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Electrochemistry

JEE Main 2025 (Online) 2nd April Evening ShiftNumericalMedium📈 Topic in 20% papers+4
Ques. 1JEE Main 2025, 2 April (Evening Shift)

0.2\% (w/v) solution of NaOH is measured to have resistivity 870.0 m m. The molar conductivity of the solution will be 102 mS dm2 mol-1. (Nearest integer)

Answer

23

Solution

0.2\% (w/v) means 0.2 g NaOH in 100 mL, i.e. 2 g NaOH per litre, so

c = 240 = 0.05 mol L-1 = 50 mol m-3

With = 870.0 m m = 0.870 m,

= 1 = 10.870 = 1.149 S m-1

m = c = 1.14950 = 2.299 10-2 S m2 mol-1

1 S m2 = 103 mS 102 dm2 = 105 mS dm2,

m = 2.299 103 mS dm2 mol-1

= 22.99 102 23 102 mS dm2 mol-1

Answer: 23

Keep practising — Electrochemistry

Ques. 2JEE Main 2025, 29 January (Morning Shift)

For a Mg | Mg2+(aq) || Ag+(aq) | Ag the correct Nernst Equation is :

  1. Ecell = E cell - RT2F [Ag+]2[Mg2+]
  2. Ecell = E cell - RT2F [Mg2+][Ag+]
  3. Ecell = E cell - RT2F [Ag+][Mg2+]
  4. Ecell = E cell + RT2F [Ag+]2[Mg2+]
Ques. 3JEE Main 2025, 29 January (Morning Shift)

The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?

  1. Molar conductivity increases sharply with increase in concentration.
  2. Molar conductivity decreases sharply with increase in concentration.
  3. A small increase in molar conductivity is observed at infinite dilution.
  4. A small decrease in molar conductivity is observed at infinite dilution.
Ques. 4JEE Main 2025, 29 January (Evening Shift)

Match List - I with List - II : Choose the correct answer from the options given below :

  1. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  3. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  4. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
Ques. 5JEE 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
Ques. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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