How much charge is required for the reduction of 1 mol of MnO-4 to Mn2+ ?
1
1 F
2
5 F
3
3 F
4
6 F
Official Solution
Correct Option: (2)
The reduction of MnO₄⁻ to Mn²⁺ involves the following balanced equation:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
From the balanced equation, we can see that 5 electrons (5e⁻) are required for the reduction of 1 mole of MnO₄⁻ to Mn²⁺.
The unit of charge, Faraday (F), represents 1 mole of electrons, which is equivalent to 96,485.3329 Coulombs.
Therefore, the charge required for the reduction of 1 mole of MnO₄⁻ to Mn²⁺ is:
5 × 96,485.3329 C = 482,426.6645 C
Rounded to the nearest whole number, the charge required is approximately 482,427 C.
Therefore, the correct answer is 482,427 C, which is approximately 5 F (Faraday).
02
PYQ 2022
easy
chemistryID: cuet-ug-
Among the following statements related to ionic conductance. choose the correct statements.
A. Ionic conductance depends on the nature of electrolyte
B. Ionic conductance is due to the movements of electrons
C. Ionic conductance is also called electronic conductance
D. Ionic conductance depends on temperature
E. Ionic conductance also depends on the nature of solvent.
Choose the correct answer from the options given below:
1
A, B and C only
2
B, C and D only
3
B, C and E only
4
A, D and E only
Official Solution
Correct Option: (4)
Among the given statements related to ionic conductance, the correct statements are:
A. Ionic conductance depends on the nature of electrolyte. C. Ionic conductance is also called electronic conductance. D. Ionic conductance depends on temperature. E. Ionic conductance also depends on the nature of solvent.
Therefore, the correct answer is A, D, and E only.
03
PYQ 2022
medium
chemistryID: cuet-ug-
for , and are and respectively. Calculate for
1
2
3
4
Official Solution
Correct Option: (1)
=
=
=
Hence, the correct option is (A):
04
PYQ 2023
easy
chemistryID: cuet-ug-
The cell reaction occurring at anode in the electrolysis of aqueous NaCl solution is:
1
2
3
4
Official Solution
Correct Option: (4)
In the electrolysis of aqueous NaCl solution, the anode is where oxidation occurs. In the case of NaCl electrolysis, the species that undergoes oxidation at the anode is water (H₂O), not chloride ions (Cl⁻), due to the lower oxidation potential of chloride ions compared to water. Thus, oxygen gas is produced at the anode. - The relevant half-reaction for oxidation at the anode is the oxidation of water: - Chloride ions (Cl⁻), while present, are not oxidized at the anode because the oxidation potential for chloride is less favorable than that of water. Therefore, oxygen is the product at the anode. Thus, the correct cell reaction occurring at the anode in the electrolysis of aqueous NaCl solution is:
05
PYQ 2023
hard
chemistryID: cuet-ug-
The quantity of charge required to obtain 2 mol of Mn from MnO is:
1
2 F
2
10 F
3
5 F
4
1 F
Official Solution
Correct Option: (3)
To reduce MnO to Mn , the oxidation state of manganese changes from +7 in MnO to +2 in Mn , meaning 5 electrons are required for the reduction of 1 mol of MnO to Mn . For 2 mol of Mn , 5 electrons are required for each mole, so the total number of moles of electrons required is: To calculate the quantity of charge (Q), we use the relationship: Where:
- is the number of moles of electrons (10 mol),
- is Faraday’s constant, . Thus, the total charge required is: Thus, the correct answer is 10 F.
06
PYQ 2023
hard
chemistryID: cuet-ug-
During electrolysis of acidified water. 2.8 Litre of oxygen gas is released (at STP). The charge passed through the anode is -
1
48250 C
2
24125 C
3
96500 C
4
361875 C
Official Solution
Correct Option: (1)
The correct option is (A) : 48250 C
07
PYQ 2023
medium
chemistryID: cuet-ug-
Decreasing order of reducing power of the given species as per their standard reduction potential is : A. -2.36 B. +0.54 C. +1.23 D. -0.74 E. + 1.81 Choose the correct answer from the options given below:
1
A>D>E>B>C
2
A>D>B>C>E
3
E>C>A>B>D
4
E>C>B>D>A
Official Solution
Correct Option: (2)
The correct option is(B):A>D>B>C>E
08
PYQ 2023
medium
chemistryID: cuet-ug-
A solution of CuSO4 is electrolysed for 10 min with a current of 1.5 A. What is the mass of copper deposited at the cathode (molar mass of Cu= 63 g) (1F = 96487c).
1
2.93 g
2
3.4 g
3
1.9 g
4
0.2938 g
Official Solution
Correct Option: (4)
The correct option is(D):0.2938 g
09
PYQ 2023
medium
chemistryID: cuet-ug-
Which of the following relationships are not correct ? (A) (B) (C) Cell constant = (D) (E) Choose the correct answer from the options given below:
1
(B), (C), (D) only
2
(A), (B), (D) only
3
(B), (C), (D), (E) only
4
(A), (C), (E) only
Official Solution
Correct Option: (2)
The correct option is (B) : (A), (B), (D) only
10
PYQ 2023
medium
chemistryID: cuet-ug-
Given:
Choose the correct order : (A) Ag (B) Ag (C) Ag > Hg > Cr > Mg > K (decreasing oxidising power) (D) Hg > Cr > Ag > K > Mg (increasing reactivity) (E) K > Mg > Cr > Hg > Ag (decreasing reactivity order) Choose the correct answer from the options given below:
1
(A), (C) and (E) only
2
(B), (C) and (E) only
3
(C), (D) and (E) only
4
(A) and (B) only
Official Solution
Correct Option: (1)
The correct option is (A) : (A), (C) and (E) only
11
PYQ 2023
medium
chemistryID: cuet-ug-
The standard reduction potential for Sn /Sn is +0.15 V and for Cr /Cr is -0.74 V. These two half-cells, coupled in their standard states, are connected to make a cell. The galvanic cell can be correctly represented by:
1
Sn (aq) | Sn (aq) || Cr (aq) | Cr(s)
2
Sn (aq) | Sn (aq) || Cr (aq) | Cr(s)
3
Cr(s) | Cr (aq) || Sn (aq) | Sn (aq)
4
Cr(s) | Cr (aq) | Sn (aq) || Sn (aq)
Official Solution
Correct Option: (3)
The standard reduction potential for Sn /Sn is +0.15 V and for Cr /Cr is -0.74 V. The cell potential is calculated as: In this case, the cathode (where reduction occurs) will be the Sn /Sn half-reaction, and the anode (where oxidation occurs) will be the Cr /Cr half-reaction. Based on the reduction potential, Sn will be reduced and Cr will be oxidized. Thus, the cell notation will be:
12
PYQ 2023
medium
chemistryID: cuet-ug-
In the formation of a leclanche cell, the space between electrodes is filled with: (A) ZnCl2 (B) NH4CI (C) AlCl3 (D) ZnS (E) NH4NO3 Choose the correct answer from the options given below:
1
(A) and (E) only
2
(A) and (B) only
3
(C) and (D) only
4
(D) and (E) only
Official Solution
Correct Option: (2)
The correct option is (B) : (A) and (B) only
13
PYQ 2023
medium
chemistryID: cuet-ug-
Find the value of in the missing link in following diagram
1
2
3
4
Official Solution
Correct Option: (1)
The correct option is (A) :
14
PYQ 2023
medium
chemistryID: cuet-ug-
Match List - I with List - II Choose the correct answer from the options given below:
1
(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
2
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
3
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
4
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Official Solution
Correct Option: (3)
Let’s match each cell with its correct property: - Fuel Cell (A): A fuel cell is a type of electrochemical cell that uses the reaction of hydrogen and oxygen to produce water and generate electrical energy. The reaction at the fuel cell is: Fuel cells are non-rechargeable and operate continuously as long as fuel is supplied. Therefore, the correct match is (A)-(III). - Mercury Cell (B): A mercury cell is a primary cell, which provides a steady potential but is not rechargeable. The reaction at the anode in a mercury cell involves the oxidation of zinc: Thus, the correct match is (B)-(IV). - Leclanché Cell (C): The Leclanché cell is a primary electrochemical cell and gives a steady potential, typically used for powering small devices. Thus, the correct match is (C)-(II). - Ni-Cd Cell (D): The Nickel-Cadmium (Ni-Cd) cell is a rechargeable secondary cell. This type of cell is typically used in applications requiring rechargeability. Thus, the correct match is (D)-(I). Summary:
The correct answer is Option (3): - (A)-(III): Fuel cell produces electricity via the reaction . - (B)-(IV): Mercury cell provides steady potential with the anode reaction . - (C)-(II): Leclanché cell provides steady potential but is non-rechargeable. - (D)-(I): Ni-Cd cell is rechargeable.
15
PYQ 2024
hard
chemistryID: cuet-ug-
What is the numerical value of one Faraday in Coulombs?
1
96587
2
96487
3
99500
4
6.023
Official Solution
Correct Option: (1)
One Faraday, in terms of electrochemistry, is defined as the charge of one mole of electrons. This is calculated using the charge of a single electron and Avogadro's number.
The charge of one electron is approximately Coulombs. Avogadro's number, which is the number of atoms or molecules in one mole, is approximately .
The numerical value of one Faraday is the product of these two constants:
Faraday = Charge of one electron × Avogadro's number
Faraday = C ×
This calculation yields approximately 96587 Coulombs.
Therefore, the correct answer is: 96587
16
PYQ 2024
easy
chemistryID: cuet-ug-
Read the following passage and answer the next five questions based on it.
Battery or cell converts chemical energy of the redox reaction to electrical energy. In fuel cells (a galvanic cell), the chemical energy of combustion of fuels like H2, ethanol, etc., is directly converted to electrical energy. In a fuel cell, H2 and O2 react to produce electricity, where H2 gas is oxidized at the anode and oxygen is reduced at the cathode, and the reactions involved are:
Anode reaction: H2 + 2OH- → 2H2O + 2e-
Cathode reaction: O2 + 2H2O + 4e- → 4OH-
67.2 L of H2 at STP reacts in 15 minutes.
Official Solution
Correct Option: (1)
17
PYQ 2025
easy
chemistryID: cuet-ug-
In an electrochemical cell, the standard electrode potential of Zn /Zn is and that of Cu /Cu is . What is the standard EMF of the cell?
1
1.10 V
2
-1.10 V
3
0.42 V
4
-0.42 V
Official Solution
Correct Option: (1)
Step 1: Identify the cathode and anode. In a galvanic cell, the electrode with higher (more positive) potential acts as cathode. Step 2: Use the EMF formula.
