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Which of the following reactions is possible at anode ? [AIEEE-2002]- a)F2 + 2e- → 2F-
- b)
- c)2Cr3+ + 7H2O → Cr2O72- + 14H+ + 6e-
- d)Fe2+ → Fe3+ +e-
Correct answer is option 'C'. Can you explain this answer?
Which of the following reactions is possible at anode ?
[AIEEE-2002]
a)
F2 + 2e- → 2F-
b)
c)
2Cr3+ + 7H2O → Cr2O72- + 14H+ + 6e-
d)
Fe2+ → Fe3+ +e-
|
Sushil Kumar answered |
The correct answer is Option C.
The anode is the electrode where oxidation (loss of electrons) takes place; in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.
The cathode is the electrode where reduction (gain of electrons) takes place; in a galvanic cell, it is the positive electrode, as less oxidation occurs, fewer ions go into solution, and fewer electrons are left on the electrode.
So, in the reaction
2Cr3+ +7H2O→Cr2O72−+14H+
Cr is getting oxidised as the oxidation state is changing from +3 to +6 and hence this reaction is possible at anode.
The cathode is the electrode where reduction (gain of electrons) takes place; in a galvanic cell, it is the positive electrode, as less oxidation occurs, fewer ions go into solution, and fewer electrons are left on the electrode.
So, in the reaction
2Cr3+ +7H2O→Cr2O72−+14H+
Cr is getting oxidised as the oxidation state is changing from +3 to +6 and hence this reaction is possible at anode.
Calculate the standard cell potentials of galvanic cell, ∆rG and equilibrium constant of the reactions if the reaction is
- a)0.01V, – 2.800 kJ mol–1, 3.2
- b)0.03V, – 2.895 kJ mol–1, 3.2
- c)0.02V, – 2.850 kJ mol–1, 3.2
- d)0.04V, – 2.955 kJ mol–1, 3.2
Correct answer is option 'B'. Can you explain this answer?
Calculate the standard cell potentials of galvanic cell, ∆rG and equilibrium constant of the reactions if the reaction is
a)
0.01V, – 2.800 kJ mol–1, 3.2
b)
0.03V, – 2.895 kJ mol–1, 3.2
c)
0.02V, – 2.850 kJ mol–1, 3.2
d)
0.04V, – 2.955 kJ mol–1, 3.2
|
Infinity Academy answered |
The standard emf of a galvanic cell involving 3 moles of electrons in its redox reaction is 0.59V.
Eºcell = 0.59V
n=3
The equilibrium constant for the reaction of the cell is given by the expression: ln K = RT nFEºcell
ln K = 8.314 × 298 × 3 × 96500 × 0.59 = 68.9
K≈1030
n=3
The equilibrium constant for the reaction of the cell is given by the expression: ln K = RT nFEºcell
ln K = 8.314 × 298 × 3 × 96500 × 0.59 = 68.9
K≈1030
The equilibrium constant for the reaction of the cell is 1030.
For the redox reaction :Zn (s) + Cu2+ (0.1M) → Zn2+ (1M) + Cu(s) taking place in a cell, Eºcell is 1.10 volt. Ecell for the cell will be
[AIEEE-2003]- a)1.07 volt
- b)0.82 volt
- c)2.14 volt
- d)1.8 volt
Correct answer is option 'A'. Can you explain this answer?
For the redox reaction :
Zn (s) + Cu2+ (0.1M) → Zn2+ (1M) + Cu(s) taking place in a cell, Eºcell is 1.10 volt. Ecell for the cell will be
[AIEEE-2003]a)
1.07 volt
b)
0.82 volt
c)
2.14 volt
d)
1.8 volt
|
Lavanya Menon answered |
The correct answer is Option A.
Eocell = Ecell – (0.0591/n) log (Zn2+/ Cu2+)
= 1.10 – (0.0591/n) log (1 / 0.1)
= 1.10 – 0.02955
= 1.070 Volt
Eocell = Ecell – (0.0591/n) log (Zn2+/ Cu2+)
= 1.10 – (0.0591/n) log (1 / 0.1)
= 1.10 – 0.02955
= 1.070 Volt
When during electrolysis of a solution of AgNO3 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be – [AIEEE-2003]- a)21.6 g
- b)108 g
- c)1.08 g
- d)10.8 g
Correct answer is option 'D'. Can you explain this answer?
When during electrolysis of a solution of AgNO3 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be –
[AIEEE-2003]
a)
21.6 g
b)
108 g
c)
1.08 g
d)
10.8 g
|
Om Desai answered |
The correct answer is Option D.
Ag+ + e- ---> Ag
So, no. of the moles of Ag = Q / F
= 9650 / 96500
=1 / 10 moles
∴ Mass of Ag produced :->
(1/10) × 108
= 10.8 g
Ag+ + e- ---> Ag
So, no. of the moles of Ag = Q / F
= 9650 / 96500
=1 / 10 moles
∴ Mass of Ag produced :->
(1/10) × 108
= 10.8 g
In an electrolytic cell current flows from -- a)Cathode to anode in outer circuit
- b)Anode to cathode outside the cell
- c)Cathode to anode inside the cell
- d)Anode to cathode inside the cell
Correct answer is option 'A'. Can you explain this answer?
In an electrolytic cell current flows from -
a)
Cathode to anode in outer circuit
b)
Anode to cathode outside the cell
c)
Cathode to anode inside the cell
d)
Anode to cathode inside the cell
|
Krishna Iyer answered |
In an electrolytic cell, current flows from cathode to anode in outer circuit and in daniell cell, it is the reverse direction of flow of current from anode to cathode in outer circuit.
The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96500 C mol-1 ; R = 8.314 JK-1mol-1) [AIEEE-2004]- a)1.0 × 101
- b)1.0 ×105
- c) 1.0 × 1010
- d)1.0 × 1030
Correct answer is option 'C'. Can you explain this answer?
The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96500 C mol-1 ; R = 8.314 JK-1mol-1)
[AIEEE-2004]
a)
1.0 × 101
b)
1.0 ×105
c)
1.0 × 1010
d)
1.0 × 1030
|
Geetika Shah answered |
The correct answer is option C
For a cell reaction in equilibrium at 298 K,
Eocell =0.0591/nlogKc
(Ke = equilibrium constant)
Give, Eocell = 0.591V
Now, log Kc = (Eocell x n )/ 0.0591
=(0.0591 x n )/ 0.0591
log Kc = 10
Kc = antilog 10
For a cell reaction in equilibrium at 298 K,
Eocell =0.0591/nlogKc
(Ke = equilibrium constant)
Give, Eocell = 0.591V
Now, log Kc = (Eocell x n )/ 0.0591
=(0.0591 x n )/ 0.0591
log Kc = 10
Kc = antilog 10
- Kc = 1 x 1010
On the basis of the following E° values, the strongest oxidizing agent is: 
- a)[Fe(CN)6]4-
- b)Fe2+
- c)Fe3+
- d)[Fe(CN)6]3-
Correct answer is option 'C'. Can you explain this answer?
On the basis of the following E° values, the strongest oxidizing agent is:
a)
[Fe(CN)6]4-
b)
Fe2+
c)
Fe3+
d)
[Fe(CN)6]3-
|
Dishani Kulkarni answered |
Strongest oxidizing agent is one having more positive or less negative reduction potential.
The standard electrode potential is measured by- a)Galvanometer
- b)Voltmeter
- c)Electrometer
- d)Pyrometer
Correct answer is option 'B'. Can you explain this answer?
The standard electrode potential is measured by
a)
Galvanometer
b)
Voltmeter
c)
Electrometer
d)
Pyrometer
|
Gauri Khanna answered |
Voltmeter measures the potential.
For the cell, 
and for the cell Pt(H2) | H+ (1M)| Ag, 
Thus Ecell for theAg|Ag+ (0.1M) || Zn2+ (0.1M) | Zn is ....................and cell reaction is...............- a)1.44 V ,spontaneous
- b)0.4 V ,spontaneous
- c)-1.44 V ,non-spontaneous
- d)-1.53 V ,non-spontaneous
Correct answer is option 'D'. Can you explain this answer?
For the cell, and for the cell Pt(H2) | H+ (1M)| Ag,
and for the cell Pt(H2) | H+ (1M)| Ag, Thus Ecell for the
Ag|Ag+ (0.1M) || Zn2+ (0.1M) | Zn is ....................and cell reaction is...............
a)
1.44 V ,spontaneous
b)
0.4 V ,spontaneous
c)
-1.44 V ,non-spontaneous
d)
-1.53 V ,non-spontaneous
|
|
Geetika Shah answered |
Ecell < 0, hence reaction is non-spontaneous.
For the following cell with hydrogen electrodes at two different pressure p1 and p2 
emf is given by- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
For the following cell with hydrogen electrodes at two different pressure p1 and p2
emf is given by
a)
b)
c)
d)
|
|
Krishna Iyer answered |
For SHE E°SHE = 0.00 V
Oxidation at anode (left)
Reduction at cathode (right)
Net
Oxidation at anode (left)
Reduction at cathode (right)
Net
This is the type of the cell in which electrodes at different pressures are dipped in same electrolyte and connectivity is made by a salt-bridge.
Reaction Quotient (Q) 
∵
In Daniel cell, oxidation takes place at- a)Any of the two electrodes
- b)Depends on the salts and their solutions
- c)Anode
- d)Cathode
Correct answer is option 'C'. Can you explain this answer?
In Daniel cell, oxidation takes place at
a)
Any of the two electrodes
b)
Depends on the salts and their solutions
c)
Anode
d)
Cathode
|
Rohit Shah answered |
The solutions in which the electrodes are immersed are called electrolytes. The chemical reaction that takes place in a galvanic cell is the redox reaction. One electrode acts as anode in which oxidation takes place and the other acts as the cathode in which reduction takes place.
Which of the following solutions has the highest equivalent conductance ?- a)0.01M NaCl
- b)0.050 M NaCl
- c)0.005M NaCl
- d)0.02M NaCl
Correct answer is option 'C'. Can you explain this answer?
Which of the following solutions has the highest equivalent conductance ?
a)
0.01M NaCl
b)
0.050 M NaCl
c)
0.005M NaCl
d)
0.02M NaCl
|
Preeti Iyer answered |
Higher the dilution higher will be the equivalent conductance
Correct arrangement of Al, Cu, Fe, Mg and Zn in the order in which they displace each other from the solution of their salts is- a)Mg, Al, Zn, Fe, Cu
- b)Mg, Al, Zn, Cu, Fe
- c)Mg, Al, Cu, Fe, Zn
- d)Cu, Al, Zn, Fe, Mg
Correct answer is option 'A'. Can you explain this answer?
Correct arrangement of Al, Cu, Fe, Mg and Zn in the order in which they displace each other from the solution of their salts is
a)
Mg, Al, Zn, Fe, Cu
b)
Mg, Al, Zn, Cu, Fe
c)
Mg, Al, Cu, Fe, Zn
d)
Cu, Al, Zn, Fe, Mg
|
Mehul Choudhary answered |
Reactivity series.
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10-3 S cm-1.- a)0.219 cm-1
- b)0.239 cm-1
- c)0.229 cm-1
- d)0.209 cm-1
Correct answer is option 'A'. Can you explain this answer?
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10-3 S cm-1.
a)
0.219 cm-1
b)
0.239 cm-1
c)
0.229 cm-1
d)
0.209 cm-1
|
|
Preeti Iyer answered |
Κ = G x cell constant and G = 1/R.
Given EºCr3+/Cr = – 0.72 V, EºFe2+/Fe= – 0.42 V. The potential for the cell Cr |Cr3+ (0.1 M)| |Fe2+ (0.01 M) | Fe is - [AIEEE 2008]- a)0.339 V
- b)– 0.339 V
- c)– 0.26 V
- d)0.26 V
Correct answer is option 'D'. Can you explain this answer?
Given EºCr3+/Cr = – 0.72 V, EºFe2+/Fe= – 0.42 V. The potential for the cell Cr |Cr3+ (0.1 M)| |Fe2+ (0.01 M) | Fe is - [AIEEE 2008]
a)
0.339 V
b)
– 0.339 V
c)
– 0.26 V
d)
0.26 V
|
|
Gaurav Kumar answered |
The correct answer is option D
Cr∣Cr3+(0.1 M)∣∣Fe2+(0.01 M)∣Fe
Oxidation half-cell
Cr→Cr3++3e−]×2
Reduction half-cell
Fe2++2e−→Fe]×3
Net cell reaction
2Cr+3Fe2+→2Cr3++3Fe, n=6
E∘cell= E∘oxi+E∘red
=0.72−0.42 = 0.30 V
Cr∣Cr3+(0.1 M)∣∣Fe2+(0.01 M)∣Fe
Oxidation half-cell
Cr→Cr3++3e−]×2
Reduction half-cell
Fe2++2e−→Fe]×3
Net cell reaction
2Cr+3Fe2+→2Cr3++3Fe, n=6
E∘cell= E∘oxi+E∘red
=0.72−0.42 = 0.30 V
One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done- a)by covering the surface with oil
- b)by covering the surface with salt
- c)by covering the surface with citric acid
- d)by covering the surface with paint or by some chemicals
Correct answer is option 'D'. Can you explain this answer?
One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done
a)
by covering the surface with oil
b)
by covering the surface with salt
c)
by covering the surface with citric acid
d)
by covering the surface with paint or by some chemicals
|
Rajesh Chatterjee answered |
Preventing corrosion is essential to ensure the longevity and integrity of metallic objects. One of the simplest and most effective methods to prevent corrosion is by covering the surface of the object with paint or certain chemicals. This method creates a protective barrier between the metal surface and the atmosphere, preventing contact and thereby reducing the chances of corrosion.
Here is a detailed explanation of why covering the surface with paint or chemicals is an effective method of preventing corrosion:
1. Creation of a Barrier: By covering the surface of the metallic object with paint or chemicals, a physical barrier is created between the metal and the surrounding atmosphere. This barrier prevents the metal from coming into direct contact with oxygen, moisture, and other corrosive elements present in the air.
2. Protection from Moisture: Moisture is one of the primary causes of corrosion. When metal comes into contact with moisture, it undergoes a chemical reaction known as oxidation, leading to the formation of rust. By covering the surface, the paint or chemicals act as a waterproof layer, preventing moisture from reaching the metal surface and reducing the chances of corrosion.
3. Prevention of Oxygen Exposure: Oxygen is another key component in the corrosion process. By covering the surface, the paint or chemicals act as a barrier, preventing oxygen from reaching the metal surface. This helps in inhibiting the oxidation reaction and reducing the chances of corrosion.
4. Chemical Protection: Some paints and chemicals used for covering the surface of metallic objects contain corrosion inhibitors. These inhibitors are chemicals that actively work to prevent or slow down the corrosion process. They form a protective layer on the metal surface, hindering the reaction between the metal and corrosive elements.
5. Flexibility and Ease of Application: Paints and chemicals offer the advantage of being flexible and easy to apply on different types of metal surfaces. They can be brushed, sprayed, or dipped, allowing for complete coverage and protection against corrosion.
It is important to note that the choice of paint or chemical coating depends on the specific requirements of the metal object, the environment it will be exposed to, and the type of corrosion it is susceptible to. Consulting with experts or conducting thorough research is recommended to ensure the most suitable protective coating is applied.
Here is a detailed explanation of why covering the surface with paint or chemicals is an effective method of preventing corrosion:
1. Creation of a Barrier: By covering the surface of the metallic object with paint or chemicals, a physical barrier is created between the metal and the surrounding atmosphere. This barrier prevents the metal from coming into direct contact with oxygen, moisture, and other corrosive elements present in the air.
2. Protection from Moisture: Moisture is one of the primary causes of corrosion. When metal comes into contact with moisture, it undergoes a chemical reaction known as oxidation, leading to the formation of rust. By covering the surface, the paint or chemicals act as a waterproof layer, preventing moisture from reaching the metal surface and reducing the chances of corrosion.
3. Prevention of Oxygen Exposure: Oxygen is another key component in the corrosion process. By covering the surface, the paint or chemicals act as a barrier, preventing oxygen from reaching the metal surface. This helps in inhibiting the oxidation reaction and reducing the chances of corrosion.
4. Chemical Protection: Some paints and chemicals used for covering the surface of metallic objects contain corrosion inhibitors. These inhibitors are chemicals that actively work to prevent or slow down the corrosion process. They form a protective layer on the metal surface, hindering the reaction between the metal and corrosive elements.
5. Flexibility and Ease of Application: Paints and chemicals offer the advantage of being flexible and easy to apply on different types of metal surfaces. They can be brushed, sprayed, or dipped, allowing for complete coverage and protection against corrosion.
It is important to note that the choice of paint or chemical coating depends on the specific requirements of the metal object, the environment it will be exposed to, and the type of corrosion it is susceptible to. Consulting with experts or conducting thorough research is recommended to ensure the most suitable protective coating is applied.
The standard reduction potentials for Zn2+/Zn, Ni2+ /Ni, and Fe2+/Fe are – 0.76, – 0.23 and – 0.44 V respectively. The reaction X + Y2+ → X2+ + Y will be spontaneous when – [AIEEE 2012]- a) X = Ni, Y= Zn
- b)X = Fe, Y = Zn
- c)X = Zn, Y = Ni
- d)X = Ni, Y = Fe
Correct answer is option 'C'. Can you explain this answer?
The standard reduction potentials for Zn2+/Zn, Ni2+ /Ni, and Fe2+/Fe are – 0.76, – 0.23 and – 0.44 V respectively. The reaction X + Y2+ → X2+ + Y will be spontaneous when – [AIEEE 2012]
a)
X = Ni, Y= Zn
b)
X = Fe, Y = Zn
c)
X = Zn, Y = Ni
d)
X = Ni, Y = Fe
|
Vivek Rana answered |
The correct answer is option C
The elements with high negative value of standard reduction potential are good reducing agents and can be easily oxidised. ltbRgt Thus X should have a high negative value of standard potential then Y, so that it will be oxidised to X2+ by reducing Y2+ to Y.
X= Zn, Y= Ni
Zn + Ni2+ → Zn2+ + Ni
Alternatively, for a spontaneous reaction E∘ must be positive.
E∘=E∘reduced−E∘oxidised
=−0.23−(0.76)
implies E∘=+0.53V
The elements with high negative value of standard reduction potential are good reducing agents and can be easily oxidised. ltbRgt Thus X should have a high negative value of standard potential then Y, so that it will be oxidised to X2+ by reducing Y2+ to Y.
X= Zn, Y= Ni
Zn + Ni2+ → Zn2+ + Ni
Alternatively, for a spontaneous reaction E∘ must be positive.
E∘=E∘reduced−E∘oxidised
=−0.23−(0.76)
implies E∘=+0.53V
Given the data at 25ºC,Ag + I- → AgI + e- Eº = 0.152 VAg → Ag+ + e- Eº = -0.800 VWhat is the value of log Ksp for AgI ?(2.303 RT/F = 0.059 V) [AIEEE 2006]- a)+8.612
- b)–37.83
- c)–16.13
- d) –8.12
Correct answer is option 'C'. Can you explain this answer?
Given the data at 25ºC,
Ag + I- → AgI + e- Eº = 0.152 V
Ag → Ag+ + e- Eº = -0.800 V
What is the value of log Ksp for AgI ?
(2.303 RT/F = 0.059 V) [AIEEE 2006]
a)
+8.612
b)
–37.83
c)
–16.13
d)
–8.12
|
Neha Sharma answered |
The correct answer is option C
AgI (s) + e− ⇌ Ag(s) + I−;
E∘ = −0.152 + Ag(s) → Ag+ + e−
E∘ = −0.8
Agl (s) → Ag+ + I−
E∘ = −0.952
AgI (s) + e− ⇌ Ag(s) + I−;
E∘ = −0.152 + Ag(s) → Ag+ + e−
E∘ = −0.8
Agl (s) → Ag+ + I−
E∘ = −0.952
For a cell reaction involving a two-electron change the standard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25º C will be – [AIEEE-2003]- a)10
- b)1 × 1010
- c) 1 × 10-10
- d)29.5 × 10-2
Correct answer is option 'B'. Can you explain this answer?
For a cell reaction involving a two-electron change the standard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25º C will be – [AIEEE-2003]
a)
10
b)
1 × 1010
c)
1 × 10-10
d)
29.5 × 10-2
|
|
Geetika Shah answered |
The correct answer is Option B.
Eocell = (0.0591/n) log Kc
0.259 = (0.0591/2) log Kc
0.259 = 0.0295 log Kc
Kc = antilog 10
Kc = 1 x 1010
Eocell = (0.0591/n) log Kc
0.259 = (0.0591/2) log Kc
0.259 = 0.0295 log Kc
Kc = antilog 10
Kc = 1 x 1010
Standard reduction electrode potentials of three metals A, B and C are respectively +0.5 V, – 3.0V and – 1.2 V. The reducing powers of these metals are -[AIEEE-2003]- a)C > B > A
- b) A > C > B
- c) B > C > A
- d) A > B > C
Correct answer is option 'C'. Can you explain this answer?
Standard reduction electrode potentials of three metals A, B and C are respectively +0.5 V, – 3.0V and – 1.2 V. The reducing powers of these metals are -
[AIEEE-2003]
a)
C > B > A
b)
A > C > B
c)
B > C > A
d)
A > B > C
|
|
Jyoti Aiims Aspirant answered |
More negative the electrode potential higher will the reducing potential of metal
EMF of the following cell is 0.2905 V 
The equilibrium constant for the cell reaction is [IIT JEE 2004]- a)100.32/0.059
- b)100.32/0.0295
- c)100.26/0.0295
- d)100.32/0.295
Correct answer is option 'B'. Can you explain this answer?
EMF of the following cell is 0.2905 V
The equilibrium constant for the cell reaction is
[IIT JEE 2004]
a)
100.32/0.059
b)
100.32/0.0295
c)
100.26/0.0295
d)
100.32/0.295
|
Dilip Chaurasiya answered |
E0 cell = .0591/n log k. k= equilibrium constant
The variation of equivalent conductance vs decrease in concentration of a strong electrolyte is correctly given in the plot -- a) A
- b) A
- c)A
- d) A
Correct answer is option 'A'. Can you explain this answer?
The variation of equivalent conductance vs decrease in concentration of a strong electrolyte is correctly given in the plot -
a)
A
b)
A
c)
A
d)
A
|
|
Om Desai answered |
On decreasing the value of M will increase but increase will be hyberbolic.
For a cell given below Ag | Ag+ || Cu2+ | Cu — + Ag+ + e- → Ag, Eº = x Cu2+ +2e- → Cu, Eº = y Eº cell is – [AIEEE-2002]- a)x + 2y
- b) 2x + y
- c) y – x
- d) y – 2x
Correct answer is option 'C'. Can you explain this answer?
For a cell given below Ag | Ag+ || Cu2+ | Cu
— +
Ag+ + e- → Ag, Eº = x
Cu2+ +2e- → Cu, Eº = y
Eº cell is –
[AIEEE-2002]
a)
x + 2y
b)
2x + y
c)
y – x
d)
y – 2x
|
|
Vivek Rana answered |
The correct answer is Option C.
The left portion of the salt bridge(| |) is anode and the right portion of the salt bridge(| |) is cathode.
Since, Ag | Ag+ is half-cell oxidation and Cu2+ | Cu is half-cell reduction. Thus,
Eo cell = E cathode - E anode
= y - x
Eo cell = E cathode - E anode
= y - x
Refining of impure copper with zinc impurity is to be done by electrolysis using electrods as –[AIEEE-2002]- a)Pure Copper(Cathode) Pure Zinc(Anode)
- b)Pure Zinc(Cathode) Pure Cathode(Anode)
- c)Pure Copper(Cathode) Impure Copper(Anode)
- d)Pure Zinc(Cathode) Impure Zinc(Anode)
Correct answer is option 'C'. Can you explain this answer?
Refining of impure copper with zinc impurity is to be done by electrolysis using electrods as –
[AIEEE-2002]
a)
Pure Copper(Cathode) Pure Zinc(Anode)
b)
Pure Zinc(Cathode) Pure Cathode(Anode)
c)
Pure Copper(Cathode) Impure Copper(Anode)
d)
Pure Zinc(Cathode) Impure Zinc(Anode)
|
|
Vivek Rana answered |
Cathode is always taken of pure substance because cathode is for reduction. at anode oxidation takes place and the metal oxidises itself into cation and moves to cathode, and gets deposited there. Whereas the impurities settle down below anode as anode mud.
The passage of current through a solution of certain electrolye results in the evolution of H2 at cathode and Cl2 at anode. The electrolytic solution is -- a)Water
- b)H2SO4
- c)A NaCl
- d)A CuCl2
Correct answer is option 'C'. Can you explain this answer?
The passage of current through a solution of certain electrolye results in the evolution of H2 at cathode and Cl2 at anode. The electrolytic solution is -
a)
Water
b)
H2SO4
c)
A NaCl
d)
A CuCl2
|
Manjunath Madari answered |
Aq NaCl because in aqueous condition at cathode first group and second group element are not deposited so H2 will deposit at cathode.
Two half-cells are given
For a spontaneous cell reaction, cell set up is - a)Ag| AgCI| KCI (0.2M) 11 KBr (0.001 M)| AgBr| Ag
- b)Ag| AgBr| KBr (0.001 M) 11 KCI (0.2 M) | AgCI | Ag
- c)Both (a) and (b)
- d)None of the above
Correct answer is option 'A,B'. Can you explain this answer?
Two half-cells are given
For a spontaneous cell reaction, cell set up is
a)
Ag| AgCI| KCI (0.2M) 11 KBr (0.001 M)| AgBr| Ag
b)
Ag| AgBr| KBr (0.001 M) 11 KCI (0.2 M) | AgCI | Ag
c)
Both (a) and (b)
d)
None of the above
|
Riya Banerjee answered |
Ksp (AgCI) = 2.8 x 10-10
[Ag+] [Cl-] = 2 .8 x 10-10
[Ag+] [Cl-] = 2 .8 x 10-10
∴ [Ag+]left = 
Ksp (AgBr) = 3.3 x 10-12
[Ag+][Br-] = 3.3 x 10-13
Ksp (AgBr) = 3.3 x 10-12
[Ag+][Br-] = 3.3 x 10-13
∴ [Ag+]left = 
Net
= -0.037 V
Thus, cell reaction is non-spontaneous
(b) Cell is reversed of (a), thus spontaneous.
Thus, cell reaction is non-spontaneous
(b) Cell is reversed of (a), thus spontaneous.
If mercury is used as cathode in the electrolysis of aqueous NaCl solution, the ions discharged at cathode are-- a)H+
- b)Na+
- c)OH¯
- d)Cl¯
Correct answer is option 'B'. Can you explain this answer?
If mercury is used as cathode in the electrolysis of aqueous NaCl solution, the ions discharged at cathode are-
a)
H+
b)
Na+
c)
OH¯
d)
Cl¯
|
|
Geetika Shah answered |
If mercury is used as cathode in the electrolysis of aqueos NaCl solution then the metal is discharged at mercury to from amalgam.
A solution of Fe2+ is titrated potentiometrically using Ce4+ solution.Fe2+ → Fe3+ + e- , E0 = -0.77 Vemf of the Pt | Fe2+ , Fe3+ pair at 50% and 90% titration of Fe2+ are - a)0.77 V , 0.826 V
- b)-0.826 V , -0.77 V
- c)-0.77 V ,-0.826 V
- d)0.00 V , 0.00 V
Correct answer is option 'C'. Can you explain this answer?
A solution of Fe2+ is titrated potentiometrically using Ce4+ solution.
Fe2+ → Fe3+ + e- , E0 = -0.77 V
emf of the Pt | Fe2+ , Fe3+ pair at 50% and 90% titration of Fe2+ are
a)
0.77 V , 0.826 V
b)
-0.826 V , -0.77 V
c)
-0.77 V ,-0.826 V
d)
0.00 V , 0.00 V
|
|
Lavanya Menon answered |
When Fe2+ is 50% titrated
=
where Fe2+ is 90% titrated
The density of Cu is 8.94 g cm–3. The quantity of electricity needed to plate an area 10 cm × 10cm to a thickness of 10-2 cm using CuSO4 solution would be- a)13586 C
- b)27172 C
- c)40758 C
- d)20348 C
Correct answer is option 'B'. Can you explain this answer?
The density of Cu is 8.94 g cm–3. The quantity of electricity needed to plate an area 10 cm × 10cm to a thickness of 10-2 cm using CuSO4 solution would be
a)
13586 C
b)
27172 C
c)
40758 C
d)
20348 C
|
|
Lavanya Menon answered |
Volume : 10 × 10 × 10–2 = 1 cm3
mass of Cu = 8.94 g
= 27172 C
mass of Cu = 8.94 g
= 27172 C
A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of 2.768 × 10-3 ohm-1 cm-1. If its resistance is 82.4 ohm at 25ºC, the cell constant is-- a)0.2182 cm-1
- b)0.2281 cm-1
- c)0.2821 cm-1
- d)0.2381 cm-1
Correct answer is option 'B'. Can you explain this answer?
A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of 2.768 × 10-3 ohm-1 cm-1. If its resistance is 82.4 ohm at 25ºC, the cell constant is-
a)
0.2182 cm-1
b)
0.2281 cm-1
c)
0.2821 cm-1
d)
0.2381 cm-1
|
|
Vivek Rana answered |
K = G. L/A
10–3 × 2.768 = 1/R ×L/A
L/A = 228.08 × 10–3
= 0.2281 cm–1
10–3 × 2.768 = 1/R ×L/A
L/A = 228.08 × 10–3
= 0.2281 cm–1
Aluminium oxide may be electrolysed at 1000ºC to furnish aluminium metal (At. Mass=27 amu ; 1 Faraday = 96,500 Coulombs). The cathode reaction isAl3+ + 3e-→ Alº To prepare 5.12 kg of aluminium metal by this method would require - [AIEEE-2005]- a)1.83 × 107 C of electricity
- b)5.49 × 107 C of electricity
- c)5.49 × 101 C of electricity
- d)5.49 × 104 C of electricity
Correct answer is option 'B'. Can you explain this answer?
Aluminium oxide may be electrolysed at 1000ºC to furnish aluminium metal (At. Mass=27 amu ; 1 Faraday = 96,500 Coulombs). The cathode reaction is
Al3+ + 3e-→ Alº
To prepare 5.12 kg of aluminium metal by this method would require -
[AIEEE-2005]
a)
1.83 × 107 C of electricity
b)
5.49 × 107 C of electricity
c)
5.49 × 101 C of electricity
d)
5.49 × 104 C of electricity
|
Anjali Gupta answered |
Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively are- a)3, 1 and 2
- b)1, 3 and 2
- c)3, 1 and 3
- d)2, 3 and 2
Correct answer is option 'B'. Can you explain this answer?
Salts of A (atomic weight = 7), B (atomic weight = 27) and C (atomic weight = 48) were electrolysed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively are
a)
3, 1 and 2
b)
1, 3 and 2
c)
3, 1 and 3
d)
2, 3 and 2
|
|
Roshni Desai answered |
If x = 1 ⇒ y = 3, z = 2
A concentration cell reversible to anion (Cl-) is set up 
cell reaction is spontaneous ,if - a)C1 > C2
- b)C1 < C2
- c)C1 = C2
- d)C1 = 0
Correct answer is option 'A'. Can you explain this answer?
A concentration cell reversible to anion (Cl-) is set up
cell reaction is spontaneous ,if
a)
C1 > C2
b)
C1 < C2
c)
C1 = C2
d)
C1 = 0
|
|
Om Desai answered |
This is a type of concentration cell with gas electrodes at the same pressure (1 bar) but dipped in aqueous solution of different concentration. Hence, a potential difference is set up.
At anode
At cathode
= 
To make cell reaction spontaneous, Ecell > 0, hence C, > C2
One mole of electron passes through each of the solution of AgNO3, CuSO4 and AlCl3 when Ag, Cu and Al are deposited at cathode. The molar ratio of Ag, Cu and Al deposited are- a)1 : 1 : 1
- b)6 : 3 : 2
- c)6 : 3 : 1
- d)1 : 3 : 6
Correct answer is option 'B'. Can you explain this answer?
One mole of electron passes through each of the solution of AgNO3, CuSO4 and AlCl3 when Ag, Cu and Al are deposited at cathode. The molar ratio of Ag, Cu and Al deposited are
a)
1 : 1 : 1
b)
6 : 3 : 2
c)
6 : 3 : 1
d)
1 : 3 : 6
|
Anisha Bose answered |
Molar ratio
All have the same equivalent
All have the same equivalent
When KMnO4 acts as an oxidizing agent and ultimately forms MnO42-, MnO2, Mn2O3, and Mn2+ then the number of electrons transferred in each case - a)1, 3, 4, 5
- b)4, 3, 1, 5
- c)3, 5, 7, 1
- d)1, 5, 3, 7
Correct answer is option 'A'. Can you explain this answer?
When KMnO4 acts as an oxidizing agent and ultimately forms MnO42-, MnO2, Mn2O3, and Mn2+ then the number of electrons transferred in each case
a)
1, 3, 4, 5
b)
4, 3, 1, 5
c)
3, 5, 7, 1
d)
1, 5, 3, 7
|
Nidhi Yadav answered |
No. of electrons = change in oxidation number of active elemnt.
For a spontaneous reaction the ΔG, equilibrium constant (K) and Ecell will be respectively - [AIEEE-2005]- a)+ve, > 1, –ve
- b)–ve, > 1, +ve
- c)–ve, > 1, –ve
- d)– ve, < 1, –ve
Correct answer is option 'B'. Can you explain this answer?
For a spontaneous reaction the ΔG, equilibrium constant (K) and Ecell will be respectively -
[AIEEE-2005]
a)
+ve, > 1, –ve
b)
–ve, > 1, +ve
c)
–ve, > 1, –ve
d)
– ve, < 1, –ve
|
|
Anaya Patel answered |
The correct answer is option B
we know ΔG=−RTlnK=−nFE∘
We know for a spontaneous reaction
ΔG<0.
∴−RTlnK<0
∴lnK>0
∴K>1
and −nFE∘<0
∴Ecell∘>0
we know ΔG=−RTlnK=−nFE∘
We know for a spontaneous reaction
ΔG<0.
∴−RTlnK<0
∴lnK>0
∴K>1
and −nFE∘<0
∴Ecell∘>0
In the electrolysis of CuSO4, the reaction :Cu2+ + 2e¯ → Cu, takes place at :- a)Anode
- b)Cathode
- c)In solution
- d)None
Correct answer is option 'B'. Can you explain this answer?
In the electrolysis of CuSO4, the reaction :
Cu2+ + 2e¯ → Cu, takes place at :
a)
Anode
b)
Cathode
c)
In solution
d)
None
|
|
Ananya Singh answered |
Cu²⁺ reduces to Cu. Because of reduction of Cu²⁺, reaction takes place at cathode.Read NCERT book for more clear
When an aqueous solution of H2SO4 is electrolysed, the ion discharged at anode is-
- a)H¯
- b)OH¯
- c)SO42-
- d)O2
Correct answer is option 'B'. Can you explain this answer?
When an aqueous solution of H2SO4 is electrolysed, the ion discharged at anode is-
a)
H¯
b)
OH¯
c)
SO42-
d)
O2
|
Deepak Kapoor answered |
Water oxidation at anode can be represented and SO42– can't be discharged
so
2H2O → 4H+ + O2 + 4e–
so ion which will be discharged at anode will be OH–.
so
2H2O → 4H+ + O2 + 4e–
so ion which will be discharged at anode will be OH–.
The reduction potential values of M, N and O are +2.46, -1.13 and -3.13 V respectively. Which of the following order is correct regarding their reducing property?- a)M>N>O
- b)O>M>N
- c)M>O>N
- d)O>N>M
Correct answer is option 'D'. Can you explain this answer?
The reduction potential values of M, N and O are +2.46, -1.13 and -3.13 V respectively. Which of the following order is correct regarding their reducing property?
a)
M>N>O
b)
O>M>N
c)
M>O>N
d)
O>N>M
|
|
Geetika Shah answered |
Reduction potential means the tendency to reduce itself, i.e SO if we need reducing agent, then we need elements having more oxidising potential or reversing reduction potential order.
The order of reduction potential - O < N < M
So the order of reducing agent = M < N < O
The order of reduction potential - O < N < M
So the order of reducing agent = M < N < O
Consider the following cell reaction,
E°= 1.67 VAt [Fe2+] = 1 x 10-3 M. 
and pH = 3,the cell potential at 298 K is - a)1.47 V
- b)1.77 V
- c)1.87 V
- d)1.57 V
Correct answer is option 'D'. Can you explain this answer?
Consider the following cell reaction,
E°= 1.67 VAt [Fe2+] = 1 x 10-3 M. and pH = 3,the cell potential at 298 K is
and pH = 3,the cell potential at 298 K is a)
1.47 V
b)
1.77 V
c)
1.87 V
d)
1.57 V
|
|
Nikita Singh answered |
pH = 3, [H+] = 10-3M
Electrons involved n = 4
Electrons involved n = 4
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