Cls Jeead-15-16 Xii Che Target-5 Set-2 Chapter-3
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Transcript of Cls Jeead-15-16 Xii Che Target-5 Set-2 Chapter-3
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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)
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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
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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
74 Electrochemistry Solution of Assignment (Set-2)
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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
75Solution of Assignment (Set-2) Electrochemistry
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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)
Applying the equation
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
Applying the equation
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
76 Electrochemistry Solution of Assignment (Set-2)
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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.
77Solution of Assignment (Set-2) Electrochemistry
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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–
78 Electrochemistry Solution of Assignment (Set-2)
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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
79Solution of Assignment (Set-2) Electrochemistry
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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
80 Electrochemistry Solution of Assignment (Set-2)
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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
81Solution of Assignment (Set-2) Electrochemistry
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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.
82 Electrochemistry Solution of Assignment (Set-2)
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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
83Solution of Assignment (Set-2) Electrochemistry
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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
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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
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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.
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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–
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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
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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
Sol. Answer (2, 3, 4)
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
Sol. Answer (2, 3, 4)
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
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Sol. Answer (1, 3, 4)
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
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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
Sol. Answer (1, 2, 4)
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
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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
Sol. Answer (2, 3, 4)
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
Sol. Answer (1, 2, 3)
At cathode always reduction takes place and at anode always oxidation takes place.
Hence (4) will not the correct statement.
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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
Sol. Answer (2, 3, 4)
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
Sol. Answer (1, 2, 3)
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
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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
Sol. Answer (2, 3, 4)
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
Sol. Answer (1, 2, 4)
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.
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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
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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
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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)
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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]
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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 )
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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.
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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.
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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.
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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.
2. Match the following
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
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(D)11.2
22.4 =
1
2
2Hn
w
E =
it
F it =
w
E F
= w
2M
⎛ ⎞⎜ ⎟⎝ ⎠
F = 1
22
F = F
3. Match the following
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
4. Match the following
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)
5. Match the following
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)
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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)
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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.
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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.
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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⇒
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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
109Solution of Assignment (Set-2) Electrochemistry
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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
110 Electrochemistry Solution of Assignment (Set-2)
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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
111Solution of Assignment (Set-2) Electrochemistry
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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
112 Electrochemistry Solution of Assignment (Set-2)
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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 ]
113Solution of Assignment (Set-2) Electrochemistry
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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
114 Electrochemistry Solution of Assignment (Set-2)
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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.