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Electrochemistry Practice Questions - DPP for NEET

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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,  
Page 2


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
Page 3


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
Page 4


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
Page 5


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
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FAQs on Electrochemistry Practice Questions - DPP for NEET

1. What is electrochemistry?
Ans. Electrochemistry is a branch of chemistry that deals with the study of the interconversion of electrical and chemical energy. It involves the study of chemical reactions that produce electricity and the use of electricity to drive non-spontaneous chemical reactions.
2. What are the applications of electrochemistry?
Ans. Electrochemistry has various applications in our daily lives. Some of the important applications include batteries and fuel cells, electrolysis processes, corrosion prevention, electroplating, and sensing devices such as pH meters and biosensors.
3. What is a galvanic cell?
Ans. A galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts chemical energy into electrical energy. It consists of two half-cells, each containing an electrode and an electrolyte. The two half-cells are connected by a salt bridge or a porous barrier to allow the flow of ions and maintain electrical neutrality.
4. What is the difference between anode and cathode in electrochemistry?
Ans. In electrochemistry, the anode is the electrode where oxidation occurs, meaning it loses electrons and undergoes a chemical reaction. On the other hand, the cathode is the electrode where reduction occurs, meaning it gains electrons and undergoes a chemical reaction. The movement of electrons from the anode to the cathode creates an electric current.
5. How does electrolysis work?
Ans. Electrolysis is a process that uses an electric current to drive a non-spontaneous chemical reaction. It involves the decomposition of an electrolyte into its constituent elements or ions. The positive ions migrate towards the cathode, where they gain electrons and undergo reduction, while the negative ions migrate towards the anode, where they lose electrons and undergo oxidation.
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