ni deposited

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
0.0622
or wNi = 0.0622 × 58.71/2 = 1.825 g
(Ni2+ + 2e → Ni)

Ex.16 A current of 3.7 ampere is passed for 6 hrs. between Ni electrodes in 0.5
litre of 2M solution of Ni(NO3)2. What will be the molarity of solution at the
end of electrolysis ?
Sol. The electrolysis of Ni(NO3)2 in presence of Ni electrode will bring in following changes :
At anode : Ni → Ni2+ + 2e
At cathode : Ni2+  + 2e → Ni

Eq. of Ni2+  formed = Eq. of Ni2+  lost
Thus, there will be no change in conc. of Ni(NO3)2 solution during electrolysis i.e. ,
It will remain 2M

Ex.17 How long a current of 3 ampere has to be passed through a solution of AgNO3 to coat a metal surface of 80 cm2 with a thickness of 0.005 mm ?Density of Ag is 10.5 g cm-3
Sol. Given , i = 3 ampere
Also Volume covered by
Ag = 80 × 0.005 × 10-1 cm3 = 0.04 cm3
Therefore,Weight of Ag used = 0.04 × 10.5 g
wAg = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

0.04 × 10.5 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Therefore,t = 125.09 sec.

Ex.18 Calculate e.m.f of half cells given below :
(a) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes EºOP = 0 V
(b) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes EºOP = 0.44 V
(c) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes EºOP =- 1.36 V
Sol. (a) H2 → 2H+ +  2e
Therefore,EOP = EºOP- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes logPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes [H  = 0.02 × 2M]
= 0- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.100 V
(b) Fe → Fe 2e Therefore,EOP = EºOP- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog [Fe]

= 0.44- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog[0.2] Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.4606 volt.
(c) 2Cl- → Cl2 2e Therefore,EOP = EºOP- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

=-1.36- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=- 1.49 volt

Ex.19 If the oxidation of oxalic acid by acidified MnO4- solution were carried out in a reversible cell, what would be the electrode reaction ? Also calculate the equilibrium constant of the reaction. Given

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 1.51 V and Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =- 0.49 V.

Sol.Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 1.51 V  Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =- 1.51 V
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =- 0.49 V Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.49 V
More is EºOP, more is the tendency to get oxidise

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes → 2CO2 2e; EºOP = 0.49 v Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes 8H  5e → Mn 4H2O; EºRP = 1.51 V

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes 16H  → 10CO2 2Mn  8H2O; n = 10 Q 10 electron are used in redox change.
Therefore,EºCell = EºOPRP = 0.49 1.51 = 2.0 V Also Eº = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes log KC
Therefore,2 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog KTherefore,KC = 10338.98

Ex.20 The e.m.f. of cell, Ag|AgI(s), 0.05 MKI || 0.05 M AgNO3 | Ag, is 0.788 V.

Sol. Ksp of AgI = [Ag ] [I-] = [Ag ] [0.05]
For given cell ECell = EOPAg ERPAg ….(i)
= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes-Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog [Ag ]L.H.S.Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog[Ag ]R.H.S. ECell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes 0.788 = logPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes Therefore,[Ag ]L.H.S. = 2.203 × 10-15 
By equation (i) Ksp = [2.203 × 10-15] [0.05] Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 1.10 × 10-16

Ex.21 Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in 2.0 MFe and 0.02 M Fe solution. Given Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.771 V.
Sol. The half cell reaction is : Fe3+ + e → Fe(or take Fe2+  → Fe3+  + e)
Thus Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Therefore,Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 0.771 + Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.771 + Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog 10-2 = 0.653 V

Ex.22 An electrochemical cell has two half cell reactions as, A 2e- → A; E° = 0.34 V, X → X2 + 2e- ; E° = 2.37 V. The cell voltage will be  
(A) 2.71 V 
(B) 2.03 V 
(C) - 2.71 V 
(D) - 2.03 V
Sol. (A) Ecell = 0.34 2.37 = 2.71 V

Ex.23 Eº of some oxidants are given as :
I2 + 2e → 2I- , Eº = 0.54 V
MnO4- 8H+ +  5e → Mn2+ + 4H2O,  Eº= 1.52V
Fe3+ + e → Fe2+  ,Eº = 0.77V
Sn4+  + 2e → Sn2+  Eº = 0.1 V
(a) Select the strongest reductant and oxidant in these.
(b) Select the weakest reductant and oxidant in these.
(c) Select the spontaneous reaction from the changes given below :
(i) Sn4+  + 2Fe2+  → Sn2+  + 2Fe3+  
(ii) 2Fe2+  + I2 → 2Fe3+  + 2I-
(iii) Sn4+  + 2I- → Sn2+  + I2 
(iv) Sn2+  + I2 → Sn4+  + 2I-
Sol. (a) More or ve the EºOP, more is the tendency for oxidation or stronger is reductant.
Therefore, since maximum EºOP stands for
Sn2+  → Sn4+  + 2e , EºOP =-0.1 V
Therefore,strongest reductant : Sn2+  , and weakest oxidant : Sn4+
(b) More or ve is EºRP, more is the tendency for reduction or stronger is oxidant.
Therefore, since maximum EºRP stands for :
MnO4- + 8H+ +  5e → Mn2+  + 4H2O, EºRP = 1.52 V
Therefore,strongest oxidant : MnO4- , and weakest reductant: Mn2+
Note : Stronger is oxidant, weaker is its conjugate reductant and vice-versa.
(c) For (i) EºCell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

=-0.77 + 0.1
Fe2+  oxidizes and Sn4+  reduces in change, EºCell =-0.67 V, EºCell is negative
Therefore,(i) Is non-spontaneous change
For (ii) EºCell =Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =-0.77 0.54 =- 0.23 V
Therefore,(ii) Is non-spontaneous change
For (iii) EºCell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =-0.54 0.1 =-0.44 V
Therefore,(iii) Is non-spontaneous change
For (iv) EºCell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =-0.1 0.54 = 0.44 V
(iv) Is spontaneous change 

Ex.24 Calculate the standard cell potentials of galvanic cell in which the following reactions take place :(Given EºOP Cr, Cd, Fe2+ , Ag are 0.74V, 0.40
V,- 0.77 V and- 0.80 V respectively)
(a) 2Cr(s) + 3Cd2+ (aq.) → 2Cr3+ (aq) + 3Cd 
(b) Fe2+(aq) + Ag (aq.) → Fe3+ (aq.) + Ag(s)
Calculate the Dr,Gº and equilibrium constant of the reactions.
Sol. (a) Cell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
[2Cr → 2Cr3+ + 6e; 3Cd2+  + 6e → 3Cd]
= 0.74 - 0.40) = 0.34 V
Six electrons (n = 6) are used in redox change
rGº=nEºF = 6 × 0.34 × 96500 J = 196860 J
or ΔrGº =-196.86 kJ
Also-ΔrGº = 2.303 RT log K
Therefore,196860 = 2.303 × 8.314 × 298 log K
K = 3.17 × 1034
(b) Cell =Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
[Fe → Fe e; Ag  e → Ag]
= -0.77 + 0.80 = 0.03 V
Also -ΔrGº = nEºF = 1 × 0.03 × 96500
or ΔrGº =-2895 J
Also -ΔrGº = 2.303 RT log K
2895 = 2.303 × 8.314 × 298 log K
K= 3.22

Ex.25 A cell is constituted as follows  Pt, H2 (1 atm) | HA1 | HA2 | H2 (1 atm), Pt
The pH of two acids solutions HA1 and HA2 are 5 and 3 respectively. The emf of the cell is
(A) 0.059 V
(B) 0.0295 V
(C) 0.118 V
(D)- 0.118 V
Sol. (C)pH1 = 3 Therefore,[H ]c = 1 × 10-3 M ; pH2 = 5 Therefore,[H ]a = 1 × 10-5
Now, Ecell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes log Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.059 × 2 = 0.118 V 

Ex.26 The standard EMF of the cell reaction

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes Cu(s) + Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes Cl2(g) → Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesCu2+  + Clis 1.02 V. The value of DG° will be
(A) unpredictable
(B)- 98.43 kJ
(C)- 196.86 kJ
(D)- 98.43 J
Sol. (B) ΔG° =- nFE° = - 1 × 96500 × 1.02
=- 98430 J
or =- 98.43 kJ

Ex.27 A current of 1.70 ampere is passed through 300 mL of 0.160 M solution of ZnSO4 for 230 sec with a current efficiency of 90%. Find the molarity of Zn after the deposition of Zn. Assume the volume of the solution remains constant during electrolysis.
Sol. We know, i = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notesampere
Therefore,Eq. of Zn2+  lost = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 3.646 × 10-3
Therefore,Meq. of Zn2+  lost = 3.646
Initial Meq. of Zn2+  = 300 × 0.160 × 2
[M × 2 = N for Zn2+ , Meq. = N × V(in mL)] = 48 × 2 = 96
Therefore,Meq. of Zn2+  left in solution = 96- 3.646 = 92.354
Therefore,[ZnSO4] = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 0.154 M

Ex.28 If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm-1 then its molar conductance in ohm-1 cm2 mol-1 will be :
(A) 104
(B) 103
(C) 102
(D) 10
Sol. (B) K = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes × cell const. = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 10-2

Lm = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 1000 = 103 ohm-1 cm2 mol-1

Ex.29 Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if current efficiency is 50%. If the potential drops across the cell is 3.0 volt, how much energy will be consumed 
Sol. C6H5NO2 + 6H+ + 6e → C6H5NH2 + 2H2O

N3+  + 6e → N3-
Therefore,Eq. wt. of nitrobenzene = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, Since current efficiency is 50% Therefore,i = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Now w = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes 12.3 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes Therefore,i × t = 115800 coulomb
Now energy used = Q × V = 115800 × 3 = 347.4 kJ
Ex.30 A cell is containing two H electrodes. The negative electrode is in contact with a solution of 10-6 M H  ion. The e.m.f. of the cell is 0.118 volt at 25ºC. Calculate [H ] at positive electrode.
Sol. Anode : H2 → 2H+ + 2e
(negative polarity) [H ] = 10-6 M

Cathode : 2H+ + 2e → H2
(positive polarity). [H ] → aM
Therefore,Ecell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10[H ] 2AnodePractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10[H ] 2Cathode
= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes , 0.118=Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, Therefore,[H ]Cathode = 10-4 M

Ex.31 A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is :
(A) 1
(B) 2
(C) 3
(D) 4
Sol. (C) E = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes × 96500 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes × 96500 = 59.5

Oxi. state = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 3

Ex.32 Cost of electricity for the production of x L H2 at NTP at cathode is Rs x; then cost of production of x L O2 at NTP at anode will be (assume 1 mole of electron as one unit of electricity)
(A) 2x
(B) 4x
(C) 16x
(D) 32x
Sol. (A) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 2, Volume of O2 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

Thus, Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes L O2 requires Rs x for its production. i.e., x L O2 will require Rs x for its production.

Ex.33 In which direction can the reaction, 2Hg(l) + 2Ag+ (aq.) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes 2Ag(s) + Hg22+ (aq.) proceed spontaneously at the following concentrations of the ions participating in the reactions (i) and (ii) ?
(i) [Ag+ ] = 10-4 mol L-1 and [Hg22+ ] = 10-1 mol L-1
(ii) [Ag+]=10-1 mol L-1 and [Hg22+ ] = 10-4 mol L-1
Given : Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.79 V; Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 0.80 V

Sol. (i) Q = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 107, Eº = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.80- 0.79 = 0.01 V
E = Eº- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog Q = 0.01- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog 107 =- 0.1965 V
Negative value shows that the reaction will proceed from right to left, i.e. in backward direction.
(ii) Q = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 10-2, Eº = 0.01 volt, E = Eº- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes log10 Q = 0.01- Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10 10-2
= 0.01 0.059 V = 0.069 V
Since, the value of cell potential is positive, the reaction will proceed spontaneously in forward direction. 

Ex.34 Two students use same stock solution of ZnSO4 and a solutions of CuSO4. The e.m.f. of one cell is 0.03 V higher than the other. The conc. of CuSO4 in the cell with higher e.m.f. value is 0.5 M. Find out the conc. of CuSO4 in the other cell [(2.303 RT/F) = 0.06].
Sol. As given,
Cell I : ZnPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesCu, Ecell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes….(i)
Cell II : ZnPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesCu E¢cell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes….(ii)
If Ecell>E¢cell , then Ecell-E¢cell=0.03 V and C2=0.5 M
By Eq. (i) and (ii)
Ecell- E¢cell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, 0.03 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes , C¢2 = 0.05 M

Ex.35 How much will the reduction potential of a hydrogen electrode change when its solution initially pH = 0 is neutralized to pH = 7.
(A) Increase by 0.059 V 
(B) Decrease by 0.058 V 
(C) Increase by 0.41 V 
(D) Decrease by 0.41 V
Sol. (D) Eº = E Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog 10-7 = Eº Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Eº- 0.41 V

Ex.36 The time required to coat a metal surface of 80 cm2 with 5 × 10-3 cm thick layer of silver (density 1.05 g cm-3) with the passage of 3A current through a silver nitrate solution is.
(A) 115 sec 
(B) 125 sec 
(C) 135 sec 
(D) 145 sec
Sol. (B) Weight of Ag required
= 80 × 5 × 10-3 × 1.05 (wt. = V × d) = 0.42 g
Q w = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes Therefore,0.42 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes Therefore,t = 125 sec

Ex.37 Standard electrode potentials are
Fe2+/Fe (E° =- 0.44 V), Fe3+ /Fe2+  (E° = 0.77 V)
Fe2+, Fe3+  and Fe blocks are kept together, then :
(A) Fe3+  increases 
(B) Fe3+  decreases
(C) Fe2+ /Fe3+  remains unchanged 
(D) Fe2+  decreases
Sol. (B) Fe 2+ → Fe + Fe+3, Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =- 0.44- 0.77 = -0.121
So spontaneous reaction is Fe + Fe3+ → Fe 2+

Ex.38 Conductivity (Unit : siemen's S) is directly proportionally to the area of the vessel and the concentration of the solution in it and is inversely proportional to the length of vessel, then the unit of constant of proportionality is
(A) S m mol-1 
(B) S m2 mol-1  
(C) S-2 m2 mol 
(D) S2 m2 mol-2
Sol. (B) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

Ex.39 Passage of three faraday of charge through aqueous solution of AgNO3, CuSO4, Al(NO3)3 and NaCl respectively will deposit the metals in the ratio (molar).
(A) 1 : 2 : 3 : 1 
(B) 6 : 3 : 2 : 6 
(C) 6 : 3 : 0 : 0 
(D) 3 : 2 : 1 : 0
Sol. (C) 3 eq. of Ag and Cu, zero equivalent of Na and Al will be deposited.

Note electrolysis of NaClaq. and AlCl3aq. does not give Na and Al metal at cathode.

Thus molar ratio is : for Ag : Cu

or 3 : Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes or 6 : 3 : 0 : 0 for Ag : Cu : Al : Na

Ex.40 Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notesand Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notesare -0.441 V and 0.771 V respectively, Eº for the reaction.
Fe + 2Fe3+  → 3Fe2+ , will be.
(A) 1.212 V 
(B) 0.111 V 
(C) 0.330 V 
(D) 1.653 V

Sol. (A) Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

= 0.441 + 0.771 = 1.212 V

Ex.41 Efficiency of a cell with cell reaction under standard conditions,
A(s) + B+  → A+ + B(s); DHº =-300 kJ is 70%. The standard electrode potential of cell is.
(A) 2.176 V 
(B) 2.876 V 
(C) 1.248 V 
(D) 1.648 V
Sol. (A) Efficiency = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes,  Eº = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 2.176 V

Ex.42 Eº for F2 2e Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes 2F- is 2.7 V. Thus Eº for F-Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesF2 e is.
(A) 1.35 V 
(B)- 1.35 V 
(C)- 2.7 V 
(D) 2.7 V
Sol. (C) OP =- EºRP

Ex.43 Salts of A (atomic weight 7). B (atomic weight 27) and C (atomic weight 48) were electrolysed under identical condition using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively.
(A) 3, 1 and 2 
(B) 2, 6 and 3 
(C) 3, 1 and 3 
(D) 2, 3 and 2
Sol. (B) gm EqA = gm Eq B = gm EqC

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, 0.3 x = 0.1 y = 0.15 z
3x = y = 1.5z
x = 2, y = 6, z = 3

Ex.44 Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below:
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes + 8H +(aq.) + 5e →
Mn2+ (aq.) + 4H2O(l); Eº = 1.51 V
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes + 14H +(aq.) + 6e →
2Cr3+ (aq.) + 7H2O(l) ; Eº = 1.38 V
Fe3+(aq.) + e- → Fe2+(aq.); Eº = 0.71 V
Cl2(g) + 2e- → 2C-(aq.); Eº = 1.40 V
Identify the only incorrect statement regarding the quantitative estimation of
aqueous Fe(NO3)2.
(A) MnO4- can be used in aqueous HCl 
(B) Cr2O72_ can be used in aqueous HCl
(C) MnO4- can be used in aqueous H2SO4 
(D) Cr2O72_ can be used in aqueous H2SO4
Sol. (A) MnO4- will oxidise Cl- ion according to equation

Mn7+  + 5e → Mn2+  , 2Cl- → Cl2 + 2e,

Thus Eºcell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=-1.40 1.51 = 0.11 V

or reaction is feasible.

MnO4- will oxidize Fe2+  to Fe3+

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesMn7+  + 5e → Mn2+  , Fe3+  → Fe2+  + e, Eºcell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = -0.77 + 1.51 = 0.74 V

or reaction is feasible

Thus MnO4- will not oxidize only Fe2+ to Fe3+  in aqueous HCl but it will also oxidise Cl- to Cl2. Suitable oxidant should not oxidise Cl- to Cl2 and should oxidise only Fe2+  to Fe3+  in redox titration.
Ex.45 The emf of the cell Zn | Zn2+  (0.01M)|| Fe2+  (0.001M)| Fe at 298 K is 0.2905, then the value of equilibrium constant for the cell reaction is.
(A) e0.32/0.0295 
(B) 100.32/0.0295 
(C) 100.26/0.0295 
(D) 100.32/0.0591
Sol. (B) Zn + Fe2+  → Fe + Zn2+

Ecell = Eºcell  + Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, 0.2905 = Eºcell + Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NoteslogPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes

Therefore,Eºcell = 0.2905 + 0.0295 = 0.32 V, Now Eºcell =Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10 Kc , 0.32 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Noteslog10 Kc

Therefore,Kc = 100.32/0.0295

Ex.46 The rusting of iron takes place as follows :
2H+ + 2e + ½O2 → H2O(l); Eº = 1.23 V
Fe2+  + 2e → Fe(s); Eº =-0.44 V
The ΔGº for the net process is.
(A)-322 kJ mol-1 
(B)-161 kJ mol-1 
(C)- 152 kJ mol-1 
(D)-76 kJ mol-1
Sol. (A) cell = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes + Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.44 +1.23 = 1.67 V

Therefore,ΔGº =-nEº F =-2 × 1.67 × 96500 J =-322.31 kJ mol-1

Passage : (Q.47 to 52) 
Electrolysis involves electronation and de-electronation at the respective electrodes. Anode of electrolytic cell is the electrode at which de-electronation takes place whereas at cathode electronation is noticed. If two or more ions of same charge are to be electronated or de-electronated, the ion having lesser discharge potential is discharged. Discharge potential of an ion refers for EºOP or EºRP as the case may be. The products formed at either electrode is given in terms of Faraday's laws of electrolysis i.e., w =Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes.

Ex.47 During electrolysis of CH3COONa(aq.), the mole ratio of gases formed at cathode and anode is.
(A) 1 : 2 
(B) 2 : 1 
(C) 3 : 1 
(D) 1 : 3
Sol. (D) Anode :
2CH3COO- → C2H6 + 2CO2 + 2e (3 mole)
Cathode : 2H+ +  2e → H(1 mole)

Ex.48 During electrolysis of HCOONa(aq.), the gas liberated at anode and cathode are respectively.
(A) H2, CO2 and H2 
(B) H2, CO2 and O2 
(C) H2 and O2
(D) O2 and H2
Sol. (A) Anode : 2HCOO- → H2 + 2CO2 + 2e;
Cathode : 2H + + 2e → H2

Ex.49 During electrolysis of CuSO4(aq.), the pH of solution becomes.
(A) < 7 
(B) > 7 
(C) = 7 
(D) ³ 7
Sol. (A) Anode : 2OH- → H2O + ½O2 + 2e;
Cathode : Cu2+  + 2e → Cu
[OH-] decreases, thus pH decrease i.e. < 7.

Ex.50 5 litre solution of 0.4 M CuSO4(aq.) is electrolysed using Cu electrode. A current of 482.5 ampere is passed for 4 minute. The concentration of CuSO4 left in solution is.
(A) 0.16 M 
(B) 0.32 M 
(C) 0.34 M 
(D) 0.40 M
Sol. (D) It is the case of attacked electrodes that is
Anode : Cu → Cu2++ 2e;
Cathode : Cu2+ + 2e → Cu
Thus no change in conc. of Cu ions.

Ex.51 5 litre solution of 0.4 M Ni(NO3)2 is electrolysed using Pt electrodes with 2.4125 ampere current for 10 hour.
(A) 0.31 M 
(B) 0.22 M 
(C) 0.26 M 
(D) 0.40 M
Sol. (A) Eq. of Ni(NO3)2 = 5 × 0.4 × 2 = 4;
Eq. of Ni lost = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.9
Therefore,Eq. of Ni(NO3)2 left = 4- 0.9 = 3.1, Therefore,Molarity of Ni(NO3)2 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 0.31

Ex.52 The volume of octane required to be used for its combustion by the oxygen liberated during electrolysis of an NaNO3(aq.) by passing 9.65 ampere current for 1 hr. is.
(A) 322.56 mL 
(B) 32.256 mL 
(C) 3.22 mL 
(D) 1.612 × 102 mL
Sol. (D) Anode : 2OH- → H2O+ ½O2 + 2e;
Cathode : 2H+ + 2e → H2
Eq. of O2 formed = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.36
Therefore,Mole of O2 formed = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 0.09, C8H18Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes O2 → 8CO2 9H2O
Therefore,Mole of C8H18 = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes× 0.09 = 7.2 × 10-3 , Therefore,Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 7.2 × 10-3 × 22400 161.2 mL

Ex.53 The specific conductivity of 0.02 M KCl solution at 25°C is 2.768 × 10-3 ohm-1 cm-1. The resistance of this solution at 25°C when measured with a particular cell was 250.2 ohm. The resistance of 0.01 M CuSO4 solution at 25°C measured with the same cell was 8331 ohm. Calculated the molar conductivity of the copper sulphate solution.
Sol. Cell constant = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes= 2.768 × 10-3 × 250.2
For 0.01 M CuSO4 solution
Sp. conductivity = Cell constant × Conductance = 2.768 × 10-3 × 250.2 × Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Molar conductance = Sp. cond. × Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes =Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes×Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=8.312 ohm-1 cm2 mol-1

Ex.54 A decinormal solution of NaCl has specific conductivity equal to 0.0092. If ionic conductances of Na and Cl- ions at the same temperature are 43.0 and 65.0 ohm-1 respectively. Calculate the degree of dissociation of NaCl solution
Sol. Equivalent conductance of N/10 NaCl solution
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Sp. conductivity × dilution = 0.0092 × 10,000 = 92 ohm-1
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 43.0 65.0 = 108 ohm-1
Degree of dissociation, a = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 0.85

Ex.55 The specific conductance of saturated solution of AgCl is found to be 1.86 × 10-6 ohm-1 cm-1 and that of water is 6 × 10-8 ohm-1 cm-1. The solubility of AgCl is ….
Given, Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 137.2 ohm-1 cm2 eq-1
(A) 1.7 × 10-3 M (B) 1.3 × 10-5 M (C) 1.3 × 10-4 M (D) 1.3 × 10-6 M
Sol. kAgCl = kAgCl (Solution)Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 1.86×10-6-6×10-8=1.8 × 10-6 ohm-1 cm-1 

Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes, S=Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes=1.31×10-5 M

Ex.56 Resistance of a solution (A) is 50 ohm and that of solution (B) is 100 ohm, both solution being taken in the same conductivity cell. If equal volumes of solution (A) and (B) are mixed, what will be the resistance of the mixture, using the same cell ? Assume that there is no increase in the degree of dissociation of (A) and (B) mixing
Sol. Let us suppose k1 and k2 are the specific conductance of solutions `A' and `B' respectively and cell constant is `y'. We know that,
Specific conductance=Conductance × Cell constant
For (A), Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes For (B), Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
When equal volume of (A) and (B) are mixed, the volume becomes double. Then, Specific conductance of mixture = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Therefore,Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
R = 200/3 = 66.66 ohm 

Ex.57 A big irregular shaped vessel contained water, specific conductance of which was 2.56 × 10-5 mho cm-1, 500 g of NaCl was then added to the water and the specific conductance after the addition of NaCl was found to be 3.1 × 10-5 mho cm-1. Find the capacity of the vessel if it was fully filled with water. ( NaCl = 149.9 ohm-1 cm2 eq-1)
Sol. Let us suppose the volume of vessel is V mL
Volume containing 1 equivalent = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
Specific conductance of NaCl = Specific conductance of NaCl solution- Specific conductance of water
= 3.1 × 10-5- 2.56 × 10-5 = 0.54 × 10-5 mho cm-1 
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = k × volume containing 1 equivalent of electrolyte ….(i)
For very dilute solution, when the big vessel is fully filled
Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes = 149.9 ohm-1 cm2 eq-1 , Thus, from eq. (i) 149.9 = 0.54 × 10-5 × Practice Questions(Solved) - Electrochemistry, Class 12, Chemistry | EduRev Notes
V = 237258.38 L

Ex.58 Maximum Conductivity would be of-
(A) K3Fe(CN)6 [0.1 M solution] 
(B) K2Ni(CN)4 [0.1M solution]
(C) FeSO4.Al2(SO4) 3.24H2O [0.1 M solution] 
(D) Na[Ag(S2O2)3] [0.1 M solution]  
Sol. (C)
Double salt on ionization gives more ions. One molecule of the salt gives Fe 2, 2Al 3, 4SO4-2 ions. Hence its conductance would be highest.

Discussion Questions
(i) For the reaction given below, apply Le-Chatelier principle to justify the results recorded by you and also bring out mathematical rationalisation of your results.
Zn(s) Cu (aq) Zn (aq) Cu(s),
(ii) Determine the slope of the graph. Match experimental value with the theoretical value. On what factors does the value of slope depend?
(iii) Devise another experiment to study the variation in cell potential with concentration of one of the ions involved in a cell reaction.
(iv) What factor is kept in mind while selecting an electrolytic solution for the construction of a salt bridge?
(v) Is it possible to measure the single electrode potential?

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Dynamic Test

Content Category

Related Searches

Chemistry | EduRev Notes

,

video lectures

,

Extra Questions

,

Sample Paper

,

study material

,

Practice Questions(Solved) - Electrochemistry

,

Free

,

Semester Notes

,

Important questions

,

pdf

,

Previous Year Questions with Solutions

,

Practice Questions(Solved) - Electrochemistry

,

Summary

,

mock tests for examination

,

Viva Questions

,

MCQs

,

Chemistry | EduRev Notes

,

Class 12

,

Class 12

,

Practice Questions(Solved) - Electrochemistry

,

shortcuts and tricks

,

Exam

,

practice quizzes

,

ppt

,

past year papers

,

Objective type Questions

,

Class 12

,

Chemistry | EduRev Notes

;