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Electrochemistry

JEE Main 2025 (Online) 29th January Evening ShiftMCQMedium📈 Topic in 8% papers+4−1
Ques. 1JEE Main 2025, 29 January (Evening Shift)

O2 gas will be evolved as a product of electrolysis of : Choose the correct answer from the options given below :

  1. (B) and (C) only
  2. (B) and (D) only
  3. (A) and (D) only
  4. (A) and (C) only

Answer

(1) (B) and (C) only

Solution

(A) Ag anode dissolves ⇒ no O2

(B) Pt: 2H2O O2+4H++4e- at anode ✓

(C) dilute H2SO4: same anode reaction ✓

(D) conc. H2SO4: 2SO42- S2O82-+2e-, no O2

Answer: (1)

Keep practising — Electrochemistry

Ques. 2JEE Main 2025, 28 January (Morning Shift)

Given below is the plot of the molar conductivity vs concentration for KCl in aqueous solution. [figure] If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100 , then the resistance of the same cell with the dilute solution is 'x' . The value of x is (Nearest integer)

Ques. 3JEE Main 2025, 28 January (Evening Shift)

Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is . (Nearest integer).

Ques. 4JEE Main 2025, 24 January (Morning Shift)

For the given cell Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) The standard cell potential of the above reaction is Given: Ag+ + e- Ag E = xV Fe2+ + 2e- Fe E = yV Fe3+ + 3e- Fe E = zV

  1. x + 2y - 3z
  2. x + 2y
  3. y - 2x
  4. x + y - z
Ques. 5JEE Main 2025, 24 January (Evening Shift)

Based on the data given below : E (Cr2O72-/Cr3+)=1.33 V E (Cl2/Cl-)=1.36 V E (MnO4-/Mn2+)=1.51 V E (Cr3+/Cr)=-0.74 V the strongest reducing agent is :

  1. Cl-
  2. MnO4-
  3. Cr
  4. Mn2+
Ques. 6JEE Main 2025, 23 January (Morning Shift)

FeO42- +2.0 V Fe3+ 0.8 V Fe2+ -0.5 V Fe0 In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of E (FeO42-/Fe2+) is :

  1. 1.2 V
  2. 2.1 V
  3. 1.4 V
  4. 1.7 V
Ques. 7JEE Main 2025, 23 January (Evening Shift)

Standard electrode potentials for a few half cells are mentioned below : E Cu2+/Cu = 0.34 V, E Zn2+/Zn = -0.76 V E Ag+/Ag = 0.80 V, E Mg2+/Mg = -2.37 V Which one of the following cells gives the most negative value of G ?

  1. Zn | Zn2+(1M) || Ag+(1M) | Ag
  2. Ag | Ag+(1M) || Mg2+(1M) | Mg
  3. Zn | Zn2+(1M) || Mg2+(1M) | Mg
  4. Cu | Cu2+(1M) || Ag+(1M) | Ag
Ques. 8JEE Main 2025, 22 January (Morning Shift)

Which of the following electrolyte can be used to obtain H2S2O8 by the process of electrolysis ?

  1. Concentrated solution of sulphuric acid
  2. Dilute solution of sodium sulphate.
  3. Acidified dilute solution of sodium sulphate.
  4. Dilute solution of sulphuric acid
Ques. 9JEE Main 2025, 22 January (Morning Shift)

A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A. The amount of the aluminium deposited at the cathode is . [Given : molar mass of aluminium and chlorine are 27 g mol-1 and 35.5 g mol-1 respectively. Faraday constant = 96500 C mol-1]

  1. 1.660 g
  2. 1.007 g
  3. 0.336 g
  4. 0.441 g
Ques. 10JEE Main 2025, 22 January (Evening Shift)

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true

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