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Electrochemistry

JEE Main 2025 (Online) 2nd April Morning ShiftNumericalHard📈 Topic in 29% papers+4
Ques. 1JEE Main 2025, 2 April (Morning Shift)

Consider the following electrochemical cell at standard condition. Au(s) | QH2, Q | NH4X(0.01 M) || Ag+(1 M) | Ag(s) Ecell = +0.4 V The couple QH2/Q represents quinhydrone electrode, the half cell reaction is given below: [figure] [Given : E0 Ag+/Ag = +0.8 V and 2.303RTF = 0.06 V] The pKb value of the ammonium halide salt (NH4X) used here is . (nearest integer)

Answer

6

Solution

Anode (left, oxidation): QH2 Q + 2H+ + 2e-

Cathode (right, reduction): 2Ag+ + 2e- 2Ag

The reduction potential of the quinhydrone electrode is

EQ/QH2 = E Q/QH2 - 0.062 1[ H+]2

= 0.7 - 0.06 pH

and Ecell = E Ag+/Ag - EQ/QH2 (with [ Ag+] = 1 M, E Ag+/Ag = 0.8 V), so

0.4 = 0.8 - (0.7 - 0.06 pH)

0.06 pH = 0.3

pH = 5

NH4X is a salt of a weak base and a strong acid, so on hydrolysis

pH = 7 - 12pKb - 12 C

5 = 7 - 12pKb - 12 (10-2)

5 = 8 - 12pKb

pKb = 6

Answer: 6

Keep practising — Electrochemistry

Ques. 2JEE 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)

Ques. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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