Electrochemistry Practice Questions - DPP for JEE

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1. (c) At infinite dilution each ion makes a definite contribution towards
molar conductance which is given by
2. (d) Degree of dissociation,
3. (a)
4. (d) Standard Gibbs free energy is given as ?G° = – nE°F
If  < 0 i.e., – ve
?G° > 0
Further ?G°  = – RT ln K
eq
? ?G° > 0 and K
eq
 < 1
5. (b)     ;    Eº = – 0.44
The metals having higher negative electrode potential values can
displace metals having lower values of negative electrode potential
from their salt solutions.
6. (d)
7. (b)
For concentration cell,  
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1. (c) At infinite dilution each ion makes a definite contribution towards
molar conductance which is given by
2. (d) Degree of dissociation,
3. (a)
4. (d) Standard Gibbs free energy is given as ?G° = – nE°F
If  < 0 i.e., – ve
?G° > 0
Further ?G°  = – RT ln K
eq
? ?G° > 0 and K
eq
 < 1
5. (b)     ;    Eº = – 0.44
The metals having higher negative electrode potential values can
displace metals having lower values of negative electrode potential
from their salt solutions.
6. (d)
7. (b)
For concentration cell,  
In it R, T, n and F are constant
So E is based upon 
Now  = 
= –RTlnC
2
/C
1
At constant temperature  is based upon ln (C
2
/C
1
).
8. (b) For Zn
2+
 ? Zn
9. (a) The given order of reduction potentials (or tendencies) is Z > Y >
X. A spontaneous reaction will have the following characteristics
Z reduced and Y oxidised
Z reduced and X oxidised
Y reduced and X oxidised
Hence, Y will oxidise X and not Z.
10. (b) For, , 0.44 – 0.33 = 0.11V  is positive,
hence reaction is spontaneous.
11. (a) will oxidise  ion according to the equation
 
The cell corresponding to this reaction is as follows:
1.51 -1.40 = 0.11 V
being +ve,  will be -ve and hence the above reaction is
feasible.  will not only oxidise  ion but also  ion
simultaneously.
12. (b) Conductivity (X) = conductance (c) × cell constant
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1. (c) At infinite dilution each ion makes a definite contribution towards
molar conductance which is given by
2. (d) Degree of dissociation,
3. (a)
4. (d) Standard Gibbs free energy is given as ?G° = – nE°F
If  < 0 i.e., – ve
?G° > 0
Further ?G°  = – RT ln K
eq
? ?G° > 0 and K
eq
 < 1
5. (b)     ;    Eº = – 0.44
The metals having higher negative electrode potential values can
displace metals having lower values of negative electrode potential
from their salt solutions.
6. (d)
7. (b)
For concentration cell,  
In it R, T, n and F are constant
So E is based upon 
Now  = 
= –RTlnC
2
/C
1
At constant temperature  is based upon ln (C
2
/C
1
).
8. (b) For Zn
2+
 ? Zn
9. (a) The given order of reduction potentials (or tendencies) is Z > Y >
X. A spontaneous reaction will have the following characteristics
Z reduced and Y oxidised
Z reduced and X oxidised
Y reduced and X oxidised
Hence, Y will oxidise X and not Z.
10. (b) For, , 0.44 – 0.33 = 0.11V  is positive,
hence reaction is spontaneous.
11. (a) will oxidise  ion according to the equation
 
The cell corresponding to this reaction is as follows:
1.51 -1.40 = 0.11 V
being +ve,  will be -ve and hence the above reaction is
feasible.  will not only oxidise  ion but also  ion
simultaneously.
12. (b) Conductivity (X) = conductance (c) × cell constant
 Cell constant = 
Conductivity of NaOH = .Z
m (NaOH) = 
13. (c) From  the given data we find Fe
3+
 is strongest oxidising agent.
More the positive value of E°, more is the tendency to get
oxidized. Thus correct option is (c).
14. (a) In case of molar conductance of strong electrolyte there is little
increase with dilution.
15. (d) A device that converts energy of combustion of fuels, directly into
electrical energy is known as fuel cell.
16. (d) E = 
= 
E =  0.0592 × 2
= 0. 118 × 1 = 0.118V
17. (b) In  fuel cell, the combustion of H
2
 occurs to create
potential difference between the two electrodes
18. (a) H
2
 2H
+
 + 2e
–
1 atm        10
–10
 = 0 –  log 
 = +0.59 V
19. (b) HCl completely dissociates to give H
+
 and  ions, hence act as
very good electrolyte. While others are non- electrolytes.
20. (d) Kohlrausch’s Law states that at infinite dilution, each ion migrates
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1. (c) At infinite dilution each ion makes a definite contribution towards
molar conductance which is given by
2. (d) Degree of dissociation,
3. (a)
4. (d) Standard Gibbs free energy is given as ?G° = – nE°F
If  < 0 i.e., – ve
?G° > 0
Further ?G°  = – RT ln K
eq
? ?G° > 0 and K
eq
 < 1
5. (b)     ;    Eº = – 0.44
The metals having higher negative electrode potential values can
displace metals having lower values of negative electrode potential
from their salt solutions.
6. (d)
7. (b)
For concentration cell,  
In it R, T, n and F are constant
So E is based upon 
Now  = 
= –RTlnC
2
/C
1
At constant temperature  is based upon ln (C
2
/C
1
).
8. (b) For Zn
2+
 ? Zn
9. (a) The given order of reduction potentials (or tendencies) is Z > Y >
X. A spontaneous reaction will have the following characteristics
Z reduced and Y oxidised
Z reduced and X oxidised
Y reduced and X oxidised
Hence, Y will oxidise X and not Z.
10. (b) For, , 0.44 – 0.33 = 0.11V  is positive,
hence reaction is spontaneous.
11. (a) will oxidise  ion according to the equation
 
The cell corresponding to this reaction is as follows:
1.51 -1.40 = 0.11 V
being +ve,  will be -ve and hence the above reaction is
feasible.  will not only oxidise  ion but also  ion
simultaneously.
12. (b) Conductivity (X) = conductance (c) × cell constant
 Cell constant = 
Conductivity of NaOH = .Z
m (NaOH) = 
13. (c) From  the given data we find Fe
3+
 is strongest oxidising agent.
More the positive value of E°, more is the tendency to get
oxidized. Thus correct option is (c).
14. (a) In case of molar conductance of strong electrolyte there is little
increase with dilution.
15. (d) A device that converts energy of combustion of fuels, directly into
electrical energy is known as fuel cell.
16. (d) E = 
= 
E =  0.0592 × 2
= 0. 118 × 1 = 0.118V
17. (b) In  fuel cell, the combustion of H
2
 occurs to create
potential difference between the two electrodes
18. (a) H
2
 2H
+
 + 2e
–
1 atm        10
–10
 = 0 –  log 
 = +0.59 V
19. (b) HCl completely dissociates to give H
+
 and  ions, hence act as
very good electrolyte. While others are non- electrolytes.
20. (d) Kohlrausch’s Law states that at infinite dilution, each ion migrates
independently of its co-ion and contributes to the total equivalent
conductance of an electrolyte a definite share which depends only
on its own nature.
From this definition we can see that option (d) is the correct answer.
21. (212.3) ?G = –nF E°
cell
 = –2 × 96500 × 1.1 J = 212.3 kJ.
22. (1.096) Writing the equation for pentane-oxygen fuel cell at respective
electrodes and overall reaction, we get
At Anode:
At Cathode:
Calculation of ?G° for the above reaction
?G° = [5×(–394.4) + 6× (–237.2)]
– [–8.2]
= – 1972.0 – 1423.2 + 8.2 = – 3387.0 kJ
= – 3387000 Joules.
From the equation we find n = 32
Using the relation, ?G° = – and substituting various values, we
get
– 3387000 = –32×96500× (F = 96500C)
or 
= or V = 1.0968 V
23. (97) CH
3
OH (l) + O
2 
(g) ? CO
2
 (g) + 2H
2
O (l)
 
= – 394.4 + 2 (–237.2) – (–166.2) – 0
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1. (c) At infinite dilution each ion makes a definite contribution towards
molar conductance which is given by
2. (d) Degree of dissociation,
3. (a)
4. (d) Standard Gibbs free energy is given as ?G° = – nE°F
If  < 0 i.e., – ve
?G° > 0
Further ?G°  = – RT ln K
eq
? ?G° > 0 and K
eq
 < 1
5. (b)     ;    Eº = – 0.44
The metals having higher negative electrode potential values can
displace metals having lower values of negative electrode potential
from their salt solutions.
6. (d)
7. (b)
For concentration cell,  
In it R, T, n and F are constant
So E is based upon 
Now  = 
= –RTlnC
2
/C
1
At constant temperature  is based upon ln (C
2
/C
1
).
8. (b) For Zn
2+
 ? Zn
9. (a) The given order of reduction potentials (or tendencies) is Z > Y >
X. A spontaneous reaction will have the following characteristics
Z reduced and Y oxidised
Z reduced and X oxidised
Y reduced and X oxidised
Hence, Y will oxidise X and not Z.
10. (b) For, , 0.44 – 0.33 = 0.11V  is positive,
hence reaction is spontaneous.
11. (a) will oxidise  ion according to the equation
 
The cell corresponding to this reaction is as follows:
1.51 -1.40 = 0.11 V
being +ve,  will be -ve and hence the above reaction is
feasible.  will not only oxidise  ion but also  ion
simultaneously.
12. (b) Conductivity (X) = conductance (c) × cell constant
 Cell constant = 
Conductivity of NaOH = .Z
m (NaOH) = 
13. (c) From  the given data we find Fe
3+
 is strongest oxidising agent.
More the positive value of E°, more is the tendency to get
oxidized. Thus correct option is (c).
14. (a) In case of molar conductance of strong electrolyte there is little
increase with dilution.
15. (d) A device that converts energy of combustion of fuels, directly into
electrical energy is known as fuel cell.
16. (d) E = 
= 
E =  0.0592 × 2
= 0. 118 × 1 = 0.118V
17. (b) In  fuel cell, the combustion of H
2
 occurs to create
potential difference between the two electrodes
18. (a) H
2
 2H
+
 + 2e
–
1 atm        10
–10
 = 0 –  log 
 = +0.59 V
19. (b) HCl completely dissociates to give H
+
 and  ions, hence act as
very good electrolyte. While others are non- electrolytes.
20. (d) Kohlrausch’s Law states that at infinite dilution, each ion migrates
independently of its co-ion and contributes to the total equivalent
conductance of an electrolyte a definite share which depends only
on its own nature.
From this definition we can see that option (d) is the correct answer.
21. (212.3) ?G = –nF E°
cell
 = –2 × 96500 × 1.1 J = 212.3 kJ.
22. (1.096) Writing the equation for pentane-oxygen fuel cell at respective
electrodes and overall reaction, we get
At Anode:
At Cathode:
Calculation of ?G° for the above reaction
?G° = [5×(–394.4) + 6× (–237.2)]
– [–8.2]
= – 1972.0 – 1423.2 + 8.2 = – 3387.0 kJ
= – 3387000 Joules.
From the equation we find n = 32
Using the relation, ?G° = – and substituting various values, we
get
– 3387000 = –32×96500× (F = 96500C)
or 
= or V = 1.0968 V
23. (97) CH
3
OH (l) + O
2 
(g) ? CO
2
 (g) + 2H
2
O (l)
 
= – 394.4 + 2 (–237.2) – (–166.2) – 0
= – 394.4 – 474.4 + 166.2 = – 702.6 k J
% efficiency = 
24. (1) 112 ml of H
2
 at STP = 
(Since 22400 ml at STP = M.wt)
Amount deposited 
25. (1.567) Here n = 4, and [H
+
] = 10
– 3 
(as pH = 3)
Applying Nernst equation
E = Eº  – 
 = 1.67 – 0.103 = 1.567 V