Cls Jeead-16-17 Xii Che Target-5 Set-2 Chapter-3 (1)
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jee advanced...
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Chapter
3
Electrochemistry 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
⎛W⎞ it it ⎜M⎟ = = (Ag+ + e– Ag) ⎝ ⎠ Ag nF F ⎛W⎞ it ⎜M⎟ = (Ca2+ + 2e– Ca) ⎝ ⎠Ca 2F ⎛W⎞ it (Al3+ + 3e Al) and ⎜ M ⎟ = 3F ⎝ ⎠ Al
Molar ratio is 1 : 2.
1 1 : 3 2
or
6:3:2
The molar conductances at infinite dilution of BaCl 2, 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) BaCl = Ba2 2Cl– 2
...(i)
= Na Cl– NaCl
...(ii)
= Na HO– NaOH
...(iii)
for Ba(OH)2 (i) + 2(iii) –2(ii) (Ba(OH)2 ) = (280 × 10–4) + 2(248 × 10–4) – 2(126.5 × 10–4) = 523×10–4 Sm2 mol–1. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
76 3.
Electrochemistry
Solution of Assignment (Set-2)
During electrolysis of aqueous solution of a salt under inert electrodes, 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) E
Cu2 /Cu
When Cu2+
E'
2
Cu
/Cu
1 0.0591 log 2 [Cu ] 2 solution is diluted to 100 times [Cu2+] decreases to 1/100
= Eº –
= Eº –
E’ = Eº –
100 0.0591 log [Cu2 ] 2
0.0591 [[log 100 – log [Cu2+]] 2
E’ = Eº –
1 0.0591 0.0591 ×2– log [Cu2 ] 2 2
E’ = Eº –
1 0.0591 log – 0.0591 [Cu2 ] 2
E’ = E – 0.0591, Hence, Potential decreases by 59 mV. 5.
The Ecell of the reaction
MnO 4 Fe 2 H Mn2 Fe 3 H2O 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)
Eocell = 0.59 V MnO4– + Fe2+ + H+ Mn2+ + Fe3+ + H2O 0.0591 log Qc 5 At equilibrium, E = 0; Qc = Kc
E = Eocell –
Eocell =
0.0591 log Kc 5
5 0.59 = log Kc 0.59 50 = log Kc
Kc = 1050. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
6.
Electrochemistry
77
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 Ait w= F nF
n=
177 2 5 60 60 A i t = = 2.97 22.2 96500 wF
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 + O2 + 4e–
⎛w⎞ it ⎛ w ⎞ it ⎜M⎟ = ; ⎜ ⎟ = ⎝ ⎠H2 2F ⎝ M ⎠O2 4F
nH2 : nO2 = VH2 2.8
=
2 1
it it : 2:1 2F 4F
Volume ratio VH2 : VO2 = 2 : 1
VH2 = 2.8 × 2 = 5.6 L
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78 9.
Electrochemistry
Solution of Assignment (Set-2)
The equilibrium constant for the reaction Sr (s) + Mg+2 (aq) 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
Sr+2 (aq) + Mg (s) is 2.69 × 1012 at 25°C.
(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
0.0591 log Kc 2 0.0591 0.0591 Eº = log Kc = log (2.69 × 1012) = 0.3667 V 2 2 0 = Eº –
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. In which of the following pair, first specie is a better oxidising agent than second specie under standard conditions? (1) Br2 & Au3+
(2) H2 & Ag+
(3) Cr3+ & Cd2+
(4) O2 in acidic medium & O2 in basic medium
Sol. Answer (4) 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: o o = H + Cl– HCl
...(i)
o CH = CH3COO– Na 3 COONa
...(ii)
o o = Na + Cl– NaCl
...(iii)
Doing the operation Equation (i) + (ii) – (iii) o o + CH3COONa – NaCl = CH3COO– + H HCl (426.1 + 91.0 – 126.45) = CH3COOH
CH = 390.6 cm2 –1/eq 3 COOH
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Solution of Assignment (Set-2)
Electrochemistry
79
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 Eit w ⇒ F F E
w=
13.5 13.5 ⎛ i × t = ⎜ 27 ⎞⎟ × F = F = 1.5 F 9 ⎝ 3 ⎠
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 it 0.5F ⎜ ⎟ = = = 0.25 moles F 2F 2F ⎝M⎠
w=
V × molarity = No. of moles 500 × x × 10–3 = 0.25 x=
0.25 103 = 0.5 500
Molarity = 0.5 M 15. M+ is not stable and undergoes disproportionation to form M and M2+. Calculate E for M+ disproportionation
E 2 M
/M
0.153 V, E
(1) +0.683 V
M /M
0.53 V (2) –0.367 V
(3) 0.754 V
(4) +0.3415 V
Sol. Answer (3) Given EoCu2 /Cu = +0.153 V and EoCu /Cu = 0.53 V The reaction 2Cu+ Cu + Cu2+
o o E oCell = ECu /Cu2 + ECu /Cu
o we required ECu /Cu2
Cu2+ + 2e Cu; 0.153 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
80
Electrochemistry
Solution of Assignment (Set-2)
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)
EoCu2 /Cu = – 0.224
or
EoCu /Cu2 = 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 =
n=
100 5 4 60 60 n 96500
100 5 4 60 60 =3 (96500 25)
Hence, valency = 3 17. A well stirred (1 L) solution of 0.1 M CuSO4 is electrolysed at 25°C using copper 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 NH 3 and 1 M NH4+ 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 + O2 + 4e– 4OH– (R.H.S.) 2H2O 4H+ + O2 + 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. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Electrochemistry
81
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
⎛w⎞ it it it ⎜M⎟ = + = ⎝ ⎠T 2F 2F F 3 50 10
25 12 60 60 = nT 100 96500
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
570 100 10 3 = 96500 n 51
n=
570 51 100 10 –3 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 1000 K (1) xy
(2)
⎛ 1000 K ⎞ ⎜⎜ ⎟⎟ ⎝ xy ⎠
2
1000 143.5 K (3) xy
⎛ 10 3 143.5 K ⎞ ⎟ (4) ⎜⎜ ⎟ xy ⎠ ⎝
2
Sol. Answer (2) AgCl Ag+ + Cl– Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Electrochemistry
Solution of Assignment (Set-2)
Specific conductance = K–1cm–1
K 1000 C
o o AgCl = Ag Cl = (x + y)
(x + y) =
K 1000 C
1000K C = (x y)
Solubility product =
C2
⎡ 1000K ⎤ = ⎢ ⎥ ⎣ (x y) ⎦
2
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) o o = Na + Cl– = 126.5 ...(i) NaCl o o = K + Cl– = 150 KCl
…(ii)
o o = K + Br – = 152 KBr
…(iii)
Adding (i) & (iii) subtract (ii)
o
Na
o o o o o o o + Cl– + K + Br – – K – Cl– = Na + Br –
o = (126.5 + 152 – 150) = (76) + 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 (1) 5.3
(2) 4.3
(3) 2.3
(4) 1.3
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Solution of Assignment (Set-2)
Electrochemistry
83
Sol. Answer (4) o o CH = CH3 COO– Na 3 COONa
o o o = H Cl– HCl
o NaCl = Na Cl– = CH3COONa HCl– – NaCl CH 3 COOH 91 + 426 + (–126) = 391 = CH3COOH
C 19.55
= [H+]
C
=
19.55 = 0.05 391
= C = 1 × 0.05 M
pH = – log [H+] = – log (5 × 10–2) 2–log 5 = 2 – 0.7 =1.3 25.
E(Na /Na) 2.71 V, E(Mg 2 /Mg) 2.37 V E(Fe 2 /Fe) 0.44 V, E(Cr 3 /Cr) 0.41 V 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.
Vol. of KCl
(4)
Vol. of KCl
Conductance
Vol. of KCl
(3)
Conductance
(2)
Conductance
(1)
Conductance
26. Which of following type of plot would you expect from the titration of AgNO3 against KCl solution?
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); E1o = 0.337 V (ii)... Cu2+ + e– Cu+ Eo2 = 0.153 V Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
84
Electrochemistry
Solution of Assignment (Set-2)
Reversing equation (ii); we get Cu+ Cu2+ + e–
...(iii)
Adding equation (i) and (iii) We get, Cu+ + e– Cu
o Go G1o Go2 = (–1 × F × Eº) = –2 F E1
+ –1 F E o 2
– 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+ ECu 2 /Cu 0.34 V, E Ag /Ag 0.8 V (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 o E oCell = E Ag/ Ag ECu2 /Cu
E oCell = (– 0.8) + (0.34) = – 0.46
Gº = – n F × Eº = – 2 × F × (–0.46) = – 2 × –0.46 × 96500 = 88780 J or Gº = + 89 kJ. o
29. For the reduction half reaction EQuinhydrone 1.30 V
O–H
O +
–
+ 2H + 2e O–H
O
At pH = 3 at 298 K, electrode reduction potential is (Consider quinone and hydroquinone have identical concentration) (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 = E oCell –
0.0591 log [H+]2 2
Ecell = 1.30 –
0.0591 log (10–3)2 2
=–
0.0591 × (–6) log 10 + 1.30 = 0.0591 × 3 + 1.30 = 1.477 1.48 2
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Solution of Assignment (Set-2)
Electrochemistry
85
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 ⎡ Zn2 ⎤ ⎣ ⎦
E = Eº –
0.0591 log 2 ⎡Cu2 ⎤ ⎣ ⎦
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 – – – – SO2– 4 < HO < Cl < Br < I
33. Which metal oxide is thermally unstable? (1) Al2O3
(2) Na2O
(3) BaO
(4) Ag2O
Sol. Answer (4) Ag2O decomposes as 1 Ag2O 2Ag + 2 O2.
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. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Electrochemistry
Solution of Assignment (Set-2)
35. H2(1 atm) | 2.26 M HCOOH || 0.222 M CH3COOH| (1 atm) H2 Ka(HCOOH) = 1.77 × 10–4, Ka(CH3COOH) = 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
(3) –418.424
(4) 229.284
Sol. Answer (2) HCOOH HCOO– + H+
C1(1 – ) C1 C1 [H+]L = C1 =
(Ka)1 C1
CH3 COOH CH3COO– + H+ C2(1 – )C2 [H+]R =
C2 (Ka)2 C2 [H ]L 0.0591 log [H ]R 1
E = Eº –
E=0–
or E = –
E=–
K1 C1
0.0591 log 1
K 2 C2
1.77 10 4 2.26 0.0591 log 2 1.8 10 5 0.222
0.0591 log 100 = – 0.0591 2
36. Given that NiO2 + 4H+ + 2e– Ni2+ + 2H2O, E° = 1.678 V NiO2 + 2H2O + 2e– Ni(OH)2 + 2OH–, E° = –0.49 V For the following reaction Ni(OH)2 + 2H+ Ni2+ + 2H2O Gibb’s free energy change (in kJ mol–1) is (1) 418.424
(2) –229.284
Sol. Answer (3)
NiO2
4H 2e –
Ni 2 2H2O
, E1o 1.678 V
Ni(OH)2 2OH – 2e –
NiO2 2H2O 2e – , Eo2 0.49 V
Ni(OH)2 2H
Ni 2 2H2O
, Eo x V
Go = G1o Go2 – nF(E1o Eo2 ) = –2 × 96500 × (1.678 + 0.49) J mol–1 = –418.424 kJ mol–1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Electrochemistry
87
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 20% by weight? (Atomic mass, Zn = 65.4 g) (1) 5.6 A
(2) 7.2 A
(3) 8.85 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 25 9x
x = 3 gram Equivalent of Zn Current
32 65.4
6 96500 8.85 A 65.4 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 EoZn2 / Zn is less than Eo H /H
2
39. Emf of cell given, Ag(s), AgCl(s)||KCl(aq)|Hg2Cl2(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)
Hg22+ + 2e– 2Hg
:
(cathode)
∵ n=2 ⎛ Ecell ⎞ H = –nFEcell nFT ⎜ ⎟ ⎝ T ⎠P
= 2 × 96500(300 × 3.34 × 10–4 – 0.05) = 9650 J mol–1 40. Given : Ag+ + e– Ag ; E°red = 0.799 V Dissociation constant for [Ag(NH3)2]+ into Ag+ and NH3 is 6 × 10–14. Then for the following half-cell reaction: [Ag(NH3)2]+ + e– Ag + 2NH3, calculate E° at 298 K. (1) 0.019 V
(2) 0.03 V
(3) 0.014 V
(4) 0.19 V
Sol. Answer (1)
Ag e – ;
EoOP – 0.799 V
Ag(NH3 )2 e –
Ag NH3 ;
o ERP ?
Ag(NH3 )2
Ag
Ag 2NH3
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Electrochemistry
o Ecell Ecell
Solution of Assignment (Set-2)
[Ag(NH3 )2 ] 0.0591 0 at equilibrium log10 1 [Ag ] [NH3 ]2
Eocell 0.0591 log10 K C 0.0591 log10 (6 10–14 ) = –0.780 V o EoOP( Ag/ Ag ) ERP( Ag(NH
3 )2 / Ag)
EoAg(NH
– 0.780 0.799
3 )2 / Ag
= +0.019 V 41. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time [IIT-JEE 2008] required to liberate 0.01 mol of H2 gas at the cathode is (1 faraday = 96500 C mol–1) (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
W 10 103 t 0.01 2 E 96500 t = 19.3 × 104 s 42. Consider the following cell reaction: 2 2H O(l) 2Fe(s) O2(g) 4H(aq) 2Fe(aq) ; 2
E° = 1.67 V
At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25°C is (1) 1.47 V
(2) 1.77 V
(3) 1.87 V
[IIT-JEE 2011] (4) 1.57 V
Sol. Answer (4) Ecell 1.67
0.0591 [Fe2 ]2 log 4 pO2 [H ]4
1.67
0.0581 (103 )2 log 4 0.1 (103 )4
1.67
0.0591 106 log 4 1013
1.67
0.0591 log107 4
1.67
0.0591 7 1.57 4
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Solution of Assignment (Set-2)
Electrochemistry
89
43. AgNO3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. [IIT-JEE 2011] The plot of conductance () versus the volume of AgNO3 is
volume (P)
volume (Q)
(1) (P)
volume (R)
(2) (Q)
(3) (R)
volume (S) (4) (S)
Sol. Answer (4) Ag+ and K+ have nearly same ionic mobility AgNO3 + KCl AgCl(s) + KNO3
conc. of KCl
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– Cl2 + 2e–
w it w 96.50 10 60 60 10 –3 = = = = 0.018 2 96500 M nF M For Cl– = 0.036; Molarity = 0.1 – 0.036 = 0.064
VCl2 =
0.018 22.4 = 0.202 L or 202 ml. 2
K+ will not discharge. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
90 2.
Electrochemistry
Solution of Assignment (Set-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 Ptelectrode (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 = w=
n=
E it F
63.5 9.65 2 60 60 × = 22.86 96500
W = 0.36 M
2 – 0.36 = 1.64 Hence, molarity = 1.64 M using Pt electrode w= 3.
75 (9.65) 2 60 60 63.5 × = 17.15 g 100 96500 2
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 SO2– 4 Using Pt electrode At cathode; Cu2+ + 2e Cu Using Cu electrodes At anode : Cu Cu2+ + 2e– 4.
Daniell cell : Zn|Zn 2(aq) || Cu 2(aq)|Cu operates as electrolysis cell for 60 min and a current of ( 50 ml 1M)
( 50 ml 1M)
0.965 amp is passed. Which is/are correct?
(E Cu 2 /Cu 0.34 V, E Zn 2 /Zn 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
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Solution of Assignment (Set-2)
Electrochemistry
91
Sol. Answer (1, 2) WCu (deposited) =
31.75 × 0.965 × 60 × 60 = 1.143 g 96500
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 = 0.64 M 63.5 50
Molarity of Zn2+ solution = 1 + 0.36 = 1.36 M 5.
Which compounds have maximum conductivity? (1) 0.2 M [Cr(NH3)3Cl3]
(2) 0.15 M [Cr(NH3)4Cl2]Cl
(3) 0.1 M [Cr(NH3)5Cl]Cl2
(4) 0.07 M [Cr(NH3)6]Cl3
Sol. Answer (2, 3) [Cr(NH3)4Cl2]Cl [Cr(NH3)4Cl2]2+ +Cl– 0.15 × 2 = 0.30 and for [Cr(NH3)5Cl]Cl2 [Cr(NH3)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 H2A 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 1.0 (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 as per H2 A 2H A 2 is 2.56 × 10–4
Sol. Answer (1, 3)
m 0 m
10 1 2.5% 400 40
H2 A 2H A 2 At equation, C(1 ) 2C C K eq
(2C )2 C C(1 )
Putting C = 2 M, = 0.025 Keq = 2.56 × 10–4 [H+] = 2C = 0.1 = 10–1 pH = – log 10–1 = 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
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92
Electrochemistry
Solution of Assignment (Set-2)
(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 AxBy 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 By 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)
l A
(2)
R
(3) = (G)–1
(4)
G K
Sol. Answer (1, 3) R=
1 1 l l RA A
K = C
l A
RA l ; and (G)1 l A
Vol. of NaOH
Vol. of NaOH
(4)
Vol. of NaOH
Conductance
(3)
Conductance
(2)
Conductance
(1)
Conductance
10. Which of following plots will not be obtained for a conductometric titration of HCl and NaOH?
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
E Zn 2 /Zn 0.75 V, ENi 2 /Ni 0.24 V 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 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Electrochemistry
93
(4) Cell reaction is non-spontaneous when concentration of Ni+2 is less than 5.5 × 10–18 M 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.
[Zn2 ] 0.0591 E = Eº – log [Ni2 ] 2 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+ + SO2– 4 H2O H+ + HO– At cathode : Cu2+ + 2e– Cu Anode : 4HO– 2H2O + O2 + 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, when electrolysed among inert electrodes? (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– H2. 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 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
94
Electrochemistry
Solution of Assignment (Set-2)
Sol. Answer (1, 3) It is known fact that H and temperature coefficient are related as,
⎛ E ⎞ H > 0 for ⎜ ⎟ >0 ⎝ T ⎠ ⎛ E ⎞ and H < 0 for ⎜ ⎟ 0 is spontaneous for Fe3+ + Ni Ni2+ + Fe o o Eº = ENi/Ni2 EFe3 /Fe = (0.24) + (–0.04) > 0 i.e. spontaneous
and for reaction Cd2+ + Zn Zn2+ + Cd o o Eo = EZn/Zn2 ECd2 /Cd = (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)|H2(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|H2(1 atm), Pt
Sol. Answer (1, 2, 3) o For Ecell Ecell , K C 1.
22. In which of the following cells, reaction quotient is equal to one? (1) Pb|PbC2O4, CaC2O4, CaCl2(0.1 M)||CuSO4(0.1 M) | Cu (2) Zn|ZnSO4(0.1 M)||CuSO4(0.1 M)|Cu (3) Zn|ZnSO4(0.1 M)||Hg2Cl2, KCl(0.1 M)|Hg, Pt (4) Cu|CuSO4(0.1 M)||SnCl2(0.1 M)|SnCl4(0.1M), Pt Sol. Answer (1, 2) In (3), = 10 In (4), = 0.1 23. Which solution(s) become(s) more acidic after the electrolysis using inert electrodes? (1) NaCl solution
(2) CuSO4 solution
(3) AgNO3 solution
(4) Na2SO4 solution
Sol. Answer (2, 3) In the electrolysis of CuSO4 solution and AgNO3 solution, H2SO4 and HNO3 are formed respectively. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Electrochemistry
97
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) H2SO4 is consumed
(2) Pb is formed
(3) SO2 is evolved
(4) PbSO4 is consumed
Sol. Answer (2, 3, 4) Pb is consumed and PbSO4 is formed. SO2 is not evolved. 26. For the reduction of NO3– 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 NO3– 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. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
98
Electrochemistry
Solution of Assignment (Set-2)
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 [E Cu 2 /Cu 0.34 V, E Zn 2 /Zn 0.76 V]
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) EoCu2 /Cu = 0.34 V and EoZn2 / Zn = – 0.76 V
ECell
[Zn2 ] 0.0591 = Eº – log [Cu2 ] 2
[Zn2+] = [Cu2+] = 1M o o E = Eº = EZn/Zn2 ECu2 /Cu
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) CuSO4 solution
(3) Both (1) & (2)
(4) Can't say
Sol. Answer (1) When NH3 is added to ZnSO4 solution, NH3 reacts with Zn2+ in the following manner : 2+ Zn2+ + 4 NH3 [Zn(NH3)4]
i.e., [Zn2+] decreases. In the equation
E = Eo –
0.0591 [Zn2 ] log 2 [Cu2 ]
If [Zn2+] decreases then log
3.
[Zn2 ] [Cu2 ]
decreases hence, EMF of cell increases.
In a separate experiment, 50 ml of 1.5 M NH3 is added to CuSO4 solution. Emf of the cell is [Kf ([Cu(NH3)4]+2) = 5.88 × 1013] (1) 0.933 V
(2) 1.327 V
(3) 1.467 V
(4) 0.696 V
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Solution of Assignment (Set-2)
Electrochemistry
99
Sol. Answer (4) Due to the complex formation, [Cu2+] decreases & it can be calculated by the reaction, Cu2 4HN3 [Cu(NH3)4]2+
50 0.1
50 1.5
[Cu2+] =
50 0.1 100 = = 9.3 × 10–15 M (0.55)4 5.88 1013
[Cu (NH3 )4 ]2 [NH3 ]4 k f
Ecell = 1.1 –
⎛ 0.0591 0.1 log ⎜⎜ –15 2 ⎝ 9.3 10
⎞ ⎟⎟ = 0.715 V ⎠
Thus, the e.m.f. of cell decreases. Comprehension-II Molar conductivity of ions are given as product of charge on ions to their ionic mobilities and Faraday constant
An
n
An
F (here is the ionic mobility of An+). For electrolytes say A B , molar conductivity is given by x y
m(A xBy ) x n
1.
An
F ym
Am
F.
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
The equivalent conductance of CaSO4 at infinite dilution is (2) 28.51 × 10–4
(1) 279
(3) 31.82 × 10–4
(4) 306
Sol. Answer (1) Equivalent conductance of CaSO4 is the sum of ionic conductance of Ca2+ & SO2– 4 . CaSO = Ca2 SO2– 4 4
= UCa2
= USO2–
Ca2
SO2– 4
U
Ca2
4
F
F
& USO24 are ionic mobilities
CaSO = F {12.33 + 16.58} × 10–4 4 CaSO = 96500 × 10–4 × 28.91 = 278.98 279 4
Equivalent conductance of CaSO4 is 279 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
100 2.
Electrochemistry
Solution of Assignment (Set-2)
If degree of dissociation of CaSO4 solution is 10% then equivalent conductance of CaSO4 is (2) 2.851 × 10–4
(1) 27.9
(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 Comprehension-III Given below are a set of half-cell reactions (in acidic medium) alongwith their E° (in volt) values.
1.
I2 2e 2I
E 0.54
Cl2 2e 2CI
E 1.36
Mn3 e Mn2
E 1.50
Fe3 e Fe2
E 0.77
O2 4H 4e 2H2 O
E 1.23
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 O2 (4) Mn+3 oxidises H2O to O2
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. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Electrochemistry
101
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 = 2.
4 500 2 1000
nCl2 1.
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. 1.
[IIT-JEE 2010]
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
M M (1 M)
(0.05 M)
Apply Nernst equation
E Eº –
0.059 0.05 log 1 1
E–
0.059 log 5 10 –2 1
E–
0.059 ⎡ –2 log5 ⎤⎦ 1 ⎣
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102 2.
Electrochemistry
Solution of Assignment (Set-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)
E1 log0.05 E2 log0.0025 E1 log5 10 –2 E2 log25 10 –4
E1 70(given) 70 –1.3 1 E2 –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) 0.059
log
0.059 0.001 log 2 2 (M )
0.001 2 [M2 ]
0.001 100 K sp = 4 (10 5 )3 = 4 10 15 [M2 ] [M2 ] 105 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 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Electrochemistry
103
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 Br2 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(s) Zn 2 (aq) Zn 2 (aq) Zn for spontaneous cell reaction C1 < C2. C1
C2
and STATEMENT-2 : For concentration cell, E cell
C RT loge 2 for spontaneous reaction Ecell = +ve C2 > C1. nF C1
Sol. Answer (1) The given cell is Zn|Zn2+(C1) ||Zn2+ (C2)|Zn Zn(s) Zn2+ + 2e– E
E
Zn/Zn2
o = E Zn/ Zn2 –
Zn2 /Zn
= EZn2 /Zn –
0.0591 log (C1) 2
⎛ 1 ⎞ 0.0591 log ⎜ C ⎟ 2 ⎝ 2⎠
o o E = (E Zn/ Zn2 E Zn2 / Zn ) –
⎛ C1 ⎞ 0.0591 log ⎜ C ⎟ 2 ⎝ 2⎠
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104
Electrochemistry
Solution of Assignment (Set-2)
EMF of cell E=
⎛C ⎞ – 0.0591 log ⎜ 1 ⎟ 2 ⎝ C2 ⎠
⎛ C1 ⎞ log ⎜ C ⎟ < 0 for spontaneity ⎝ 2⎠ ⎛ C1 ⎞ log ⎜ C ⎟ < log 1 ⎝ 2⎠
C1 < C2. 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 cations and anions. and STATEMENT-2 : Kohlrausch’s law is applicable for strong electrolytes.
Sol. Answer (2) 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 o of Cu+2/Cu is lesser than E o STATEMENT-2 : ERP
Ag / Ag
.
Sol. Answer (1) Cu + AgNO3 Cu2+ + Ag Eo for reaction is positive because Eo
Cu2 /Cu
7.
Eo
Ag / Ag
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. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
8.
Electrochemistry
105
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 Cl2 gas is evolved. 10. STATEMENT-1 : H2 + O2 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 + O2 fuel cell, Anode : 2H2(g) + 4OH–(aq) 4H2O() + 4e– Cathode : O2(g) + 2H2O() + 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 Maximum molar conductivity (S cm2 mol–1)
Complex (with coordination number of Co3+ as equal to six) (A) CoCl3.6NH3
(p)
97
(B) CoCl3.5NH3
(q)
0
(C) CoCl3.4NH3
(r)
404
(D) CoCl3.3NH3
(s)
229
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106
Electrochemistry
Solution of Assignment (Set-2)
Sol. Answer A(r), B(s), C(p), D(q) [Co(NH 3 ) 6 ] Cl 3 will give maximum number of ions(4) because of which conductivity is maximum i.e. 404. In [Co(NH3)3Cl3] no ions are given Hence molar conductivity is zero. [Co(NH3)5Cl]Cl2 & [Co(NH3)4Cl2] 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)
3F
(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 it wF = it = E F E
⎛ wF ⎞ wF it = 3 ⎜ = 3F ⎟ = E ⎝ M ⎠ (B) 2.3 g Na+
it =
2.3 w F= F = 0.1 F 23 E
(C) 3.6 g of Mg2+ it =
(D)
3.6 F = 0.3 F 12
11.2 1 = 22.4 2
nH 2
w it w = it = F E F E
⎛w⎞ = 2⎜ ⎟ F ⎝M⎠ = 2
1 F=F 2
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Solution of Assignment (Set-2)
3.
Electrochemistry
107
Match the following Column-I
Column-II c m
(A) , specific conductance
(p)
(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
m
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 and decreases with dilution m
(D) Resistance 4.
l and decreases with dilution A
Match the following Column-I
Column-II
(A) Calomel electrode
(p)
Electrolyte concentration cell
(B) Zn-Cd(C1) |CdCl2| Zn-Cd(C2)
(q)
Metal-insoluble anion half cell
(C) Quinhydrone electrode
(r)
Electrode concentration cell
(s)
Redox half cell
(D) Pt|H2(1
atm)|H+(C1)||H+(C2)|H2(1
atm)|Pt
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
(q)
Chlorine gas evolved at anode
(C) CuSO4 using copper electrodes
(r)
Oxygen gas evolved at anode
(D) 50% H2SO4 solution
(s)
A compound with peroxide bond is formed
electrodes (B) Very dilute aqueous solution of NaCl using mercury cathode
Sol. Answer A(q), B(r), C(p), D(s) 6.
The standard reduction potential data at 25°C is given below. E°(Fe3+, Fe2+) = + 0.77 V; E°(Fe2+, Fe) = – 0.44 V E°(Cu2+, Cu) = + 0.34 V; E°(Cu+, Cu) = + 0.52 V
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108
Electrochemistry
Solution of Assignment (Set-2)
E°[O2(g) + 4H+ + 4e– 2H2O] = + 1.23 V; E°[O2(g) + 2 H2O + 4e– 4OH–] = + 0.40 V E°(Cr3+, Cr) = – 0.74 V; E°(Cr2+, Cr) = – 0.91 V Match Eº of the redox pair in Column-I with the values given in Column-II and select the correct answer using the code given below the Columns : [JEE(Advanced)-2013] Column-I
Column-II
(A) Eº(Fe3+, Fe)
(p)
–0.18 V
(B) Eº(4H2O 4H+ + 4OH–)
(q)
–0.4 V
(C) Eº(Cu2+ + Cu 2Cu+)
(r)
–0.04 V
(D) Eº(Cr3+, Cr2+)
(s)
–0.83 V
Codes : A
B
C
D
(1) s
p
q
r
(2) q
r
s
p
(3) p
q
r
s
(4) r
s
p
q
Sol. Answer (4) (A)
Fe
+3
+2
+ e
Fe
ºG1 = – 1F × 0.77
Fe
+2
+ 2e
Fe
ºG2 = + 2F × 0.77
Fe
+3
+ 3e
Fe
ºG3 = – 3F × EºFe+3/Fe
ºG3 = ºG1 + ºG2
– 3F × EºFe+3 /Fe = – 0.77 F + 0.88 F – 3EºFe+3 /Fe = 0.11 (V) EºFe+3 /Fe = –
0.11 (V) = – 0.036 (V) 3 +
2H2O
(B)
O2 + 4H + 4e –
4e + O2 + 2H2O
OH
+
4 H2O (C) Cu2+ + 2e
2Cu 2+
Cu
+ Cu
Eº = + 0.40 V –
4H + 4OH Cu
Eº = – 1.23 V
Eº = – 0.83 V
Eº = + 0.34 V
2+
Cu + 2eu +
2Cu
Eº = – 0.52 V Eº = – 0.18 V
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Solution of Assignment (Set-2) +3
(D) Cr
Electrochemistry
+ 3e
Cr
Cr
Cr
+ e
Cr
+3
Cr
+2
109
ºG1 = + 3 × F × 0.74 + 2e
ºG2 = + 2 × F ×0.91
+2
EºCr+3 /Cr+2 = – 0.4 V 7.
An aqueous solution of X is added slowly to an aqueous solution of Y as shown in Column-I. The variation in conductivity of these reactions is given in Column-II. Match Column-I with Column-II and select the correct answer using the code given below the Columns : Column-I (A) (C2H5)3N
[JEE(Advanced)-2013] Column-II
+
CH3COOH
X
(p) Conductivity decreases and then increases
Y
(B) KI (0.1M)
+ AgNO3 (0.01M)
X
Y
change much
(C) CH3COOH + KOH X
(r) Conductivity increases and then does not
Y
(D) NaOH
change much
+ HI
X
(q) Conductivity decreases and then does not
(s) Conductivity does not change much and
Y
then increases
Codes : A
B
C
D
(1) r
s
q
p
(2) s
r
q
p
(3) q
r
s
p
(4) p
s
r
q
(A)
Conductivity
Sol. Answer (1)
(C2H5)3N Firstly it decreases due to nutrilization of CH3COOH and replacement of H+ by but thereafter buffer formation takes place and [H+] becomes constant and increases hence conductivity increases but after equivalence point (C2H5)3N is not ionized due to much higher concentration of in solution. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Electrochemistry
Solution of Assignment (Set-2)
Conductivity
110
(B)
VKI
AgNO3 KI AgI(S) KNO3
Conductivity
Initially only Ag+ is replaced by K+ hence conductivity remain the same thereafter equivalence point [K+] increases hence conducitity increases.
(C)
VCH3COOH
(D)
Conductivity
Initially conducitity decreases due to replacement of OH– by CH3COO– and then almost constant due to buffer formation
VNaOH Decreases due to removal of H+ by Na+ then increases due to OH–.
SECTION - F Integer Answer Type Questions 1.
The half cell potentials of a half cell | A x n , A x | 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) A
x n
ne – A x
o 0.101 ERP
0.059 ⎛ 75.6 ⎞ log ⎜ ⎟ n ⎝ 24.4 ⎠
o 0.115 ERP
0.059 ⎛ 51.2 ⎞ log ⎜ ⎟ n ⎝ 48.8 ⎠
⇒n2
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Solution of Assignment (Set-2)
2.
Electrochemistry
111
o o The standard reduction potential of EBi3 /Bi and ECu2 /Cu 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 log Cu2 does Bismuth starts to deposit during electrolysis. Sol. Answer (4) The passage of current would initially deposit Cu2+ till ECu2 /Cu becomes 0.266 V because then only Bi3+ will be deposited. o Thus, ECu2 /Cu Ecu 2 /Cu
0.266 0.344
3.
0.059 log Cu2 2
0.059 log Cu2 ⇒ log Cu2 4 2
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) Ecell 0.059log
H cathode 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
0.42
Eit 96500
108 3 t ⇒ t 125.09 s y 3 96500
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112 6.
Electrochemistry
Solution of Assignment (Set-2)
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 7.
All the energy released from the reaction X Y, rG° = –193 kJ mol–1 is used for oxidizing M+ as M+ M3+ + 2e–, E0 = –0.25 V Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F = 96500 C mol–1]
[JEE(Advanced)-2015]
Sol. Answer (4) X Y;
rG° = – 193 kJ mol–1
M+ M3+ + 2e–
E° = – 0.25 V
Hence G° for oxidation will be G° = – nFE° = –2× 96500 × (–0.25) = 48250 J = 48.25 kJ 48.25 kJ energy oxidises one mole M+ 193 kJ energy oxidizes
8.
193 mole M 4 mole M 48.25
The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity 0 0, of a solution of a weak acid HY (0.10 M). If X Y the difference in their pKa values, pKa(HX) – pKa(HY),
is (consider degree of ionization of both acids to be
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