Electrochemistry
12.1
Section A
Level - I
1. (a) Pt | Co2+aq|Co3+aq || Zn2+(aq)| Zn(s)
E° cell = –0.76–(+1.82) = –2.58 V
(b) Pt(s)| Br2 (l) | Br—(aq) || Cl—(aq)|Cl2 (g)|Pt
E° cell = +1.36– (+1.08) = +0.28 V
2. (a) Copper as EoCu2+ | Cu > EoPb2+ | Pb
(b)
(c)
(d)
(e)
3. (a)
(b)
Pb (s) + Cu 2+ (aq) ® Pb 2+ (aq) + Cu (s)
Copper cathode
From anode (Pb) to cathode (Cu)
Towards Cu electrode half-cell.
Strongest reducing agent ® Al
Strongest oxidising agent ® Fe3+
4. E °cell = E °I
2 /I
–
– E °Br /Br –
2
= + 0.54 – (+1.08) = –0.54 V
As E°cell is negative, the cell is non-spontaneous. Br– can not reduce I2 to I–.
5. (a) E °cell = E °Fe3+ /Fe2+ – E °Ag + /Ag
= + 0.77 – (+0.80)
= – 0.03V (non– spontaneous)
(b) E °cell = +0.34 – 0
= +0.34 V (non-spontaneous)
6. (a) D G ° = – nFE °cell
D G ° = –1 ´ 96500 ´ (– 0.03)
D G ° = 2895 J
(b) DG ° = –2 ´ 96500 ´ (+ 0.34)
D G ° = – 65620 J
7. E °cell = E °Fe3+ /Fe2+ – E °Ni 2+ /Ni
E °cell = + 0.77 – (–0.25) = +1.02 V
æ nE ° ö
K = antilog ç
è 0.0591÷ø
æ –2 ´ 1.02 ö
= 3.0×10–35
K = antilog ç
è 0.0591 ÷ø
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12.2
Physical Chemistry for IIT-JEE
8. The cell is
Pt | H 2 (g) | H + (7.58 ´ 10 –8 M) || Ag + (1M)|Ag (s)
E °cell= +0.80 V
0.0591
[H + ]
log
n
[Ag + ]2
E = +0.80 – 0.0591 log (7.58 × 10—8)
E = +1.22 V
E = E° –
9. E = E ° – 0.0591 log
[H + ]
[Ag + ]
0.98 = +0.80 – 0.0591 log [H+]
log[H+] = –3.0457
[H+] = 9.0 ´ 10–4 M
10. The cell is
Zn (s) | Zn 2+ (0.01M)||Cr 3+ (0.001M)|Cr(s)
E °cell = –0.74 –(–0.76)
E °cell = +0.02 V
E = E° –
0.0591
[Zn 2+ ]3
log
6
[Cr3+ ]2
E = + 0.02 –
0.0591
(10 –2 )3
log
6
(10–3 )2
Ecell = +0.02 V
DG = –nFE
DG = – 6 ´ 96500 ´ 0.02
DG = –11580 J
DG = –11.58 kJ
11. (a) 2Br ® Br2 + 2e –
(b) 2H2O + 2e– ® H2 + 2OH –
–
12. Cr3+ + 3e– ® Cr
1 mol Cr3+ requires 3F
\ 0.20 mol Cr3+ will require 3 ´ 0.20 = 0.6 F
or 0.6 ´ 96500 = 57900 C
13. Q = It = 5.0 ´ 10 ´ 3600
Q = 180000 C
180000
Faradays required =
= 1.865F
96500
2+
–
14. Ni + 2e ® Ni
58 g of Ni requires 2F
Chapter-12 OLC.p65
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12.3
Electrochemistry
\
29.4 g Ni will require
2 ´ 29.4
= 1.00 F
58.7
15. Cr 2+ + 2e– ® Cu
w = zIt
(63.5 / 2)
´ 20.0 ´ t
96500
t = 3799.25 s
t = 63.3 min
25.0 =
16. Cd 2+ + 2e– ® Cd
w = zIt
(112 / 2)
mass of Cd =
´ 0.80 ´ 2.5 ´ 3600 = 4.18g
96500
17. The cell is
Pt | H 2 (g) | H + (aq) || Cu 2+aq |Cu (s)
E = + 0.34 –
0.0591
[H + ]2
log
2
[Cu 2+ ]
Addition of NaOH to hydrogen half-cell decreases the conc. of H+ and decreases the value of QC and
0.0591
logQC and increases E value.
2
18. 2H2O + 2e– ® H2(g) + 2OH –
Eo = – 0.8277 V
H2(g) ® 2H + + 2e–
Eo = 0
logQC and
2H2O ® 2H+ + 2OH –
Eo = – 0.8277
æ nE ° ö
K = antilog ç
è 0.0591 ÷ø
æ – 2 ´ 0.8277 ö
K = antilog ç
è 0.0591 ÷ø
K = 9.727 ´ 10–29
For the reaction
XXV H + + OH –
H2O WXX
KW = (9.77 ´ 10–29)1/2
KW = 9.88 ´ 10–15
» 1.0 ´ 10–14
2+
+
19. Pb (s) | Pb (aq) || H aq (1M)| H 2(g) |Pt
E o = 0 – (– 0.13) = + 0.13 V
E = E° –
Chapter-12 OLC.p65
3
0.0591
[Pb2+ ]
log + 2
2
[H ]
12/7/07, 2:13 AM
12.4
Physical Chemistry for IIT-JEE
0.0591
log[Pb2+ ]
2
[Pb2+] = 3.01 ´ 10—13 M
2
Ksp (PbCrO4) = [Pb2+] [CrO4—
]
0.50 = +0.13 –
Ksp = 3.01 ´ 10–13 ´ 0.1
Ksp = 3.01 ´ 10–14
20. Current strength =
Power 4.0
=
A
voltage 1.5
Pb ® Pb2+ + 2 e –
w = zIt
208 / 2 40
´
´t
96500 1.5
t = 695.9 s
t = 0.193 hr.
20 =
Level - II
1. (i) WCr = zIt
=
Equivalent weight
.It
96500
=
52 / 3
´ 15 ´ 45 ´ 60
96500
WCr = 7.27 g
(ii) WCl2 = Equi. wt . .I ´ t
96500
=
71/ 2
´ 15 ´ 45 ´ 60
96500
Wcl2
15 ´ 45 ´ 60
= n Cl =
2
71
2 ´ 96500
or
n Cl = 0.209 mol
2
VCl =
2
nRT 0.209 ´ 0.082 ´ 546
=
P
1
VCl = 9.4 L
2
2. During reduction of gold from AuCl4— by electrolysis, it is essential to stir the solution rapidly
because:
(i) AuCl4—moves towards cathode which is negatively charged. So, like charges repel each other.
Hence, to overcome repulsion, stirring is important
Chapter-12 OLC.p65
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12.5
Electrochemistry
(ii) Au3+ ions are very heavy and many tend to settle at the bottom of the cell rather than moving
towards the cathode. But, if the solution is stirred rapidly, these ions get pushed towards the
cathode and hence reduction occur.
3. (i) w = zQ
w=
Equi. wt .
.Q
96500
Q=
1.314× 96500
= 1931C
197 / 3
1931
I = 3.22 A
10× 60
4. (a) Reactions involved are :
(ii) I = Q/t =
Anode :
2Cl – ® Cl2 + 2 e –
Cathode: 2 H + + 2e – ® H 2
(b) Current actually being passed
=
62
´ 25 = 15.5 A
100
3
Wcl2 = 10 =
71 ´ 15.5 ´ t
2 ´ 96500
t = 175375 s
t = 48.72 hr
(c) According to the equation given
2Cl – (aq) + 2H 2 O ® 2OH – (aq) + H2(g) + Cl 2(g)
mol of OH – = mol of Cl–
1000
= 28.17
35.5
volume = 20 L
\ Molarity w.r.t. OH – ions
=
28.17
= 1.41 M
20
5. Current actually passed
=
95
× 2.68 = 2.546 A
100
Reaction : LiCl + H 2 O ® LiOH + HCl
In electrolysis of LiCl solution, H+ and Cl – ions are liberated.
Therefore faradays laws are applicable only to these two ions :
=
\
Chapter-12 OLC.p65
WCl – =
35.5
´ 2.546 ´ 3600
1× 96500
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12/7/07, 2:13 AM
12.6
Physical Chemistry for IIT-JEE
2.546 ´ 3600
= 0.095
1 ´ 96500
Moles of LiOH = 0.095
Mass of LiOH = 0.095 ´ 24 = 2.27 g
6. By Faraday’s second law i.e.,
w1 w 2
=
E1 E 2
moles of Cl– =
wH
2
E H2
=
w Cu 0.635
=
´2
E Cu
63.5
w H = 0.02g or 0.01 mol H 2
2
V=
nRT 0.01 ´ 0.082 ´ 273
=
= 0.224 L
P
1
Also,
1
O + H2
2 2
H2 O ®
1
´ 0.01 = 0.005
2
Volume of O2 = 0.005 ´ 22.4 = 0.112 L
7. Since, the same quantity of electricity is passed through both, Faraday’s second law can be applied :
n O2 =
\
WAg
E Ag
=
WAu
E Au
2.158 1.314
=
108
E Au
1.314×108
= 65.76
2.158
Atomic weight of gold = 197
E Au =
E=
Atomic wt .
oxidation state
oxidation state of gold =
197
= 3.0
65.76
8. 2 NH 4 HSO4 ® H 2 + (NH 4 )2 S2 O8
(NH 4 )2 S2 O8 + 2 H 2 O ® 2 NH 4 HSO 4 + H 2 O2
........(1)
........(2)
Let the molar mass of (NH4)2 S2O8 = M
\ From equation (2)
34 g H2O2 comes from Mg
Chapter-12 OLC.p65
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12.7
Electrochemistry
\
100 g of H 2 O2 comes from
100 M
34
50 M
g of intermediate
17
Applying first law of electrolysis to (1) :
=
Wintermediate =
I=
M I ´ 3600 50 M
=
´
2
96500
17
50 ´ 2 ´ 96500
= 157.67 A
3600 ´ 17
Since current efficiency is 50%
50
´ current passed = 157.67
100
current passed = 315.34 A
9. Weight of NaClO4 deposited = 245 g
Equivalent weight of NaClO4 =
122.5
= 61.25g
2
w = zQ
245 =
61.25
´Q
96500
245 ´ 96500
= 386000 C
61.25
Since anode efficiency is 60%
Q=
60
Q ¢ = 386000
100
Q’ = 643333.33 C = 6.43 ´ 105 C
Therefore,
10. Case I Suppose the acidic salt is CuCl2, then using second law, we have
WCu WCl2
=
E Cu
E Cl2
WCl =
2
0.4
´ 71/ 2
63.5 / 2
WCl2 = 0.447
n Cl2 = 6.296 ´ 10–3
VCl2 = 0.141 L
When electrolysis is continued for 7 more minutes, then electrolysis of water takes place.
Chapter-12 OLC.p65
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Physical Chemistry for IIT-JEE
WH2 = z I t =
WH2
2
= n H2 =
2 ´ 7 ´ 1.2 ´ 60
2 ´ 96500
7 ´ 1.2 ´ 60
2 ´ 96500
n H = 2.61×10–3
2
VH = 0.0293 L
2
and the volume of O2 will be 0.0585 L
\ Total volume of gases evolved
= 0.141 + 0.0585 + 0.0293
= 0.2288 L or 228.8 mL
Case II Suppose the acidic salt is CuSO4 then, O2 gas evolved at cathode
WCu WO2
=
E Cu
E O2
WO2 =
WO2
32
0.4
32
´
63.5 / 2 4
= nO2 =
n O2 =
0.4 2
´
63.5 4
6.3 ´ 10 –3
2
n O = 3.15 ´ 10–3
2
VO = 0.07055 L
2
When electrolysis is continued for 7 more minutes, then electrolysis of water takes place.
wH = z I t
2
2 7 × 60×1.2
´
2
96500
=
w H2
2
= n H2 =
7 ´ 60×1.2
= 2.61 ´ 10–3
2 ´ 96500
VH = 0.0585 L
2
and VO2 = 0.0293 L
Total volume of gases evolved
= 0.07055 + 0.0585 + 0.0293 = 0.158 L
11. Actual current passed = 60 ×15 = 9 A
100
Chapter-12 OLC.p65
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12/7/07, 2:13 AM
12.9
Electrochemistry
(i) WNi = zIt
58.5
´ 9 ´ 60 ´ 60 = 9.82 g
2 ´ 96500
(ii) Area of two faces = 2 ´ (4.0) ´ (4.0) = 32 cm2
d = 8.9 g/cm3
=
volume of 9.82 g Ni =
9.82
= 1.11 cm3
8.9
1.11
= 0.035 cm
32
(iii) Current efficiency w.r.t. H2 = 40%
\ current passed = 0.4 ´ 15 = 6 A
t = 1 hr = 3600 s
Q = It = 6 ´ 3600 = 21600 C
Thickness of plating =
21600
96500
2F liberates 1 mol H2 = 22.4 L
Number of faradays =
216
22.4 ´ 216
F will liberates
= 2.51 L H 2
965
2 ´ 965
\
12. WIr = zIt
0.072 =
E Ir
´ 0.01 ´ 3 ´ 3600
96500
WIr = 64.34 g
E=
MW
charge number
MW
192
=
=3
E
64.34
13. Original mass of entire solution
= 1.261 g/mL ´ 1000 mL = 1261 g
Charge number =
34.6
´ 1261 = 436.3 g
100
\ weight of water present = 1261 ´ 436.3 = 824.7 g
As seen from the equation, 2 moles of H2SO4 use 2F and produce 2 moles H2O. Let the number of
moles of H2SO4 consumed be x \ mass of solution left
(436.3 — 98x) + 824.7 + 18 x = (1261 – 80 x) g
But this solution is 27% in acid
Weight of acid =
\
Chapter-12 OLC.p65
27
(1261– 80 x ) = 436.3 – 98x
100
9
12/7/07, 2:13 AM
12.10
Physical Chemistry for IIT-JEE
\
340.47 – 21.6x = 436.3 – 98x
Þ x = 1.25
mass of H2SO4 consumed = number of faradays
1.25 F of electricity is consumed.
14. Initial molarity of H2SO4, M1 =
10 x d
MW
10 ´1.294 ´ 39
= 5.15
98
Initial moles of H2SO4 = 5.15 ´ 3.5
10 x d
Final molarity of H2SO4, M2 =
MW
M1 =
10 ´ 20 ´ 1.139
= 2.32
98
Final moles of H2SO4 = 2.32 ´ 3.5
Number of moles of H2SO4 consumed
= 5.15 ´ 3.5 – 2.32 ´ 3.5 = 9.91
nH2SO4 = number of faradays transferred
number of faradays = 9.91
number of coulombs = 9.91 ´ 96500
= 956315 C
= 956315 Amp sec
M2 =
956315
Amp – hr
3600
= 265.64 Amp–hr
15. The increase in mass at cathode is due solely to copper
=
\
WCu deposited = 22.011 g
22.011
= 0.347 mol
63.5
22.260 – 22.011 = 0.249 g is the mass of Fe, Au and Ag.
Only the iron and copper are oxidised.
The gold and silver fall to the bottom in the anode mud. Since each of the active metal requires 2 mol
electrons per mol metal there must be
t = 483 s
I = 140 A
Q = 67620 C
nCu =
67620
= 0.7 F
96500
The reactions at anode are
No. of faradays =
Fe ® Fe2+ + 2e –
56 g Fe ® 2 F
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12.11
Electrochemistry
x g Fe ®
\
and
2x
x
=
F
56 28
Cu ® Cu 2+ + 2e –
63.5 g Cu ® 2 F
y g Cu ®
\
But
2y
F
63.5
xF 2 yF
+
= 0.7 F
28 63.5
with y = 22.011 g, we have
x 2 ´ 22.011
+
= 0.7
28
63.5
Solving for x gives
x = 0.21 g
mass of iron = 0.21 g
Total mass = 22.011 + 0.21 = 22.221
22.011
´ 100 = 99.05%
22.221
0.21
´ 100 = 0.95%
% Fe =
22.221
% Cu =
16. During electrolysis of aqueous NaCl, H2 gas liberates at cathode and 2 H + + 2e – ® H 2 and Cl2
gas liberates at anode 2 Cl – ® Cl 2 + 2 e – and NaOH is left in the solution (cell). The overall reaction is
2 NaOH + 2 H 2 O ® 2 NaOH + H 2 + Cl 2
Given : Q = It = 0.250 ´ 35.0 ´ 60
and number of faradays =
0.250 ´ 35.0 ´ 60
= 5.44 ×10 –3 F
96500
2 F ® 1 mol of H 2 ® 2 mol OH –
\
5.44 ´ 10 –3 F will produce
2 ´ 5.44 ´ 10–3
= 5.44 × 10—3 mol
2
5.44 ´ 10 –3
= 0.0136 M
0.4
p(OH) = –log [OH—] = –log(0.0136)
p(OH) = 1.87
p(OH) = 14 – 1.87 = 12.13
[OH – ] =
Chapter-12 OLC.p65
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12/7/07, 2:13 AM
12.12
Physical Chemistry for IIT-JEE
17. The reaction during electrolysis is
2 NaCl + 2 H 2 O ® 2 NaOH + Cl 2
and during titration
NaOH + HCl ® NaCl + H 2 + H 2 O
nNaOH formed = nAcid used
n H = n acid ´
2
1
2
–3
= 0.25 ´ 15.5 ´ 10 ´
1
=1.9375 ´ 10 –3
2
2 H+ + 2 e– ® H2
mol of H 2 =
It
2 ´ 95600
2 ´ 96500 ´ 1.9375 ´ 10–3
25 ´ 60
I = 0.249 A
I=
18. Pdry gas = 101 kPa = 1 atm
T = 300 K, V = 0.288 L
nH =
2
PV
1 ´ 0.288
=
= 0.119
RT 300 ´ 0.0821
2 H+ + 2 e– ® H2
1 mol of H2 = 2 mol e—= 2F
\ 0.119 mol H2 = 2 ´ 0.119 = 0.238 F = 0.238 mol e–
0.238 ´ 6.023 ´ 1023 e– = 0.238 ´ 96500 C
96500
= 1.602 ´ 10 –19 C
6.023 ´ 10 23
19. Volume of Cr deposited = area ´ thickness
\
1 electron =
= 104 cm2 ´ 0.05 ´ 10–1 cm = 50 cm3
mass of Cr deposited = (volume) ´ (density)
= 50 cm3 ´ 7.19 g/cm3 = 359.5 g
I=
W 359.5 ´ 3 ´ 96500
=
zt
52 ´ 25 ´ 60
I = 1.33 ´ 103 A
20. WAl = zIt =
Chapter-12 OLC.p65
E
´ It
96500
12
12/7/07, 2:13 AM
12.13
Electrochemistry
WAl =
9 ´ 1.3 ´ 105 ´ 60
96500
WAl = 726.9 g Al
Weight of Al deposited in 100 cells
= 7.269 ´ 104 ´ 24 ´ 60g
= 1.047×105 kg per day
Mass of carbon =
E
´ It
96500
12 / 4
×1.3×105 × 24× 60× 60
96500
= 3.49×105 g per day
=
= 349.5 kg per day
21. (i)
Anode :
H2(g) + 2Cl – (aq) ® 2 HCl (aq) + 2 e –
Cathode:
AgCl(s) + e – ® Ag(s) + Cl – (aq)
Net reaction: H 2(g) + 2 AgCl(s) ® 2 HCl (aq) + 2 Ag (s)
Fe2+ (aq) ® Fe3+ (aq) + e –
(ii) Anode :
Cathode:
Sn 4+(aq) + 2 e – ® Sn 2+(aq)
Net reaction: Sn 4+ (aq) + 2 Fe 2+ ® 2 Fe3+ (aq) + Sn 2+ (aq)
(iii) Anode :
Cu
(s)
® Cu 2+ (aq) + 2 e –
Mn 4+(aq) + 2e – ® Mn 2+(aq)
Cathode:
Net reaction: Cu (s) + Mn 4+ (aq) ® Cu 2+ (aq) + Mn 2+ (aq)
Ag(s) + HCl(aq) ® AgCl(s) + H +(aq) + e –
(iv) Anode :
AgBr(s) + H + + e – ® Ag (s) + HBr(aq)
Cathode:
Net reaction: AgBr(s) + HCl(aq) ® AgCl(s) + HBr(aq)
(v) Anode :
Pb(s) ® Pb 2+(aq) + 2e –
Cathode:
2 H + (aq) + 2 e – ® H 2(g)
Not reaction: Pb(s) + 2 H +(aq) ® Pb2+(aq) + H 2(g)
Chapter-12 OLC.p65
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12.14
Physical Chemistry for IIT-JEE
Cd (s) ® Cd 2+ (aq) + 2 e –
22. (i) Anode :
I2(s) + 2e – ® 2I – (aq)
Cathode:
Cd (s) |Cd 2+ (aq) || I – (aq) | I 2(s) | Pt (s)
Ni (s) ® Ni 2+ (aq) + 2e –
(ii) Anode :
Fe3+ (aq) + e – ® Fe 2+ (aq)
Cathode:
Ni (s) | Ni 2+(aq) || Fe 3+(aq) | Fe 2+(aq) | Pt (s)
Cr(s) ® Cr3+ (aq) + 3e –
(iii) Anode :
Cu 2+ (aq) + 2e – ® Cu (s)
Cathode:
Cr(s) |Cr 3+ (aq) || Cu 2+ (aq) |Cu (s)
1
H 2(g) + Cl – (aq) ® HCl(aq) + e –
2
(iv) Anode :
AgCl(s) + e – ® Ag (s) + Cl –
Cathode:
Pt |H 2(g) |HCl || AgCl(s) |Ag(s)
Na + OH – ® NaOH + e –
(v) Anode :
2 H + + 2 e – ® H 2(g)
Cathode:
Pt | Na (s) | H 2 O | NaOH | H 2 O |H 2(g) | Pt
23. (i) Cr2O72- will oxidise Sn2+ to Sn4+ as Cr2O72- has a higher E° value
Cr2O72– + 14H+ + 3Sn2+ ® 2Cr3+ + 3 Sn4+ + 7H2O
(ii) Ni(s) can not reduce Cr3+ as
E °Ni 2+ |Ni > E °Cr 3+ |Cr
(iii) Yes, Cd can reduce Hg22+ to Hg as E° Cd 2+ | Cd is less than E° Hg 2+ | Hg
2
2+
2+
Cd(s) + Hg2 (aq) ® Cd (aq) + 2Hg(1)
(iv) No, dilute HNO3 can not oxidise Au to Au3+.
24. Fe3+ + 3e– ® Fe
WFe =
E Fe
It
It = 56
96500
3 ´ 96500
WFe
It
4 ´ 3600
= n Fe =
=
56
3 ´ 96500 3 ´ 96500
nFe = 0.0497 mol
\
Fraction of a mole of iron
0.049
= 0.49
0.1
(originally nFe = 0.1 ´ 1 = 0.1 mol was present)
Chapter-12 OLC.p65
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12/7/07, 2:13 AM
12.15
Electrochemistry
25. Al can not be produced by electrolysis of an aqueous solution because E° Al3+ |Al is less than that of
E° H+ |H So, H2 gas will liberate at cathode instead of Al. Na3AlF6 is added to Al2O3 in
2
(i)
(ii)
(iii)
(iv)
order to lower the melting point of Al2O3
order to dissolve Al2O3
order to increase electrical conductivity of Al2O3
order to lower the fuel bill.
Fe2+ ® Fe3+ + e–
26. (i) Anode :
Cr2 O72- +14 H + + 6 e - ® 2Cr 3+ + 7 H 2 O
cathode :
E °cell = E° cathode - E °anode = +1.33 – (+0.77)
E °cell = + 0.56 V
E = + 0.56 –
0.0591
log
6
[Cr3+ ]2 [Fe3+ ]6
ïì
ïü
í 2+ 6
2+ 14 ý
ïî [Fe ] [Cr2 O 7 ][H ] ïþ
E = +0.56 –
0.0591
log
6
ìï (2)6 ×(0.71) 2 üï
í 6
14 ý
îï (1) ×(0.5)(3) þï
E = + 0.56 –
0.0591
log 1.349 ´ 10 -5
6
E = + 0.56 –
0.0591
- 4.87
6
E = + 0.61 V
(ii)
Anode :
Cathode :
Cd(s) ® Cd 2+(aq) + 2 e –
2+
Co(aq)
+ 2e – ® Co(s)
E° = E° cathode – E° anode
E° = –0.28 – (– 0.403) = + 0.123 V
E = + 0.123 –
0.0591
[Cd 2+ ]
log
2
[Co2+ ]
E = + 0.123 –
0.0591
æ 0.05 ö
log ç
è 2 ÷ø
2
E = + 0.123 –
0.0591
log ( -1.602)
2
E = +0.170 V
(iii)
Anode: 2 Cl – (aq) ® Cl2(g) + 2e –
Cathode : Hg 2 Cl 2(s) + 2e – ® 2 Hg (l) + 2 Cl – (aq)
Chapter-12 OLC.p65
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Physical Chemistry for IIT-JEE
E °cell = + 0.27 – (+1.36) = +1.09 V
0.0591
log (p Cl )
2
2
Ecell = -
0.0591
log (0.80)
2
E = – 1.087 V
XXV Zn2+(aq) + 2Ag(s)
Zn(s) + 2Ag+(aq) WXX
0.1
0
E = +1.09 –
27.
Initial
conc.
equilibrium
(0.1 – 2x)
x
æ nE ° ö
K = antilog çè
0.0591÷ø
æ 2×1.56 ö
= antilog ç
è 0.0591 ÷ø
K = 6.19 ´ 1052
As K is very high, the reaction goes to completion and 2 mol Ag+ gives 1 mol of Zn2+.
The conc. of Zn2+ at equlibrium is half of Ag+ ion i.e.
[Zn2+]aq = 0.05 M
But
[Zn 2+ ]
[Ag + ]2
= 6.19×1052
2
[Ag + ]eq
=
0.05
6.19×1052
[Ag + ]aq = 8.99×10 -28 M
mass of Zn which gets oxidised to Zn2+ is
0.05 × 0.08 × 65 = 0.26 g
mass of Zn left = 3.0 – 0.26
= 2.74 g
28. (a) E °cell = 0
E cell = 0 –
0.0591
æ 0.018 ö
log ç
è 1.20 ÷ø
1
Ecell = 0.1078 V
(b) NaCl + AgNO3 ® AgCl + NaNO3
Due to addition of NaCl solution, although moles of AgCl formed will same as those of AgNO3
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Electrochemistry
originally present, but volume of the solution will increase. As a result [Ag+] in the anodic halfcell will decrease.
Now,
E cell = - 0.0591 log
[Ag + ]
1.2
[Ag + ]
is now lesser than original Qc
1.2
\ log Q¢c will be higher negative value than log Qc
As a result, the product of this higher negative value with –0.0591 will be higher than that in 1st
case. As a result, EMF of the cell has increased.
\
Qc¢ i.e.
29. E °cell = +1.36 – (+0.34) = +1.02 V
0.0591
log[Cu 2+ ][Cl– ]2
2
1.22 = 1.02 – 0.02955 log [(0.1)2 × M1]
E cell = +1.22 =1.02 –
M1 = 1.705 ´ 10–5 M
XXV H + 2OH–
30. 2H2O + 2e– WXX
2
\
Taking other electrode as SHE :
XXV 2H+ + 2e–
H WXX
2
Eo = – 0.8277 V
E° = 0
Adding two gives
XXV 2OH – + 2H +
2H2O WXX
XXV OH– + H + + OH– + H+
2H2O WXX
XXV H3O+ + OH–
2H2O WXX
which is the required equation
\ E °cell = – 0.8277 – 0
= – 0.8277 V
Þ At equlibrium
nE °
0.0591
Also since for 2 mols of water, 2e– are used, n = 1
log Kw =
log Kw =
- 0.8277
0.0591
-0.8277 ö
Kw = antilog æç
è 0.0591 ÷ø
31.
Kw = antilog (–14.0050)
Kw = 1.01 ´ 10–14
XXV Zn2+(aq) + Ni(s)
Zn(s) + Ni2+(aq) WXX
mol at
eqm.
Chapter-12 OLC.p65
(0.5 – x)
17
x
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Physical Chemistry for IIT-JEE
and the cell producing this reaction is
Zn(s) | Zn2+(aq) || Ni2+(aq) | Ni(s)
And
E °cell = -0.25 – (–0.76)
= +0.51 V
æ nE ° ö
K = antilog ç
è 0.0591÷ø
æ 2×0.51ö
K = antilog ç
è 0.0591 ÷ø
K = 1.815 ´ 1017
Let 0.5 – x = y
x = 0.5 – y
\
[Ni2+] left at eqm =
y
0.5
[Zn 2 + ] 0.5 - y
=
=1.815 ´1017
y
[Ni 2 + ]
y = 2.75 × 10–18 moles
32. E = E ° –
0.0591
1
log + 2
2
[H ]
At equilibrium, Ecell = 0
2E°
= - log[H + ]-2
0.0591
2E°
= ( - 2) {- log[H + ]}
0.0591
– E°
= pH
0.0591
pH =
- - 0.414
0.0591
pH = 7.00
XXV Hg2+
WXX
2(aq)
2Hg(l) + 2Fe3+(aq)
33.
10–3 M
Initial conc.
at eqm.
0
0.046 ´ 10 M
-3
2
(0.477 × 10 M)
18
2Fe 2+(aq)
0
–3
é Hg 2+
ù é 2+ ù
0.477×10-3 0.954×10-3
2 û ë Fe û
K= ë
=
2
2
-3
é Fe3+ ù
0.046×10
ë
û
Chapter-12 OLC.p65
+
(0.954 ´ 10–3 M)
2
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Electrochemistry
K = 0.2052
E°=
0.0591
log K
2
0.0591
log (0.2052)
2
E° = –0.020 V
E°=
But – 0.020 = E °
Fe3+ |Fe 2+
– E°
Hg 2+
2 |Hg
= + 0.79 V
H2 (g) ® 2H+(aq) + 2e–
2Ag+(aq) + 2e– ® 2Ag (s)
34.
2
é H+ ù
0.0591
E = E° –
log ë û 2
2
é Ag + ù
ë
û
+ 0.855 = +0.80 – 0.0591 log [H+]
0.855 – 0.80
= – log[H + ]= pH
0.0591
pH = 0.931
35. Since, it is a concentration cell, E°cell = 0
\
E cell =
é 5×10-4 ù
-0.0591
log ê
ú
1
ëê M1 ûú
-0.154
= log 5×10-4 – log M1
0.0591
M1 = 0.201 M
pH = –log [H+] = – log (0.201)
pH = 0.697
36. (i) Anode :
Cathode :
Zn (s)
® Zn2+(aq) + 2e–
Cl2 (g) + 2e– ® 2Cl–(aq)
(ii) Qc = [Zn2+] [Cl–]2
Qc = (0.04) (5.0 × 10-3)2
Qc = 1.0 × 10-6
E = 2.12
= 2.12 + 0.0591 × 3
= 2.297 V
(iii) If [Cl–] = 0.50 M, then
Qc¢ = (0.04) (0.5)2 = 0.01
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Physical Chemistry for IIT-JEE
And E = 2.12 –
0.0591
log10-2
2
= 2.18 V
Ecell decreases because increase in Cl– conc. increases Qc and log Qc and decreases value.
0.0591
æ
ö
çè E – 2 log Qc ÷ø value.
(iv) Addition of NH3 leads to the formation of complex
XXV [Zn(NH ) ]2+
Zn2+ + 4NH WXX
3
3 4
and decreases the concentration of Zn2+ ion. As a result, Qc decreases and Ecell increases.
æ nE ° ö
(v) K = antilog ç
è 0.0591÷ø
æ 2×2.12 ö
K = antilog ç
è 0.0591 ÷ø
K = 5.53 ´ 1071
anode
37. Cu+ + Cu+
Cu + Cu2+
cathode
E° = 0.52 –(0.15) = +0.37 V
æ nE ° ö
K = antilog ç
è 0.0591 ÷ø
æ 1×0.37 ö
K = antilog ç
è 0.0591 ÷ø
K = 1.82 × 106
38.
Anode :
Cu(s) ® Cu+aq + e–
Cathode : Cu2+(aq) + 2e– ® Cu(s)
E° = -0.52 V
E° = +0.34 V
Adding two gives
Cu2+ + e– ® Cu+
DG for eq. (1) is
DG1 = –nFE° = +0.52 F
DG for eq. (2) is
DG2 = –0.68 F
DG for eq. (3) is
DG3 = DG1 + DG2
= –0.68 F + 0.52 F
= –0.10 F = –E°F
E° = + 0.16 V
Chapter-12 OLC.p65
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Electrochemistry
XXV Ag+ + Cl–
39. (i) AgCl(s) WXX
(aq)
(aq)
anode
AgCl(s) + Ag(s)
Ag+(aq) + Cl–(aq) + Ag+(aq)
cathode
E° = + 0.22 –(–0.80)
E° = –0.58 V
æ nE ° ö
Ksp = antilog ç
è 0.0591÷ø
æ -0.58 ö
= antilog ç
è 0.0591 ÷ø
Ksp = 1.53 × 10–10
XXV Hg2+ (aq) + 2Cl–
(ii) Hg Cl
WXX
2
2(s)
2
Hg2Cl2 (s) + 2Hg(1)
(aq)
XXV Hg 22 +( aq ) + 2Cl(aq)
+ 2Hg (l )
WXX
E° = +0.27 –(+0.79)
= – 0.52 V
æ - 2×0.52 ö
Ksp = antilog ç
è 0.0591 ÷ø
Ksp = 2.53 ´ 10–18 = 4x 3
where x is the solubility of mercurous chloride.
x = 8.58 ´ 10–7 M
40. E cell = E ° –
[Ag + ]1
0.0591
log
1
[Ag + ]2
But [Ag+]1 = [Ag+]2 = 2.1 × 10-6
Ecell = E° = 0 V
41. At anode : [Ag+] [Cl–] = Ksp(AgCl)
But [Cl–] = [KCl] = 0.2 M
Þ
2.8×10 -10
=1.4×10 -9 M
0.2
At cathoe : [Ag+] [Br–] = Ksp(AgCl)
[Ag + ]anode =
[Ag + ]cathode =
3.3×10-13
= 3.3×10-10 M
-3
10
E °cell = 0
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Physical Chemistry for IIT-JEE
\
E cell =
[Ag + ]anode
- 0.0591
log
1
[Ag + ]cathode
æ 1.4×10-9 ö
E cell = - 0.0591log ç
÷
è 3.3×10-10 ø
Ecell = -0.0371 V
Since, Ecell is -ve, the cell is non-spontaneous \ in order to make it spontaneous, the polarity of the
electrodes should be reversed. Then, the spontaneous cell would be
Ag | AgBr, KBr (0.001 M)|| KCl (0.2 M), AgCl | Ag
Ecell = 0.503 V
E °cell = 0.80 – 0 = +0.80 V, n = 1
42.
E °cell = E ° –
0.0591
1
log
1
[Ag +]
0.503 = 0.80 + 0.0591 log [Ag+]
[Ag+] = 9.432 × 10-6 M
n Ag + = 9.432×10-6 ×0.35= 3.301×10-6
mass of Ag+ = 3.301 × 10-6 × 108 = 3.565 × 10-4 g
3.565×10-4
×100 = 0.0339
1.05
Ecell = 0.459 V
E °cell = +0.80 – 0 = +0.80 V
% of Ag + =
43.
E cell = E ° – 0.0591log
[H + ]
[Ag + ]
[Ag+] » 1.8 × 10-18 M
\ [Cl–] from AgCl = 1.8 × 10-8 M
Total [Cl–] = 1.8 × 10-8 + 0.01 (from HCl)
» 0.01 M
Ksp of AgCl = [Ag+] [Cl–]
Ksp = 1.8 × 10–8 × 0.01
Ksp = 1.8 × 10–10
44.
For AgI, Ksp = x 2 and x =
8.7×10-17
x = 9.33 × 10–9 M = [Ag+]
For the cell reaction Ag+(aq) + e– ® Ag(s)
æ
ö
1
E = +0.80 – 0.0591 log ç
-9 ÷
è 9.33 ´ 10 ø
E = +0.3254 V
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Electrochemistry
XXV
AgI (s) WXX
or
Ag+(aq) + I–(aq)
–
XXV Ag+
Ag(s) + AgI(s) WXX
(aq) + I (aq) + Ag (s)
+
–
Ag(s) | Ag (aq) || I | AgI | Ag
0.0591
log Ksp = 0.0591log 8.7×10 -17
1
E° = -0.949 = E°I– | AgI | Ag – E°Ag+/Ag
E °I– | AgI | Ag = –0.949 + 0.80
= -0.149 V
45. (i) Fe (s) + Ni2O3(s) ® FeO(s) + 2NiO(s)
(ii) E °cell = + 0.40 – (–0.87) = + 1.27 V
The cell emf is independent of the concentration of KOH as KOH is not appearing is the net cell
reaction.
(iii) DG° = –nFE° = –2 ´ 96500 ´ 1.27 = – 245110 J
46. 2HOCl + 2H+ + 2e– ® Cl2 + 2H2O
E° = 1.63 V
E° =
1
Cl + H2O
E° = 1.63 V
2 2
Taking other electrode as chlorine electrode
1
Cl– ®
Cl + e–
E° = –1.36 V
2 2
Adding :HOCl + H+ + Cl– ® Cl2 + H2O
This cell has anode as chlorine electrode and cathode as given electrode.
\
E °cell = 1.63 – (+1.36)
E °cell = +0.27 V
HOCl + H+ + e– ®
\
nE ° 1×0.27
=
0.0591 0.0591
K = 3.702 × 104
log K =
\ K for required reaction =
=
1
K
1
3.702 ´ 104
= 2.7 ´ 10–5.
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