Cls Jeead-15-16 Xii Che Target-5 Set-2 Chapter-3

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solutions

SECTION - A

Objective Type Questions (One option is correct)

1. On passing 3 faradays of electricity through three electrolytic cells connected in series containing Ag+, Ca+2

and Al+3 ion respectively, the molar ratio in which three metal ions are liberated at the electrode is

(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 6 : 3 : 2 (4) 3 : 4 : 2

Sol. Answer (3)

i × t is same for all the electrolytic solutions

Ag

W

M

⎛ ⎞⎜ ⎟⎝ ⎠ =

it

nF =

it

F (Ag+ + e– Ag)

Ca

W

M

⎛ ⎞⎜ ⎟⎝ ⎠

= it

2F (Ca2+ + 2e– Ca)

and Al

W

M

⎛ ⎞⎜ ⎟⎝ ⎠

= it

3F (Al3+ + 3e Al)

Molar ratio is 1 : 1

2 :

1

3or 6 : 3 : 2

2. The molar conductances at infinite dilution of BaCl2, NaCl and NaOH are respectively 280×10–4,

126.5 × 10–4, 248 × 10–4 S m2 mol–1. The molar conductance at infinite dilution for Ba(OH)

2 is

(1) 523 × 10–4 S m2 mol–1 (2) 52.3 × 10–4 S m2 mol–1

(3) 5.23 × 10–4 S m2 mol–1 (4) 65 × 10–4 S m2 mol–1

Sol. Answer (1)

2BaCl

= 2 –Ba Cl

2 ...(i)

NaCl

= –

Na Cl

...(ii)

NaOH

= –

Na HO

...(iii)

for Ba(OH)2

(i) + 2(iii) –2(ii)

2(Ba(OH) ) = (280 × 10–4) + 2(248 × 10–4) – 2(126.5 × 10–4)

= 523×10–4 Sm2 mol–1.

Chapter 3

Electrochemistry

72 Electrochemistry Solution of Assignment (Set-2)


3. During electrolysis of aqueous solution of a salt, pH in the space near one of the electrode is increased. Which

of the following salt solution was electrolysed?

(1) KCl (2) CuCl2

(3) Cu(NO3)2

(4) CuSO4

Sol. Answer (1)

In KCl solution the reaction at the electrodes are

2H+ + 2e– H2

2Cl– Cl2 + 2e–

[H+] decreases in the solution because of which [OH] increases hence increasing the pH.

4. By how much will the potential of half cell Cu+2/Cu change, if the solution is diluted to 100 times at 298 K?

(1) Increases by 59 mV (2) Decreases by 59 mV

(3) Increases by 29.5 mV (4) Decreases by 29.5 mV

Sol. Answer (2)

For Cu2+ + 2e– Cu(s)

2Cu /Cu

E = Eº – 0.0591

2 log 2

1

[Cu ]

When Cu2+ solution is diluted to 100 times [Cu2+] decreases to 1/100

2Cu /Cu'E = Eº –

0.0591

2 log 2

100

[Cu ]

E’ = Eº – 0.0591

2 [[log 100 – log [Cu2+]]

E’ = Eº – 0.0591

2 × 2 –

0.0591

2 log 2

1

[Cu ]

E’ = Eº – 0.0591

2 log 2

1

[Cu ] – 0.0591

E’ = E – 0.0591, Hence, Potential decreases by 59 mV.

5. The

cellE of the reaction

OHFeMnHFeMnO 2

322

4 is 0.59 V at 25°C. The equilibrium constant for the reaction is

(1) 50 (2) 10 (3) 1050 (4) 105

Sol. Answer (3)

o

cellE = 0.59 V

–

4MnO + Fe2+ + H+ Mn2+ + Fe3+ + H

2O

E = o

cellE –

0.0591

5 log Q

c

At equilibrium, E = 0; Qc = K

c

o

cellE =

0.0591

5 log K

c

5 0.59

0.59

= log K

c

50 = log Kc

Kc = 1050.

73Solution of Assignment (Set-2) Electrochemistry


6. A current of 2.0 A when passed for 5 hrs through a molten salt, deposits 22.2 g of metal (of atomic weight

177). The oxidation state of metal in metal salt is

(1) +1 (2) +2 (3) +3 (4) +4

Sol. Answer (3)

i = 2A, t = 5 hrs = 5 × 60 × 60 s

wM

= 22.2 g; A = 177

Applying the equation

w = Eit

F w =

Ait

nF

n = A i t

wF

=

177 2 5 60 60

22.2 96500

= 2.97

n = 3

M3+ + 3e– M

Oxidation state is +3.

7. Some Indian scientists tried to use a metal x for electroplating iron pillar in Mehrauli but they ended up with

Ecell

of the reaction to be negative. They concluded that

(1) Reaction is spontaneous (2) Reaction is non-spontaneous

(3) Reaction is reversible (4) Reaction is non-reversible

Sol. Answer (2)

For electroplating Iron a metal ‘x’ is used.

Ecell

is negative, it means that no reaction takes place and the reaction is non-spontaneous.

8. In the electrolysis of aqueous solution of NaOH, 2.8 litre of oxygen at NTP was liberated at the anode. How

much hydrogen was liberated at cathode?

(1) 5.6 litre (2) 56 ml (3) 560 ml (4) 0.056 litre

Sol. Answer (1)

NaOH is electrolysed.

NaOH Na+ + HO–

H2O H+ + HO–

At cathode: 2H+ + 2e–H2

At anode: 4HO– 2H2O + O

2 + 4e–

2H

w

M

⎛ ⎞⎜ ⎟⎝ ⎠ =

it

2F;

2O

w

M

⎛ ⎞⎜ ⎟⎝ ⎠ =

it

4F

2H

n : 2O

n = it

2F :

it

4F 2 : 1 Volume ratio

2 2H OV : V = 2 : 1

2HV

2.8 =

2

1

2HV = 2.8 × 2 = 5.6 L



9. The equilibrium constant for the reaction Sr (s) + Mg+2 (aq) Sr+2 (aq) + Mg (s) is 2.69 × 1012 at 25°C.

The E° for a cell made up of Sr/Sr+2 and Mg+2/Mg half cells is

(1) 0.3667 V (2) 0.7346 V (3) 0.1836 V (4) 3.667 V

Sol. Answer (1)

The reaction is

Sr(s) + Mg2+(aq) Mg(s) + Sr2+(aq)

Kc = 2.69 × 1012

At equilibrium, E = 0; Q = Kc

0 = Eº – 0.0591

2 log K

c

Eº = 0.0591

2 log K

c =

0.0591

2 log (2.69 × 1012) = 0.3667 V

10. Passage of one ampere current through 0.1 M Ni(NO3)2 solution using Ni electrodes bring in the concentration

of solution to _________ in 60 seconds.

(1) 0.1 M (2) 0.05 M (3) 0.2 M (4) 0.025 M

Sol. Answer (1)

The reaction taking place at the electrodes are

Anode : Ni Ni2+ + 2e

Cathode : Ni2+ + 2e– Ni

Hence [Ni2+] does not change.

Concentration of Ni2+ is 0.1M.

11. Which species in each pair is a better oxidising agent under standard conditions?

(1) Br2 & Au+3 (2) H

2 & Ag+

(3) Cd+2 & Cr+3 (4) O2 in acidic medium & O

2 in basic medium

Sol. Answer (1)

Halogens act as an oxidising agent and in Au3+ the oxidation state of Au is maximum i.e., +3. So, Au3+ also

acts as oxidizing agent.

12. At 25°C, the equivalent conductances at infinite dilution of HCl, CH3COONa and NaCl are 426.1, 91.0 and

126.45 cm2 –1eq–1 respectively. for CH3COOH (in cm2 –1eq–1) is

(1) 391.6 (2) 390.6 (3) 380.6 (4) 309.6

Sol. Answer (2)

According to given condition:

HCl

= o

H +

–

o

Cl ...(i)

3CH COONa

= –3CH COO

o

Na ...(ii)

NaCl

= o

Na +

–

o

Cl ...(iii)

Doing the operation

Equation (i) + (ii) – (iii)

HCl

+ 3CH COONa

– NaCl

= –3

o

CH COO +

o

H

(426.1 + 91.0 – 126.45) = 3CH COOH

3CH COOH

= 390.6 cm2 –1/eq



13. When electricity is passed through a solution of AlCl3 13.5 g of Al is deposited. The number of faradays must

be

(1) 1.0 (2) 1.5 (3) 0.5 (4) 2

Sol. Answer (2)

The reaction at cathode is

Al3+ + 3e– Al (n = 3)


w = Eit w

FF E

⇒

i × t =

13.5

27

3

⎛ ⎞⎜ ⎟⎝ ⎠

× F = 13.5

9 F = 1.5 F

14. 0.5 faraday of electricity was passed to deposit all the copper present in 500 ml of CuSO4 solution. What was

the molarity of this solution?

(1) 1 M (2) 0.5 M (3) 0.25 M (4) 2.5 M

Sol. Answer (2)

i × t = 0.5 F


w = E it

F

w

M

⎛ ⎞⎜ ⎟⎝ ⎠

= it

2F =

0.5F

2F = 0.25 moles

V × molarity = No. of moles

500 × x × 10–3 = 0.25

x =

30.25 10

500

= 0.5

Molarity = 0.5 M

15. Cu+ is not stable and undergoes disproportionation

E for Cu+ disproportionation

V53.0E,V153.0ECu/CuCu/Cu

2

(1) +0.683 V (2) –0.367 V (3) 0.754 V (4) +0.3415 V

Sol. Answer (3)

Given

2

o

Cu /CuE = +0.153 V and

o

Cu /CuE = 0.53 V

The reaction

2Cu+ Cu + Cu2+

o

CellE = 2

o

Cu /CuE +

o

Cu /CuE

we required 2

o

Cu /CuE

Cu2+ + 2e Cu; 0.153



Cu Cu+ + e–; – 0.53

Cu2+ + e– Cu+

–1 × F × Eº = [–2 × F × (0.153)] + [F × 0.53]

–FEº = –2F (0.153) + 0.53 F

Eº = –0.53 + (2 × 0.153)

2

o

Cu /CuE = – 0.224

or 2

o

Cu /CuE = 0.224

Eº = 0.224 + 0.53 = 0.754 V

16. 25 g of a metal is deposited on cathode during the electrolysis of metal nitrate solution by a current of 5 A

passing for 4 hours. If atomic weight of the metal is 100. The valency of metal in metal nitrate is

(1) 1 (2) 2 (3) 3 (4) 4

Sol. Answer (3)

wM

= 25 g, i = 5 A, t = 5 hrs = 4 × 60 × 60 second

A = 100 (metal nitrate was electrolysed)

Applying the equation,

w = E it

F

25 = 100 5 4 60 60

n 96500

n = 100 5 4 60 60

(96500 25)

= 3

Hence, valency = 3

17. A well stirred solution of 0.1 M CuSO4 is electrolysed at 25°C using platinum electrodes with a current of 25

mA for 6 hours. If current efficiency is 50%. At the end of the duration what would be the concentration of

copper ions in the solution?

(1) 0.0856 M (2) 0.092 M (3) 0.0986 M (4) 0.1 M

Sol. Answer (3)

18. 50 ml of a buffer of 1 M NH3 and 1 M NH

4

+ are placed in two volatic cells separately. A current of

3.0 amp is passed through both cells for 10 min. If electrolysis of water takes place as

2H2O + O

2 + 4e– 4OH– (R.H.S.)

2H2O 4H+ + O

2 + 4e– (L.H.S.)

then pH of the

(1) L.H.S. will increase (2) R.H.S. will increase

(3) R.H.S. will decrease (4) Both side will increase

Sol. Answer (2)

Because of the reactions of electrolysis, [H+] concentration will decrease as a result of which pH will increase.



19. 1 M aqueous solution of NaCl undergo electrolysis if 50 mA current is passed for 12 hours. Assume current

efficiency is 25%. The total volume of gas produced at standard state is

(1) 137 ml (2) 68.5 ml (3) 125.44 ml (4) 62.72 ml

Sol. Answer (3)

NaCl aq. solution undergoes electrolysis

NaCl Na+ + Cl–

H2O H+ + HO–

Reactions :

At cathode : 2H+ + 2e H2(g)

At anode : 2Cl– Cl2 + 2e

T

w

M

⎛ ⎞⎜ ⎟⎝ ⎠

= it

2F +

it

2F =

it

F

3 25 12 60 6050 10

100 96500

= nT

nT = 0.005595

V = 0.005595 × 22400

125.44 ml.

20. Vanadium electrode is oxidised electrically. If the mass of electrode decreases by 100 mg during the passage

of 570 coulomb, the oxidation state of vanadium in the product is (At. wt. of V = 51)

(1) 6 (2) 5 (3) 4 (4) 3

Sol. Answer (4)

i × t = 570 C

w = 100 mg = 100 × 10–3 g

w = E it

F

3

100 10

51

=

570

96500 n

n = –3

570 51

100 10 96500

= 3

21. The specific conductance of a saturated solution of AgCl is K–1 cm–1. The limiting ionic conductances of Ag+

and Cl– are x and y, respectively. The solubility product of AgCl is

(1)yx

K1000

(2)

2

yx

K1000⎟⎟⎠

⎞⎜⎜⎝

⎛

(3)yx

K143.51000

(4)

23

yx

K143.510⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

Sol. Answer (2)

AgCl Ag+ + Cl–



Specific conductance = K–1cm–1

K 1000

C

AgCl =

o o

Ag Cl = (x + y)

(x + y) = K 1000

C

C =

1000K

(x y)

Solubility product = C2 =

2

1000K

(x y)

⎡ ⎤⎢ ⎥⎣ ⎦

22. The correct order of equivalent conductance at infinite dilution of LiCl, NaCl, KCl is

(1) KCl > NaCl > LiCl (2) LiCl > NaCl > KCl (3) LiCl > KCl > NaCl (4) LiCl ~ NaCl < KCl

Sol. Answer (1)

The ions formed are Li+, Na+ and K+, the hydration is maximum in case of Li+ because of which its mobility

is least and has least conductance.

Hence, the following order.

KCl > NaCl > LiCl

23. The limiting equivalent conductance of NaCl, KCl and KBr are 126.5, 150.0 and 152.0 S cm2 eq–1 respectively.

The limiting equivalent ionic conductance of Br– is 76 S cm2 eq–1. The limiting equivalent ionic conductance

of Na+ is

(1) 25.5 (2) 52.5 (3) 75.5 (4) 57.5

Sol. Answer (2)

NaCl

= o

Na +

–

o

Cl = 126.5 ...(i)

KCl

= o

K +

–

o

Cl = 150 …(ii)

KBr

= o

K +

–

o

Br = 152 …(iii)

Adding (i) & (iii) subtract (ii)

o

Na +

–

o

Cl +

o

K +

–

o

Br

o

K– –

–

o

Cl =

o

Na +

–

o

Br

= (126.5 + 152 – 150) = (76) + o

Na

o

Na = 52.5

Equivalent ionic conductance for Na+ is 52.5.

24. The equivalent conductances of CH3COONa, HCl and NaCl at infinite dilution are 91, 426 and 126 S cm2 eq–1

respectively at 25°C. The equivalent conductance of 1 M CH3COOH solution is 19.55 S cm2 eq–1. The pH of

solution is (pKa = 4.74)

(1) 5.3 (2) 4.3 (3) 2.3 (4) 1.3



Sol. Answer (4)

3CH COONa

= –

3

o o

CH COO Na

o

HCl =

–

o o

H Cl

NaCl

= –

o

Na Cl

3CH COOH

= –3CH COONa NaClHCl

–

91 + 426 + (–126) = 391 = 3CH COOH

C19.55

= C

= 19.55

391 = 0.05

[H+] = C = 1 × 0.05 M

pH = – log [H+] = – log (5 × 10–2)

2–log 5 = 2 – 0.7 =1.3

25. V2.37EV,2.71E/Mg)(Mg/Na)(Na

2

V0.41EV,0.44E/Cr)(Cr/Fe)(Fe 32

Based on this data, which is the poorest reducing agent?

(1) Na+ (2) Mg+2 (3) Fe+2 (4) Cr+3

Sol. Answer (4)

Cr3+ is the poorest reducing agent because of least value of oxidation potential.

26. Which of following type of plot would you expect from the titration of AgNO3 against KCl solution?

(1)

Conductance

Vol. of KCl

(2)

Conductance

Vol. of KCl

(3)

Conductance

Vol. of KCl

(4)

Conductance

Vol. of KCl

Sol. Answer (3)

Fact.

27. The standard reduction potential of Cu+2/Cu and Cu+2/Cu+ are 0.337 V and 0.153 V respectively. The standard

reduction potential of Cu+/Cu half cell is

(1) 0.521 V (2) 0.490 V (3) 0.321 V (4) 0.290 V

Sol. Answer (1)

Given (i) ... Cu2+ + 2e– Cu(s); o

1E = 0.337 V

(ii)... Cu2+ + e– Cu+ o

2E = 0.153 V



Reversing equation (ii); we get

Cu+ Cu2+ + e– ...(iii)

Adding equation (i) and (iii)

We get, Cu+ + e– Cu

o o o

1 2G G G = (–1 × F × Eº) = o1–2 F E + o

2–1 F E

– FEº = [– 2× F × (0.337)] + [–F × –0.153]

– FEº = – 2 × 0.337 × F + (0.153) F

Eº = (2×0.337) – (0.153) = 0.521 V

28. What is G° for the following reaction?

Cu+2(aq) + 2Ag(s) Cu(s) + 2Ag+

V0.8EV,0.34E/AgAg/CuCu

2

(1) –44.5 kJ (2) 44.5 kJ (3) –89 kJ (4) 89 kJ

Sol. Answer (4)

The reaction given is

Cu2+ (aq) + 2Ag(s) Cu(s) + 2Ag+

o

CellE = 2

o o

Ag/Ag Cu /CuE E

o

CellE = (– 0.8) + (0.34) = – 0.46

Gº = – n F × Eº = – 2 × F × (–0.46)

= – 2 × –0.46 × 96500 = 88780 J

or Gº = + 89 kJ.

29. For the half cell oQuinhydroneE 1.30 V

O H– O

+ 2H + 2e+ –

O H–

O

At pH = 3, electrode potential is

(1) 1.48 V (2) 1.42 V (3) 1.36 V (4) 1.3 V

Sol. Answer (1)

For the reaction, on applying Nernst equation

Ecell

= o

CellE –

0.0591

2 log [H+]2

Ecell

= 1.30 – 0.0591

2 log (10–3)2

= –0.0591

2 × (–6) log 10 + 1.30 = 0.0591 × 3 + 1.30 = 1.477 1.48



30. Emf of the cell

Zn | Zn+2(aq) || Cu+2(aq) | Cu is independent of

(1) Quantity of Cu+2 and Zn+2 in solution (2) Concentration of Cu+2

(3) Concentration of Zn+2 (4) Temperature

Sol. Answer (1)

For the given cell

E = Eº –0.0591

2 log

2

2

Zn

Cu

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

When Zn2+ & Cu2+ quantity is changed the emf does not change because EMF depends upon concentration

and not the quantity.

31. Which is correct increasing order of deposition?

(1) Na+ < Mg+2 < Zn+2 < Ag+ (2) Ag+ < Zn+2 < Mg+2 < Na+

(3) Mg+2 < Na+ < Zn+2 < Ag+ (4) Mg+2 < Zn+2 < Na+ < Ag+

Sol. Answer (1)

Increasing order of deposition is related to the order of reduction and oxidation potential (in accordance with

preferential discharge theory)

Na+ < Mg2+ < Zn2+ < Ag+

32. Which is the correct order of deposition of anion?

(1) SO4

–2 > OH– > Cl– > Br– > I– (2) SO4

–2 < OH– < CI– < Br– < I–

(3) SO4

–2 > Cl– > Br– > I– > OH– (4) Br– > Cl– > I– > SO4

–2 > OH–

Sol. Answer (2)

It is in the order of discharge potential

In anion order of deposition is

2–

4SO < HO– < Cl– < Br– < I–

33. Which metal oxide is thermally unstable?

(1) Al2O

3(2) Na

2O (3) BaO (4) Ag

2O

Sol. Answer (4)

Ag2O decomposes as

Ag2O 2Ag +

1

2 O

2.

34. Rate of corrosion is maximum when

(1) An electrolyte is present in water (2) Metal has low S.R.P.

(3) Metal has high standard oxidation potential (4) All of these

Sol. Answer (4)

When metal has high standard oxidation potential, it has more tendency to undergo oxidation. In presence

of electrolyte, rate of Corrosion is maximum.



35. H2(1 atm) | 2.26 M HCOOH || 0.222 M CH

3COOH| (1 atm) H

2

Ka(HCOOH) = 1.77 × 10–4,

Ka(CH

3COOH) = 1.8 × 10–5

Emf of the cell is (Neglect the liquid-liquid junction potential)

(1) 0.0591 V (2) –0.0591 V (3) 0.02955 V (4) –0.02955 V

Sol. Answer (2)

HCOOH HCOO– + H+

C1(1 – ) C

1 C

1

[H+]L = C

1 =

1 1(Ka) C

CH3 COOH CH

3COO– + H+

C2(1 – )C

2 C

2

[H+]R =

2 2(Ka) C

E = Eº – 0.0591

1 log

L

R

[H ]

[H ]

E = 0 – 0.0591

1 log

1 1

2 2

K C

K C

or E = – 0.0591

2 log

4

5

1.77 10 2.26

1.8 10 0.222

E = – 0.0591

2 log 100 = – 0.0591

36. Given that

NiO2 + 4H+ + 2e– Ni2+ + 2H

2O, E° = 1.678 V

NiO2 + 2H

2O + 2e– Ni(OH)

2 + 2OH–, E° = –0.49 V

For the following reaction

Ni(OH)2 + 2H+ Ni2+ + 2H

2O

Gibb’s free energy change (in kJ mol–1) is

(1) 418.424 (2) –229.284 (3) –418.424 (4) 229.284

Sol. Answer (3)

– 2 o

2 2 1

– – – o

2 2 2 2

2 o

2 2

NiO 4H 2e Ni 2H O , E 1.678 V

Ni(OH) 2OH 2e NiO 2H O 2e , E 0.49 V

Ni(OH) 2H Ni 2H O , E x V

Go = o o o o

1 2 1 2G G –nF(E E )

= –2 × 96500 × (1.678 + 0.49) J mol–1

= –418.424 kJ mol–1



37. Zn amalgam is prepared by electrolysis of aqueous ZnCl2 using 9 gram Hg cathode. How much current is to

be passed through ZnCl2 solution for 1000 seconds to prepare a Zn amalgam with 25% by weight? (Atomic

mass, Zn = 65.4 g)

(1) 5.6 A (2) 7.2 A (3) 6.64 A (4) 11.2 A

Sol. Answer (3)

Let, x gram of Zn deposited on 9 gram of Hg. % of Zn in amalgam x

100 259 x

x = 3 gram

Equivalent of Zn 3 2

65.4

Current 6 96500

8.85 A65.4 1000

38. Which of the following cannot be extracted by electrolysis from aqueous solution of their salts?

(1) Zn (2) Ag (3) Cu (4) Pt

Sol. Answer (1)

As 2

o

Zn /ZnE is less than

2

o

H /HE

39. Emf of cell given, Ag(s), AgCl(s)||KCl(aq)|Hg2Cl

2(s)|Hg(s) is 0.05 V at 300 K and temperature coefficient of the

cell is 3.34 × 10–4 VK–1. Calculate the change in enthalpy of the cell.

(1) 965 (2) 9650 (3) 96500 (4) 96.5

Sol. Answer (2)

2Ag 2Ag+ + 2e– : (anode)

Hg2

2+ + 2e– 2Hg : (cathode)

∵ n = 2

H = cell

cell

P

E–nFE nFT

T

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 2 × 96500(300 × 3.34 × 10–4 – 0.05)

= 9650 J mol–1

40. Given : Ag+ + e– Ag ; E° = 0.799 V

Dissociation constant for [Ag(NH3)2]+ into Ag+ and NH

3 is 6 × 10–14. Then for the following half-cell reaction:

[Ag(NH3)2]+ + e– Ag + 2NH

3, calculate E°.

(1) 0.019 V (2) 0.03 V (3) 0.014 V (4) 0.19 V

Sol. Answer (1)

– o

OP

– o

3 2 3 RP

3 2 3

Ag Ag e ; E –0.799 V

Ag(NH ) e Ag NH ; E ?

Ag(NH ) Ag 2NH



o 3 2

cell cell 10 2

3

[Ag(NH ) ]0.0591E E log 0

1 [Ag ] [NH ]

at equilibrium

o –14

cell 10 C 10E 0.0591 log K 0.0591 log (6 10 )

= –0.780 V

3 2

o o

OP(Ag/Ag ) RP(Ag(NH ) /Ag)E E

3 2

o

Ag(NH ) /AgE –0.780 0.799

= +0.019 V

41. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time

required to liberate 0.01 mol of H2 gas at the cathode is (1 faraday = 96500 C mol–1) [IIT-JEE 2008]

(1) 9.65 × 104 s (2) 19.3 × 104 s (3) 28.95 × 104 s (4) 38.6 × 104 s

Sol. Answer (2)

W it

E F

3W 10 10 t

0.01 2E 96500

t = 19.3 × 104 s

42. AgNO3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured.

The plot of conductance () versus the volume of AgNO3 is [IIT-JEE 2011]

volume

(P)

volume

(Q)

volume

(R)

volume

(S)

(1) (P) (2) (Q) (3) (R) (4) (S)

Sol. Answer (4)

Ag+ and K+ have nearly same ionic mobility

AgNO3 + KCl AgCl(s) + KNO

3

conc. of KCl



43. Consider the following cell reaction:

2(s) 2(g) (aq) (aq) 22Fe O 4H 2Fe 2H O(l) ; E° = 1.67 V

At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25°C is [IIT-JEE 2011]

(1) 1.47 V (2) 1.77 V (3) 1.87 V (4) 1.57 V

Sol. Answer (4)

2 2

cell 42

0.0591 [Fe ]E 1.67 log

4 pO [H ]

3 2

3 4

0.0581 (10 )1.67 log

4 0.1 (10 )

6

13

0.0591 101.67 log

4 10

70.0591

1.67 log104

0.05911.67 7 1.57

4

SECTION - B

Objective Type Questions (More than one options are correct)

1. 1.0 L of 0.1 M aqueous solution of KCl is electrolysed. A current of 96.50 mA is passed through the solution

for 10 hours. Which is/are correct? (Assume volume of solution remains constant during electrolysis)

(1) After electrolysis molarity of K+ is 0.064 and molarity of Cl– is 0.064

(2) After electrolysis molarity of K+ is 0.1 and molarity of Cl– is 0.064

(3) At S.T.P. 202 ml of Cl2 produced when current efficiency is 50%

(4) At S.T.P. 606 ml of total gases produced when current efficiency is 50%

Sol. Answer (2, 3)

i = 96.50 A, t = 10 × 60 × 60 s

Solution is 1.0 L and 0.1 M

Moles present = 1 × 0.1 = 0.1 moles

Reactions :

2H+ + 2e H2 : 2Cl– Cl

2 + 2e–

w

M =

it

nF =

w

M =

–3

96.50 10 60 60 10

2 96500

= 0.018

For Cl– = 0.036; Molarity = 0.1 – 0.036 = 0.064

2ClV =

0.018 22.4

2

= 0.202 L or 202 ml.

K+ will not discharge.



2. 1000 ml 2 M CuSO4 is electrolysed by a current of 9.65 amp for 2 hours. Which is/are correct?

(1) After electrolysis remaining concentration of Cu+2 is 1.64 M using Cu electrode

(2) After electrolysis remaining concentration of Cu+2 is 1.64 M using Pt-electrode

(3) When remaining concentration of Cu+2 is 1.822 then volume of solution is reduced by 10% using Pt-

electrode

(4) 17.15 g copper deposit when current efficiency is 75% using copper electrode

Sol. Answer (2, 3, 4)

No. of moles of CuSO4

= 1000 × 2 = 2000 millimoles

= 2 moles

i = 9.65 A; t = 2 hrs = 2 × 60 × 60 s

Cu deposited is w = E it

F

w = 63.5

× 9.65 2 60 60

96500

= 22.86

n = M

W = 0.36

2 – 0.36 = 1.64

Hence, molarity = 1.64 M using Pt electrode

w = 63.5

2 ×

75

100 (9.65) 2 60 60

96500

= 17.15 g

3. For the electrolysis of CuSO4 solution which is/are correct?

(1) Cathode reaction : 2H+ + 2e H2 using Pt electrode

(2) Cathode reaction : Cu+2 + 2e– Cu using Cu electrode

(3) Anode reaction : Cu Cu+2 + 2e– using Cu electrode

(4) Anode reaction : Cu Cu+2 + 2e– using Pt electrode

Sol. Answer (2, 3)

CuSO4(aq) forms the ions

Cu2+, H+, HO– and 2–

4SO

Using Pt electrode

At cathode; Cu2+ + 2e Cu

Using Cu electrodes

At anode :

Cu Cu2+ + 2e–



4. Daniell cell : M)1 ml 50(

2

M)1 ml 50(

2 (aq)|CuCu||(aq)Zn|Zn operates as electrolysis cell for 60 min and a current of

0.965 amp is passed. Which is/are correct?

2 2Cu /Cu Zn /Zn

(E 0.34 V, E 0.76 V)

(1) After electrolysis Zn+2 concentration is 1.36 M (2) After electrolysis Cu+2 concentration is 0.64 M

(3) After electrolysis Zn+2 concentration is 0.82 M (4) After electrolysis Cu+2 concentration is 1.18 M

Sol. Answer (1, 2)

WCu

(deposited) = 31.75

96500 × 0.965 × 60 × 60 = 1.143 g

Total weight of copper = 50 × 10–3 × 1 × 63.5 = 3.175 g

Left weight of copper = 3.175 – 1.143 = 2.032 g

Molarity of Cu2+ solution = 2.032 1000

63.5 50 = 0.64 M

Molarity of Zn2+ solution = 1 + 0.36 = 1.36 M

5. Which compounds have maximum conductivity?

(1) 0.2 M [Cr(NH3)3Cl

3] (2) 0.15 M [Cr(NH

3)4Cl

2]Cl

(3) 0.1 M [Cr(NH3)5Cl]Cl

2(4) 0.07 M [Cr(NH

3)6]Cl

3

Sol. Answer (2, 3)

[Cr(NH3)4Cl

2]Cl [Cr(NH

3)4Cl

2]2+ +Cl–

0.15 × 2 = 0.30

and for

[Cr(NH3)5Cl]Cl

2 [Cr(NH

3)5Cl]2+ + 2Cl–

0.1 × 3 = 0.30

6. Molar conductance of 2 M H2A acid is 10 S cm2 mol–1. Molar conductance of H

2A at infinite dilution is 400

S cm2 mol–1. Which statement is/are correct?

(1) Degree of dissociation is 2.5% and pH of solution is 1.3

(2) Degree of dissociation is 4 and pH of solution is 1.4

(3) Dissociation constant of H2A is 6.24 × 10–5

(4) Dissociation constant of H2A is 2.56 × 10–4

Sol. Answer (1)

7. Which of following is/are correct?

(1) The metallic conduction is due to the movement of electrons in the metal

(2) The electrolytic conduction is due to the movement of ions in the solution

(3) The metallic conduction increases with increase in temperature whereas electrolytic conduction decreases

with increase in temperature

(4) None of these



Sol. Answer (1, 2)

The metallic conduction is due to the presence of electrons in the metal and electrolytic conduction is due

to the movement of ions in the solution.

8. For electrolyte AxB

y which is/are not correct relation between molar conductivity (

M) and equivalent conductivity

(eq

)

(1) M

= xy eq

(2) eq

= xy M

(3) xM

= y eq

(4) yM

= x eq


For the electrolyte Ax B

y

n-factor = xy

M = (xy) eq

Only 1st option is correct & others are incorrect option.

9. The cell constant of a conductivity cell is defined as ( = cell constant, l = length between the electrode, A

= area, R = resistance, G = conductance, K = conductivity)

(1)A

l (2)R

(3) = (Gr)–1 (4)K

G

Sol. Answer (1, 3)

R = l

A

1 1 l

R A

K = l

CA

RA

l ;

l

A and

1(G )

10. Which of following plots will not be obtained for a conductometric titration of HCl and NaOH?

(1)

Conductance

Vol. of NaOH

(2)

Conductance

Vol. of NaOH

(3)

Conductance

Vol. of NaOH

(4)

Conductance

Vol. of NaOH


In the conductometric titration of HCl and NaOH conductance first decreases, reaches a minimum value and

then increases.

11. Zn | Zn+2 (1M) || Ni+2 (1 M) | Ni, antilog (0.7411) = 5.5

V0.24E V,0.75E/NiNi/ZnZn

22

Which statement is/are correct for above cell?

(1) Emf of cell is 0.51 V and cell reaction is spontaneous

(2) Emf of cell is –0.51 V and cell reaction is non-spontaneous

(3) Emf of cell is zero when concentration of Ni+2 is 5.5 × 10–18 M

(4) Cell reaction is non-spontaneous when concentration of Ni+2 is less than 5.5 × 10–18 M




The given cell is

Zn|Zn2+(1M)||Ni2+(1M)|Ni

Eº = (0.75) + (–0.24) = 0.51V

and cell reaction is spontaneous.

E = Eº – 0.0591

2 log

2

2

[Zn ]

[Ni ]

E = 0; [Ni2+] = 5.5 × 10–18 M

The cell reaction is

Non-spontaneous when concentration of M2+ is less than 5.5 × 10–18 M.

12. Which statement is correct about electrolysis of CuSO4?

(1) At cathode Cu will deposit and at anode O2 will be produced using Pt-electrode

(2) At cathode Cu will not deposit but Cu dissolve at anode using Cu-electrode

(3) At cathode Cu will deposit and at anode O2 will be produced using Cu-electrode

(4) At cathode Cu will deposit and at anode Cu will dissolve using Cu-electrode

Sol. Answer (1, 4)

Using Pt electrodes

CuSO4 Cu2+ +

2–

4SO

H2O H+ + HO–

At cathode : Cu2+ + 2e– Cu

Anode : 4HO– 2H2O + O

2 + 4e–

Products are Cu and O2

Using Cu electrodes

Anode : Cu Cu2+ + 2 e–

Cathode : Cu2+ + 2 e– Cu

13. Aqueous solution of which electrolyte produces H2 gas at cathode?

(1) NaCl (2) MgCl2

(3) CuCl2

(4) AgCl

Sol. Answer (1, 2)

H+ has lower discharge potential as compared to Na+ and Mg2+

Hence, in case of NaCl and MgCl2 reaction is 2H+ + 2e– H

2.

14. Which is/are correct?

(1) If temperature coefficient is greater than zero, cell reaction is endothermic

(2) If temperature coefficient is less than zero, cell reaction is endothermic

(3) If temperature coefficient is less than zero, cell reaction is exothermic

(4) If Ecell

is negative then G is negative and cell reaction is spontaneous



Sol. Answer (1, 3)

It is known fact that H and temperature coefficient are related as,

H > 0 for E

T

⎛ ⎞⎜ ⎟⎝ ⎠

> 0

and H < 0 for E

T

⎛ ⎞⎜ ⎟⎝ ⎠

< 0

15. Which is/are correct statements about salt bridge?

(1) Velocity of ions of salt bridge are almost equal (2) Salt bridge completes the electric circuit

(3) Ions of salt bridge discharge at electrode (4) Ions of salt bridge do not discharge at electrode


Salt Bridge contains electrolyte which do not participate in the electrochemical change, completes the cell

circuit and it is also necessary that velocity of ions of salt bridge are almost equal.

16. The standard emf of the cell

Fe | Fe+2(aq) || Cd+2 | Cd is 0.0372 V and temperature coefficient of emf is –0.125 VK–1. Which is/are correct

about the cell (at room temperature)?

(1) G° = 7.18 kJ, H° = –7196.43 kJ (2) G° = –7.18 kJ, H° = 7196.43 kJ

(3) G° = –7.18 kJ, H° = –7196.43 kJ (4) S° = –24.125 kJ K–1, reaction is spontaneous

Sol. Answer (3, 4)

The emf of cell

Fe|Fe2+||Cd2+|Cd, Eº = 0.0372

Gº = –nFEº = – 2 × 96500 × 0.0372

Gº = – 7179.6 J = – 7.18 kJ

(S) = nF cell

P

E

T

⎛ ⎞⎜ ⎟⎝ ⎠

(Sº) = 2 × F × (–0.125)

– 7180 = – oΔH + 298 2 × 96500 × 0.125

oΔH = – 7196.43 kJ

17. Ksp

for AgBr = 8 × 10–13

Ag, AgNO3 (1.0 M) || KBr (1.0 M), AgBr, Ag

For above cell which is/are correct?

(1) Ecell

= 0.715 V (2) Ecell

= –0.715 V

(3) G = –1 × 96500 × 0.715 (4) G = 1 × 96500 × 0.715



Sol. Answer (2, 4)

For the 1st Half cell

Anode : Ag Ag+ + e–

Cathode : Ag+ + e– Ag

E = 0.0591

1 log

L

R

[Ag ]

[Ag ]

E = –0.0591 log

1

[Ag ]

Ksp

(AgBr) = [Ag+] [Br–]

[Ag+]R =

138 10

1

= 8 × 10–13 M

E = –0.0591 log 13

1

8 10

= –0.0591 log

1310

8

0.0591 log 8 × 10–13.

0.0591 (0.6 – 13) = – 0.715 V

and G = –nFE = +1 × 96500 × 0.715

18. Which is/are correct about corrosion?

(1) Due to corrosion FeO.xH2O formed

(2) Due to corrosion Fe2O

3.xH

2O formed

(3) Presence of air and moisture increases the rate of corrosion

(4) Magnesium is used as sacrificial anode


In corrosion

Fe Fe2+ + 2e is formed and the formation of oxide i.e., Fe2O

3.xH

2O takes place and presence of air and

moisture is must.

Mg can be used as sacrificial anode.

19. Which statement is/are correct?

(1) In electrochemical cell electrons flow from anode to cathode

(2) In electrochemical cell, anode is negative electrode and cathode is positive electrode

(3) Oxidation take place at anode and reduction take place at cathode in electrochemical cell

(4) In electrolytic cell oxidation take place at cathode and reduction take place at anode


At cathode always reduction takes place and at anode always oxidation takes place.

Hence (4) will not the correct statement.



20. V0.24E V,0.76E/NiNi/ZnZn

22

V0.40E V,0.04E/CdCd/FeFe

23

Which is/are correct statements?

(1) Zn+2 + Cd Cd+2 + Zn, spontaneous (2) Ni+2 + Cd Ni + Cd+2, spontaneous

(3) Fe+3 + Ni Ni+2 + Fe, spontaneous (4) Cd+2 + Zn Zn+2 + Cd, spontaneous


For (1)

Zn2+ + Cd Cd2+ + Zn

Eº = 2 2

o o

Cd/Cd Zn /ZnE E = (0.40) + (–0.76) = –0.36 < 0

Non-spontaneous

For Ni2+ + Cd Cd2+ + Ni

Eº = 2 2

o o

Cd/Cd Ni /NiE E

= (0.40) + (– 0.24) > 0 is spontaneous

for Fe3+ + Ni Ni2+ + Fe

Eº = 2 3

o o

Ni/Ni Fe /FeE E = (0.24) + (–0.04) > 0 i.e. spontaneous

and for reaction

Cd2+ + Zn Zn2+ + Cd

Eo = 2 2

o o

Zn/Zn Cd /CdE E = (0.76) + (– 0.40) > 0 is spontaneous.

21. Which of the following cells give the cell potential to their standard values?

(1) Zn|Zn2+(0.01 M)||H3O+(0.1 M)|H

2(1 atm), Pt (2) Cu|Cu2+(0.25 M)||Ag+(0.5 M)|Ag

(3) Cd|Cd2+(0.01 M)||pH = 1|H2(1 atm), Pt (4) Zn|Zn2+(0.1 M)||pH = 1|H

2(1 atm), Pt


For o

cell cell CE E ,K 1.

22. Which solution(s) become(s) more acidic after the electrolysis using inert electrodes?

(1) NaCl solution (2) CuSO4 solution (3) AgNO

3 solution (4) Na

2SO

4 solution

Sol. Answer (2, 3)

In the electrolysis of CuSO4 solution and AgNO

3 solution, H

2SO

4 and HNO

3 are formed respectively.

23. In which of the following cells, reaction quotient is equal to one?

(1) Pb|PbC2O

4, CaC

2O

4, CaCl

2(0.1 M)||CuSO

4(0.1 M) | Cu

(2) Zn|ZnSO4(0.1 M)||CuSO

4(0.1 M)|Cu

(3) Zn|ZnSO4(0.1 M)||Hg

2Cl

2, KCl(0.1 M)|Hg, Pt

(4) Cu|CuSO4(0.1 M)||SnCl

2(0.1 M)|SnCl

4(0.1M), Pt

Sol. Answer (1, 2)

In (3), = 10

In (4), = 0.1



24. Saturated solution of KNO3 is used to make ‘salt-bridge’. Then incorrect option(s) is/are

(1) Velocity of K+ is zero

(2) Velocity of NO3

– is zero

(3) Velocity of both K+ and NO3

– are nearly the same

(4) KNO3 is highly soluble in water

Sol. Answer (1, 2)

Fact.

25. When a lead-storage battery is discharged, then incorrect option(s) is/are

(1) H2SO

4 is consumed (2) Pb is formed (3) SO

2 is evolved (4) PbSO

4 is consumed


Pb is consumed and PbSO4 is formed. SO

2 is not evolved.

26. For the reduction of –

3NO ion in an aqueous solution, Eo is +0.96 V. Values of Eo for some metal ions are

given below [IIT-JEE 2009]

V2+ (aq) + 2e– V Eo = – 1.19 V

Fe3+ (aq) + 3e– Fe Eo = –0.04 V

Au3+ (aq) + 3e– Au Eo = +1.40 V

Hg2+ (aq) + 2e– Hg Eo = +0.86 V

The pair(s) of metals that is(are) oxidized by –

3NO in aqueous solution is(are)

(1) V and Hg (2) Hg and Fe (3) Fe and Au (4) Fe and V


Oxidation of V

Eo = 0.96 – (–1.19) = 2.15 V

For Fe,

Eo = 0.96 – (–0.04) = 1.0 V

For Au,

Eo = 0.96 – 1.4 = – 0.044 V (not feasible)

For Hg

Eo = 0.96 – 0.86 = 0.1 V

27. In a galvanic cell, the salt bridge [JEE(Advanced)2014]

(1) Does not participate chemically in the cell reaction

(2) Stops the diffusion of ions from one electrode to another

(3) Is necessary for the occurrence of the cell reaction

(4) Ensures mixing of the two electrolytic solutions

Sol. Answer (1, 2)

In a galvanic cell, the salt bridge does not participate in the cell reaction, stops diffusion of ions from one

electrode to another and is not necessary for the occurrence of the cell reaction.



SECTION - C

Linked Comprehension Type Questions

Comprehension-I

An electrochemical cell is constructed by immersing a piece of copper wire in 50 ml of 0.1 M CuSO4 solution and

zinc strip in 50 ml of 0.1 M ZnSO4 solution

V]0.76E V,0.34[E/ZnZn/CuCu

22

1. The emf of cell is

(1) 1.07 V (2) 1.1 V (3) 1.3 V (4) 1.13 V

Sol. Answer (2)

2

o

Cu /CuE = 0.34 V and 2

o

Zn /ZnE = – 0.76 V

ECell

= Eº – 0.0591

2 log

2

2

[Zn ]

[Cu ]

[Zn2+] = [Cu2+] = 1M

E = Eº = 2 2

o o

Zn/Zn Cu /CuE E

E = (0.76) + (0.34)

E = Eº = 1.1 V

2. The emf of cell increases when small amount of concentrated NH3 is added to

(1) ZnSO4 solution (2) CuSO

4 solution (3) Both (1) & (2) (4) Can't say

Sol. Answer (1)

When NH3 is added to ZnSO

4 solution, NH

3 reacts with Zn2+ in the following manner :

Zn2+ + 4 NH3

[Zn(NH3)4]2+

i.e., [Zn2+] decreases.

In the equation

E = Eo –

2

2

0.0591 [Zn ]log

2 [Cu ]

If [Zn2+] decreases then

2

2

[Zn ]log

[Cu ]

decreases hence, EMF of cell increases.

3. In a separate experiment, 50 ml of 1.5 M NH3 is added to CuSO

4 solution. Emf of the cell is

[Kf ([Cu(NH

3)4]+2) = 5.88 × 1013]

(1) 0.933 V (2) 1.327 V (3) 1.467 V (4) 0.61 V



Sol. Answer (4)

Due to the complex formation, [Cu2+] decreases & it can be calculated by the reaction,

2

350 0.1

50 1.5

Cu 4HN

[Cu(NH3)4]2+

[Cu2+] =

2

3 4

4

3 f

[Cu (NH ) ]

[NH ] k

= 4 13

50 0.1

100

(0.55) 5.88 10

= 9.3 × 10–15 M

Ecell

= 1.1 – –15

0.0591 0.1log

2 9.3 10

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= 0.715 V

Thus, the e.m.f. of cell decreases.

Comprehension-II

The ionic mobility for some ions in water at 298 K is given as following

ions ionic mobility

K+ 7.616 × 10–4

Ca+2 12.33 × 10–4

Br– 8.09 × 10–4

SO4

–2 16.58 × 10–4

1. The equivalent conductance of CaSO4 at infinite dilution is

(1) 279 (2) 28.51 × 10–4 (3) 31.82 × 10–4 (4) 306

Sol. Answer (1)

Equivalent conductance of CaSO4 is the sum of ionic conductance of Ca2+ &

2–

4SO .

4CaSO

= 2 2–4Ca SO

2Ca

= 2

CaU F

2–4SO

= 2–4SO

U F

2Ca

U & 24SO

U are ionic mobilities

4CaSO

= F 12.33 + 16.58 × 10–4

4CaSO

= 96500 × 10–4 × 28.91 = 278.98 279

Equivalent conductance of CaSO4 is 279

2. If degree of dissociation is 10% then equivalent conductance of CaSO4 is

(1) 27.9 (2) 2.851 × 10–4 (3) 3.182 × 10–4 (4) 30.6



Sol. Answer (1)

We know that

= C

(0.1) = C

(279)

C = 279 × (0.1) = 27.9

Equivalent conductance = 27.9

3. If the equivalent conductance of 0.01 M CaSO4 solution is 13.95, then equilibrium constant is

(1) 5.26 × 10–4 (2) 5.26 × 10–5 (3) 2.63 × 10–4 (4) 2.63 × 10–5

Sol. Answer (4)

C = 0.01M; C

= 13.95

= C

= 13.95

279 = (0.05)

K =

2C

(1 )

=

2(0.05) (0.01)

(1– 0.05)

K = 2.63 × 10–5.

Comprehension-III

Given below are a set of half-cell reactions (in acidic medium) alongwith their E° (in volt) values.

2I 2e 2I E 0.54

2Cl 2e 2CI E 1.36

3 2Mn e Mn E 1.50

3 2

Fe e Fe E 0.77

2 2O 4H 4e 2H O E 1.23

1. Among the following, identify the correct statement

(1) Cl– is oxidised by O2

(2) Fe+2 is oxidised by iodine

(3) I– is oxidised by chlorine (4) Mn+2 is oxidised by chlorine

Sol. Answer (3)

2. While Fe+3 is stable, Mn+3 is not stable in acid solution because

(1) O2 oxidises Mn+2 to Mn+3

(2) O2 oxidises both Mn+2 to Mn+3 and Fe+2 to Fe+3

(3) Fe+3 oxidises H2O to O

2

(4) Mn+3 oxidises H2O to O

2

Sol. Answer (4)



3. The strongest reducing agent in aqueous solution is

(1) I– (2) Cl– (3) Mn+2 (4) Fe+2

Sol. Answer (1)

Comprehension-IV

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules

(approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/

molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept

has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and

radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction,

which requires a clear understanding of the mole concept.

A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to

the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200,

1 faraday = 96500 coulomb). [IIT-JEE 2007]

1. The total number of moles of chlorine gas evolved is

(1) 0.5 (2) 1.0 (3) 2.0 (4) 3.0

Sol. Answer (2)

nNaCl

= 4 500

21000

2

Cln 1.

2. If the cathode is a Hg electrode, then the maximum weight (g) of amalgam formed from this solution is

(1) 200 (2) 225 (3) 400 (4) 446

Sol. Answer (4)

nNa

deposited = 2

nNa–Hg

formed = 2

Mass = 2 × 223 = 446.

3. The total charge (coulomb) required for complete electrolysis is

(1) 24125 (2) 48250 (3) 96500 (4) 193000

Sol. Answer (4)

Total charge required = 2F = 2 × 96500 = 193000 C.

Comprehension-V

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The

resulting potential difference across the cell is important in several processes such as transmission of nerve impulses

and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is

M(s) | M+(aq; 0.05 molar) || M+ (aq; 1 molar) | M(s)

For the above electrolytic cell the magnitude of the cell potential |Ecell

| = 70 mV. [IIT-JEE 2010]



1. For the above cell

(1) Ecell

< 0; G > 0 (2) Ecell

> 0; G < 0 (3) Ecell

< 0; Gº > 0 (4) Ecell

> 0; Gº < 0

Sol. Answer (2)

Cell reaction

(1M) (0.05 M)M M

Apply Nernst equation

0.059 0.05E Eº – log

1 1

–2

0.059E – log 5 10

1

0.059E – –2 log5

1⎡ ⎤ ⎣ ⎦

2. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell

potential would be

(1) 35 mV (2) 70 mV (3) 140 mV (4) 700 mV

Sol. Answer (3)

1

2

E log0.05

E log0.0025

–2

1

–4

2

E log5 10

E log25 10

1E 70(given)

2

70 –1.3 1

E –2.6 2

Comprehension-VI

The electrochemical cell shown below is a concentration cell.

M | M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the

cell at 298 K is 0.059 V. [IIT-JEE 2012]

1. The solubility product (Ksp

; mol3 dm–9) of MX2 at 298 K based on the information available for the given

concentration cell is (take 2.303 × R × 298/F = 0.059 V)

(1) 1 × 10–15 (2) 4 × 10–15 (3) 1 × 10–12 (4) 4 × 10–12

Sol. Answer (2)

2

0.059 0.0010.059 log

2 (M )



2

0.001log 2

[M ]

5 3 15

2 sp

2 5

0.001100 K = 4 (10 ) = 4 10

[M ]

[M ] 10

2. The value of G (kJ mol–1) for the given cell is (take 1 F = 96500 C mol–1)

(1) –5.7 (2) 5.7 (3) 11.4 (4) –11.4

Sol. Answer (4)

G = –nFE

= –2 × 96500 × 0.059

= –11387 joule mol–1

–11.4 kJ mol–1

SECTION - D

Assertion-Reason Type Questions

1. STATEMENT-1 : The molar conductivity of strong electrolyte decreases with increase in concentration.

and

STATEMENT-2 : At high concentration, migration of ion is slow.

Sol. Answer (1)

Molar conductance is given by the following expression

= (K × V) = K 1000

c

Here ‘c’ is the concentration

More is the concentration lesser is the molar conductance

Hence, both statements are correct and statement-2 is the correct explanation of statement-1.

2. STATEMENT-1 : Electrolysis of molten PbBr2 using platinum electrodes produces Br

2 at anode.

and

STATEMENT-2 : Br2 is obtained in gaseous state at room temperature.

Sol. Answer (3)

PbBr2 Pb2+ + 2Br–

At cathode :

Pb2+ + 2e– Pb

At anode :

2Br– Br2 + 2e–

Br2 obtained in liquid state at room temperature.

Statement-1 is correct and statement-2 is false.



3. STATEMENT-1 : For the concentration cell, Zn)aq(Zn)aq(Zn)s(Zn

21 C

2

C

2 for spontaneous cell reaction C

1 < C

2.

and

STATEMENT-2 : For concentration cell,

1

2ecellC

Clog

nF

RTE for spontaneous reaction E

cell = +ve C

2 > C

1.

Sol. Answer (1)

The given cell is

Zn|Zn2+(C1) ||Zn2+ (C

2)|Zn

Zn(s) Zn2+ + 2e–

2Zn/Zn

E = 2

o

Zn/ZnE –

0.0591

2 log (C

1)

2Zn /Zn

E = 2Zn /Zn

E – 0.0591

2 log

2

1

C

⎛ ⎞⎜ ⎟⎝ ⎠

E = 2 2

o o

Zn/Zn Zn /Zn(E E ) –

0.0591

2log

1

2

C

C

⎛ ⎞⎜ ⎟⎝ ⎠

EMF of cell

E = 1

2

C– 0.0591log

2 C

⎛ ⎞⎜ ⎟⎝ ⎠

log 1

2

C

C

⎛ ⎞⎜ ⎟⎝ ⎠

< 0 for spontaneity

log 1

2

C

C

⎛ ⎞⎜ ⎟⎝ ⎠

< log 1

C1 < C

2.

Statement-1 and statements-2 is correct and it is also the correct explanation.

4. STATEMENT-1 : A saturated solution of KCl is used to make salt bridge in concentration cells.

and

STATEMENT-2 : Mobility of K+ and Cl– are nearly same.

Sol. Answer (1)

Mobilities of ions involved in salt bridge is same which is used in concentration cells.

5. STATEMENT-1 : The molar conductance of weak electrolyte at infinite dilution is equal to sum of molar

conductances of cation and anion.

and

STATEMENT-2 : Kohlrausch’s law is applicable for both strong and weak electrolytes.

Sol. Answer (3)

AB

= –

A B

Kohlraush law is applicable for weak electrolyte and not for strong electrolyte.



6. STATEMENT-1 : When a copper wire is placed in a solution of AgNO3, the solution acquires blue colour.

and

STATEMENT-2 : oRPE of Cu+2/Cu is lesser than

o

Ag/AgE .

Sol. Answer (1)

Cu + AgNO3 Cu2+ + Ag

Eo for reaction is positive because 2

o o

Cu /Cu Ag /AgE E

7. STATEMENT-1 : G° = –nFE°.

and

STATEMENT-2 : E° should be positive for a spontaneous reaction.

Sol. Answer (2)

Go = – nFEo

But Go does not decide the spontaneity only G decides the spontaneity of reaction.

Both statements are correct but statement-2 is not the correct explanation.

8. STATEMENT-1 : One coulomb of electric charge deposits the weight that is equal to electrochemical equivalent

of substance.

and

STATEMENT-2 : One faraday deposits one mole of substance.

Sol. Answer (3)

One faraday deposits one equivalent of substance.

9. STATEMENT-1 : If an aqueous solution of NaCl is electrolysed, the product obtained at the cathode is H2 gas

and not Na.

and

STATEMENT-2 : Gases are liberated faster than metals.

Sol. Answer (3)

NaCl

Na+ + Cl–

H2O

H+ + OH–

Among cations, hydrogen has higher standard electrode potential and among anions chlorine has low

standard electrode potential. Thus, at cathode preferentially H2 gas is evolved, and at anode Cl

2 gas is

evolved.

10. STATEMENT-1 : H2 + O

2 fuel cell gives a constant voltage throughout its life.

and

STATEMENT-2 : In this fuel cell, H2 reacts with OH– ions, yet the overall concentration of OH– ions does not

change.

Sol. Answer (1)

In H2 + O

2 fuel cell,

Anode : 2H2(g) + 4OH–(aq) 4H

2O() + 4e–

Cathode : O2(g) + 2H

2O() + 4e– 4OH–(aq)

OH– consumed is reformed, so [OH–] does not change.

Hence, fuel-cell gives constant voltage throughout its life.



SECTION - E

Matrix-Match Type Questions

1. Match the following

Column-I Column-II

Complex Molar Conductivity (–1)

(A) CoCl3.6NH

3(p) 97

(B) CoCl3.5NH

3(q) 0

(C) CoCl3.4NH

3(r) 404

(D) CoCl3.3NH

3(s) 229

Sol. Answer A(r), B(s), C(p), D(q)

[Co(NH3)6] Cl

3 will give maximum number of ions(4) because of which conductivity is maximum

i.e. 404.

In [Co(NH3)3Cl

3] no ions are given

Hence molar conductivity is zero.

[Co(NH3)5Cl]Cl

2 & [Co(NH

3)4Cl

2] Cl forms 3 & 2 ions.


Column-I Column-II

(Amount of charge used for diposition/liberation)

(A) 1 mol Al+3 (p) F

(B) 2.3 gm of Na+ (q) 3 F

(C) 3.6 gm of Mg+2 (r) 0.1 F

(D) 11.2 L H2 at S.T.P. (s) 0.3 F

Sol. Answer A(q), B(r), C(s), D(p)

(A) 1 mole Al3+

w

E =

it

F it =

wF

E

it = wF

3M

⎛ ⎞⎜ ⎟⎝ ⎠

= wF

E = 3F

(B) 2.3 g Na+

it = w

E F =

2.3

23 F = 0.1 F

(C) 3.6 g of Mg2+

it = 3.6

12 F = 0.3 F



(D)11.2

22.4 =

1

2

2Hn

w

E =

it

F it =

w

E F

= w

2M

⎛ ⎞⎜ ⎟⎝ ⎠

F = 1

22

F = F


Column-I Column-II

(A) , specific conductance (p)

c

m

m

(B) m , molar conductance (q)

m

(C) , degree of dissociation (r) Decreases with dilution

(D) Kohlrausch law (s) Decreases with increase in concentration of strong

electrolytes

Sol. Answer A(r), B(s), C(p), D(q)

(A) Specific conductance decreases with dilution

(B) Molar conductance decreases with increase in concentration of electrolyte

(C)m

m

and decreases with dilution

(D) Resistance l

A and decreases with dilution


Column-I Column-II

(A) Calomel electrode (p) Electrolyte concentration cell

(B) Zn-Cd(C1) |CdCl

2| Zn-Cd(C

2) (q) Metal-insoluble anion half cell

(C) Quinhydrone electrode (r) Electrode concentration cell

(D) Pt|H2(1 atm)|H+(C

1)||H+(C

2)|H

2(1 atm)|Pt (s) Redox half cell

Sol. Answer A(q), B(r), C(s), D(p)


Column-I Column-II

(Electrolysis) (Observation)

(A) Aqueous solution of NaCl using inert (p) Metal loss at anode

electrodes

(B) Very dilute aqueous solution of NaCl (q) Chlorine gas evolved at anode

using mercury cathode

(C) CuSO4 using copper electrodes (r) Oxygen gas evolved at anode

(D) 50% H2SO

4 solution (s) A compound with peroxide bond is formed

Sol. Answer A(q), B(r), C(p), D(s)



SECTION - F

Integer Answer Type Questions

1. The half cell potentials of a half cell x n x| A , A | Pt

were found to be as follows:

% of reduced form 24.4 48.8

Half cell potential (V) 0.101 0.115

Determine the value of 'n'.

Sol. Answer (2)

x n – x

A ne A

o

RP

o

RP

0.059 75.60.101 E log

n 24.4

0.059 51.20.115 E log n 2

n 48.8

⎛ ⎞ ⎜ ⎟⎝ ⎠

⎛ ⎞ ⇒⎜ ⎟⎝ ⎠

2. The standard reduction potential of 3

o

Bi /BiE and 2

o

Cu /CuE are 0.226 V and 0.344 V respectively. A mixture of

salts of Bi and Cu at unit concentration each is electrolysed at 25°C. At what value of 2log Cu

does Bismuth

starts to deposit during electrolysis.

Sol. Answer (4)

The passage of current would initially deposit Cu2+ till 2

Cu /CuE becomes 0.266 V because then only Bi3+ will

be deposited.

Thus, 2 2

o 2

Cu /Cu cu /Cu

0.059E E log Cu

2

2 20.0590.266 0.344 log Cu log Cu 4

2

⇒

3. A cell is containing two H electrodes. The negative electrode is in contact with a solution of pH = 6. EMF of

the cell is 0.118 V at 25°C. Calculate pH at positive electrode.

Sol. Answer (4)

cell

H cathodeE 0.059log

H anode

= 0.059 [pH anode – pH cathode]

0.118 = 0.059 [6 – pH]

pH = 4

4. How many faradays of electricity is required to deposit 2 mol copper from CuSO4 solution?

Sol. Answer (4)

Equivalent weight of copper = 63.5/2

Hence, 2 mol require 4 F electricity.

5. A current of 3 ampere has to be passed through a solution of AgNO3 solution to coat a metal surface of

80 cm2 with 0.005 mm thick layer for a duration of approximately (y)3 seconds. What is the value of y?

(Density of Ag is 10.5 g/cm3)



Sol. Answer (5)

Volume of surface = 80 × 0.0005

= 0.04 cm2

WAg

= 0.04 × 10.5 = 0.42 gram

Eit

96500

3108 3 t0.42 t 125.09 s y

96500

⇒

y 5

6. The cost at 5 paise per kWh of operating an electric motor for 8 hours, which takes 15 ampere at 110 V, is

11y paise. Calculate y.

Sol. Answer (6)

Total energy consumed for 8 hours = iVt

= 15 × 110 × 8 × 10–3 kWh

= 13.2 kWh

Total cost = 5 × 13.2 = 66 paise

11y = 66

y = 6

SECTION - G

Multiple True-False Type Questions

1. STATEMENT-1 : Corrosion of iron is essentially an electrochemical phenomenon.

STATEMENT-2 : Corrosion reaction at anode : 32Fe s 2Fe 6e

STATEMENT-3 : Corrosion reaction at cathode : 2 2

O g 4H aq 4e 2H O l

(1) T T T (2) T F T (3) T F F (4) F F T

Sol. Answer (2)

Facts about corrosion

Statement 2 : 22Fe s 2Fe 4e

2. STATEMENT-1 : Using Kohlrausch's law of independent migration of ions, it is possible to calculate 0 for

any electrolyte from the ° of individual ions.

STATEMENT-2 : Limiting molar conductivity of an electrolyte can be represented as the sum of the individual

contributions of the anion and cation of the electrolyte.

STATEMENT-3 : When concentration approaches zero, molar conductivity reaches the lowest limit.

(1) T T F (2) T T T (3) F T T (4) F F T

Sol. Answer (1)

Fact.



3. STATEMENT-1 : Electrolysis of acidulated water using inert electrodes results in evolution of gases at cathode

and anode both.

STATEMENT-2 : Al3+ discharges more readily than Zn2+ at cathode.

STATEMENT-3 : In an electrolytic cell, cations move towards anode.

(1) F T T (2) T T T (3) F T F (4) T F F

Sol. Answer (4)

In electrolysis of acidulated water, H2 and O

2 is evolved in 2 : 1 volume ratio, at cathode and anode

respectively.

3 /Al 2

o o

Al Zn / ZnE E Zn2+ discharges more readily at cathode.

In an electrolytic cell, cations move towards cathode, i.e., negative electrode.

SECTION - H

Aakash Challengers Questions

1. The standard potential of the following cell is 0.23 V at 15°C and 0.21 V at 35°C.

Pt|H2(g)|HCl (aq)||AgCl(s)|Ag(s)

(i) Write the cell reaction.

(ii) Calculate H° and S° for the cell reaction by assuming that these quantities remain unchanged in the

range 15°C to 35°C.

[Given SRP of Ag+(aq)|Ag(s) is 0.80 V at 25°C]

Sol. (i)–

2

1H (g) AgCl(s) H (aq) Ag(s) Cl (aq)

2

(ii)P

ES nF ,n 1,F 96500 coulombs

T

⎛ ⎞ ⎜ ⎟⎝ ⎠

E = 0.21 – 0.23 = – 0.02 V, T = 35 – 15 = 20°C

–1 –1

–0.02S 1 96500 –96.5 JK mol

20

⎛ ⎞ ⎜ ⎟⎝ ⎠

o –1

15G –1 0.23 96500 –22195 Jmol

o

15H G – T S

= –22195 – 288 × (–96.5)

= –49987 J mol–1

–1 –1 –1

S –96.5 J k mol , H – 49987 Jmol⇒

2. Calculate o

rG of the following reaction

Ag+(aq) + Cl–(aq) AgCl(s)

Given:o

rG (AgCl) –109 kJ mol–1;

o –

rG (Cl ) –129 kJ mol–1;

o

rG (Ag )

77 kJ mol–1

(i) Represent the above reaction in form of a cell.

(ii) Calculate E° of the cell.

(iii) Find log10

Ksp

of AgCl.



Sol. (i) Ag(s) | Ag+ | | AgCl | | Cl– | Cl2, Pt

(ii) –

o o o or AgCl Ag Cl

G G – G – G (Ag+(aq) + Cl–(aq) AgCl(s))

= –109 – (–129) – 77

= –57 kJ mol–1

o

cell

–57000G –nFE E

–1 96500 ⇒

o

cellE 0.59 V⇒

(iii) G° = –2.303 RT logK

–57000logK 10

–2.303 6.314 298⇒

K = 1010

–10

sp 10 sp

1K 10 log K –10

K ⇒

3. The following electrochemical cell has been set-up:

Pt1|Fe3+|Fe2+(1M)||Ce4+ | Ce3+ (1M)|Pt

2

3 2 4 3

o o

Fe /Fe Ce /CeE 0.77 V and E 1.61 V.

If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will

the current increase or decrease with time?

Sol. SRP of Fe > SRP of Ce

Fe acts as anode and Ce acts as cathode

o o o

cell cathode anodeE E – E 1.61– 0.77 0.84 V

Hence, flow of current is from cathode to anode (right to left)

Current will decrease with time.

4. An excess of liquid mercury is added to an acidified solution of 1.0 × 10–3 M Fe3+. It is found that 5% of Fe3+

remains at equilibrium at 25°C. Calculate 2

o

Hg /HgE assuming that the only reaction that occurs is

2Hg + 2Fe3+ Hg2

2+ + 2Fe2+ 3 2

o

Fe /Fe[Given E 0.77 V] .

Sol.3 2 2

2

–3

–3

–3 –3

2Fe 2Hg 2Fe Hg

At t 0 1 10 M 0 0

0.95 10at t(eq) 0.05 10 M 0.95 10 M M

2

22 2

2o

cell cell 10 3 2

Fe [Hg ]0.0591E E – log

n [Fe ]

⎡ ⎤⎣ ⎦

2

–3 2 –3o

10 –3 2Hg /Hg

0.0591 (0.95 10 ) (0.475 10 )At eqb., 0 0.77 – E – log

2 (0.05 10 )

2

o

Hg /HgE 0.792 V⇒



5. Chromium metal can be plated out from an acidic solution containing CrO3 according to the following equation.

CrO3(aq) + 6H+(aq) + 6e– Cr(s) + 3H

2O.

(i) How many grams of chromium will be plated out by 24000 coulombs?

(ii) How long will it take to plate out 1.5 g of chromium by using 12.5 ampere current?

Sol. 52 gram of Cr is deposited by passing 6 × 96500 coulomb

(i) Amount of Cr deposited = 52 24000

6 96500

= 2.155 g

(ii) M = Zit

521.5 12.5 t

6 96500⇒

t = 1336.15 s

6. The specific conductivity of a saturated solution of AgCl is 2.30 × 10–6 ohm–1 cm–1 at 25°C. Calculate the

solubility of AgCl at 25°C if –1 2 –1

Ag61.9 ohm cm mol and

–

–1 2 –1

Cl76.3 ohm cm mol .

Sol. Dilution =

1000

Solubility

–

o –1 2 –1AgCl Ag Cl

61.9 76.3 138.2 ohm cm mol

Sp. Conductivity × dilution = oAgCl

–6

10002.30 10 138.2

s

Solubility (gram per litre) = 2.382 × 10–3 g lt–1

7. Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at 25°C.

H2(1 atm) | 0.5 M HCOOH | | 1 M CH

3COOH | (1 atm)H

2

(Ka for HCOOH and CH

3COOH are 1.77 × 10–4 and 1.8 × 10–5 respectively)

Sol.–4 –2

HCOOH a[H ] K C 1.77 10 0.5 0.9407 10 M

3

–5 –3

CH COOH[H ] 1.8 10 1 4.2426 10 M

–3

RHS

cell 10 10 –2

LHS

[H ] 4.2426 10E 0.0591 log 0.0591 log

[H ] 0.9407 10

Ecell

= –0.0204 V

8. For the cell reaction, Mg | Mg2+(aq) | | Ag+(aq) | Ag.

Calculate the equilibrium constant at 25°C and maximum work that can be obtained by operating the cell.

2

o o

Mg /Mg Ag /Ag[ Given E –2.37 V, E 0.80 V]

Sol.o

cellE 0.80 2.37 3.17 V

o

cell

c

nE 2 3.17logK 107.2758

0.0591 0.0591

Kc = 1.89 × 10107

Maximum work = o

cell– G nFE = 2 × 96500 × 3.17 = 611.81 kJ



9. Under standard conditions

(i) Will Cu reduces Ag+ to Ag? 2

o o

Ag /Ag Cu /Cu(E 0.799 V, E –0.337 V)

(ii) Will Fe3+ be reduced to Fe2+ by Sn2+? 3 2 2 4

o o

Fe /Fe Sn /Sn(E 0.771V, E –0.13V)

(iii) Would you use a silver spoon to stir a solution of Cu(NO3)2?

Sol. (i) Yes, o

cellE of Cu | Cu2+ | | Ag+ | Ag is positive

(ii) Yes, o

cellE of Pt, Sn4+ | Sn2+ | | Fe3+ | Fe2+, Pt is positive

(iii) Yes, the reaction between silver and Cu2+ does not occur

10. A weak monobasic acid is 5% dissociated in 0.01 mol/lt solution. The limiting molar conductivity at infinite

dilution is 4 × 10–2 ohm–1 cm2 mol–1. Calculate conductivity of 0.05 molar solution of acid.

Sol. Ka = C2 = 0.01 × (0.05)2 = 2.5 × 10–5

For C = 0.05 M,

5

aK 2.5 10

C 0.05

= 0.0223

2M

Mo

M

0.0223 4 10

⇒

11. 500 ml CuSO4 solution was electrolysed using a current of 2 amp (efficiency = 75%) for 60 min. Calculate

the pH of solution at the end of electrolysis. (Assume initial pH = 7)

Sol. When CuSO4 solution is electrolysed then H

2SO

4 is formed.

Equivalents of H2SO

4 =

1 752

96500 100 × 60 × 60 = 0.0559

2 4H SON =

0.0559

500

1000

= 1.12 × 10–1

Now, [H+] = 1.12 × 10–1

pH = – log [H+] = – log 1.12 × 10–1

pH = 1 – 0.05 = 0.95

12. A constant current flowed for 30 min through a solution of KI oxidising the iodide ion to iodine. At the end of

experiment, the iodine was titrated with 10 ml 0.075 M Na2S

2O

3 solution. Calculate the strength of current.

Sol. t = 30 min = 30 × 60 = 1800 s

KI K++I–

H2O H++HO–

(Cathode) 2H+ + 2e H2

(Anode) 2I– I2 + 2e–

eq. (I2) = eq. (Na

2S

2O

3) = 10–3 × 10 × 0.075 = 0.75 × 10–3

w

E =

it

F i =

wF

Et =

–3(0.75 10 ) (96500)

30 60

i = 0.04 A



13. A direct current of 3.0 amp (efficiency 75%) was passed through 400 ml 0.2 M Fe2(SO

4)3 solution for a period

of 60 min. The resulting solution in cathode chamber was analysed by titrating against acidic KMnO4 solution

20 ml of KMnO4 required to reach the end point. Determine the molarity of KMnO

4 solution.

Sol. Equivalents of Fe2+ formed = 1 75

396500 100

× 60 × 60 = 0.0839

Equivalents of Fe2+ = Equivalents of KMnO4

0.0839 = 4KMnO

N × 20 × 10–3

4KMnON =

20

0839.0× 103

4KMnON = 4.195

In acidic medium, n-factor for KMnO4 is 5.

4KMnOM =

4.195

5 = 0.84 M

14. H2O

2 can be produced by successive reactions

2NH4HSO

4 H

2 + (NH

4)2S

2O

8

2244

60%

28224 OHHSO2NHO2HOS)(NH

The first reaction is an electrolytic reaction and second is steam distillation. What amount of current

would have to be used in first reaction to produce enough (NH4)2S

2O

8 to yield 100 gm H

2O

2 per hour?

(Assume current efficiency is 75%)

Sol. Given

(NH4)2S

2O

8 + H

2O 2NH

4HSO

4 + H

2O

2

228 g 34g

34 gm of H2O

2 is produced by 228 g (NH

4)2 S

2O

8

100 g of H2O

2 will be produced by

228

34× 100 = 670 g

Since efficiency is 60% hence mass of (NH4)2 S

2O

8 required will be =

670 100

60

= 1116.66 gm

Equivalent mass of (NH4)2 S

2O

8 may be calculated

2NH4

–

4SO (NH

4)2S

2O

8+2e–

4 2 82NH S O

E = M

2 =

228

2 = 114

From I- law

w = E i t

F

1116.66 =

75i 3600 114

100

96500

i = 350.09 A



15 During discharge of a lead storage battery the density of sulphuric acid fell from 1.3 to 1.15 gm/ml. Sulphuric

acid of density 1.3 gm/ml is 40% H2SO

4 by wt and that of density 1.15 gm/ml is 20% by wt. The battery holds

4.0 L of the acid and volume remained practically constant during the discharge. Calculate the number of amp-

hours for which the battery must have been used.

Pb + SO4

–2 PbSO4 + 2e– (discharging)

PbO2 + 4H+ + SO

4

–2 + 2e– PbSO4 + 2H

2O (discharging)

Sol. Adding the charging and discharging reactions.

We get

Pb + PbO2 + 4H+ +

2–

42SO 2PbSO

4 + 2H

2O

2 4H SON =

2 4H SOM (since

2–

42SO requires 2-electrons)

i.e., Normality = Molarity

Before discharge

2 4H SOM =

40 1.3 1000

98 100

= 5.3 M

Moles of H2SO

4 = 5.3 × 4 = 21.22

After discharge

2 4(ii)H SOM =

20 1.15 1000

98 100

= 2.34 M

Moles of H2SO

4 = (2.34) (4) = 9.38

Moles or equivalents of H2SO

4 Used = 21.22 – 9.38 = 11.84

i × t = (11.84) × 96500 = 317 ampere s.

16. A dilute solution of NaCl was placed between two Pt-electrodes 8 cm apart, across which a potential of 4 V

was applied. How far would the Na+ move in 2.5 hours? Ionic conductance of Na+ at infinite dilution at 25°C

is 50.11 mho cm2.

Sol. º Na+ = 50.11 cm2/eq

Ionic mobility = ºK

F

=

50.11

96500

5.19×10–4 cm2 sec–1 volt–1

Potential gradient applied = 4

8 = 0.5 volt/cm

Speed of Na+ = ionic mobility × potential gradient

= (5.19 × 10–4) × 0.5

= 0.259 × 10–3 cm/sec

Distance traveled by Na+ in 2.5 hr

= 0.259 × 10–3 × 2.5 × 60 × 60

= 2.33 cm



17. The resistance of a conductivity cell filled with 0.01 N KCl at 25°C was found to be 500 . The specific

conductance of 0.01 N KCl at 25°C is 1.41 × 10–3 –1 cm–1. The resistance of same cell filled with

0.3 N ZnSO4 at 25°C was found to be 69 . Calculate the cell constant, equivalent and molar conductivities

of ZnSO4 solution.

Sol. K = 1

R×

l

A

(1.41 × 10–3) = 1

500 ×

l

A

l

A = 1.41 × 10–3 × 500 = 0.705

m

= K 1000

M

=

1

R ×

l

A ×

1000

M

= 1

69 × 1.41 × 10–3 × 500 ×

1000

0.3

= 34.1 –1 cm–1 eq–1.

Molar conductivity = 68.2 –1cm–1mol–1.

18. The equivalent conductance of 0.1 N of H3PO

4 at 18°C is 96.5 –1 cm2 eq–1. If °

HCl = 378.3, °

NaCl = 109,

–12–1

PONaH eqcmΩ70Λ 42

respectively, calculate the degree of dissociation and dissociation constant for

the reaction :

H3PO

4 H+ + H

2PO

4

–

Sol.3 4

o

H PO =

2 4

o o o

HCl NaH PO NaCl– = 378.3 + 70 – 109 = 339.3

= 3 4

3 4

H PO

o

H PO

= 96.5

339.3 = 0.2844

K =

2C.

(1– )

=

20.1 (.2844)

(1– .28844)

= 1.13 × 10–2

19. Ag(s) | AgCl (saturated salt), KCl (C = 0.025) || KNO3, AgNO

3 (C = 0.2) | Ag

The emf of above cell is 0.43 V.

(a) Write down the cell reaction.

(b) Calculate the solubility product of AgCl.

(Antilog (4.2758) = 1.887 × 104)

Sol. The given cell

Ag(s) |AgCl (Saturated Salt), KCl (C = 0.025 M)||AgNO3(C=0.2)|Ag

E = 0.43V

(a) Reaction in L.H.C. and R.H.C.

Ag Ag+ + e– (Anode)

Ag+ + e– Ag (Cathode)

(b)Ag/Ag

E = oAg/AgE –

0.0591

1 log [Ag+]

L

Ag /AgE =

o

Ag /AgR

0.0591 1E – log

1 [Ag ]



EMF of cell = Ag/Ag Ag /Ag

E E

E =

o o

Ag/Ag Ag /Ag(E E )

0

⇓ – 0.0591

1 log

L

R

[Ag ]

[Ag ]

E = 0.0591

1

log

L[Ag ]

(0.2)

– 0.43

0.0591 = log

[Ag ]

(0.2)

0.43

or0.0591

= log

0.2

[Ag ]

0.2

[Ag ] = 1.887 × 106 ; [Ag+] = 1.05 × 10–10 M.

Therefore, [Ag+] = 1.05 × 10–10 M

KCl –

0.025 0.025

K Cl

Ksp

(AgCl) = (0.025) (1.05 × 10–10)

Ksp

= 2.64 × 10–10

20. Pt | H2(1 bar), H+ || KCl (1.0 M saturated) | Hg

2Cl

2 | Hg was used to measure the pH of 0.05 M acetic acid

in 0.04 M CH3COONa. Calculate the cell potential.

–3 2 2

5

a(CH COOH) Hg Cl /Hg, ClK 1.8 10 , E 0.28 V

Sol. [H+] = Ka ×

[Acid]

[Salt] = 1.8 × 10–5 ×

0.05

0.04 = 2.25 × 10–5

ECell

= –2 2 2

o o

Hg Cl /Hg, Cl H /H

0.059 [H ](E – E ) – log

1 1

ECell

= 0.28 – 0.059

1 log 2.25 × 10–5

ECell

= 0.28 + 0.275 = 0.555 V

21.

Ag ZnSalt bridge

0.5 M AgNO3

Zn(NO )3 2

(A) (B)

V

(a) If the cell emf is – 1.58 V, what is the concentration of Zn+2?

(b) If NH3 added to half cell A, how emf of cell will change?

V0.76E V,0.8E/ZnZn/AgAg 2 Antilog (0.6768) = 1.4768

Sol. (a) [Zn+2] = 0.054 M; (b) emf increases



22. 2 2Cu /Cu Zn/Zn

E 0.34 V, E 0.76 V

A cell formed by the combination of Cu and Zn.

(a) When CuSO4 is added to Cu+2 compartment what is the effect on emf of cell?

(b) When NH3 is added to Cu+2 compartment what is the effect on emf of cell?

(c) When ZnSO4 is added to Zn+2 compartment what is the effect on emf of cell?

(d) When Zn+2 is diluted what is the effect on emf of cell?

Sol. 2

o

Cu /CuE = 0.34V; 2

o

Zn/ZnE = 0.76 V

EMF of the cell is

Eº = 2

o

Zn/ZnE + 2

o

Cu /CuE

(0.34) + (0.76) = 1.1V

(a) According to Nernst equation

E = Eº – 0.0591

2 log

2

2

[Zn ]

[Cu ]

[Cu2+] increases

Hence log terms will decreases and therefore EMF increases.

(b) When NH3 is added it combines with Cu2+ and, hence, [Cu2+] decreases

4NH3 + Cu2+ [Cu (NH

3)4]2+.

With decrease in [Cu2+], log tem increase hence EMF decreases.

(c) [Zn2+] will increase as result of which EMF decreases.

(d) Zn2+ dilution means concentration decreases and EMF increases.

Cls Jeead-15-16 Xii Che Target-5 Set-2 Chapter-3

Documents

Transcript of Cls Jeead-15-16 Xii Che Target-5 Set-2 Chapter-3