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Electrochemistry

JEE Main 2025 (Online) 24th January Evening ShiftMCQEasy📈 Topic in 7% papers+4−1
Ques. 1JEE 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+

Answer

(3) Cr

Solution

A reducing agent is the reduced form of a couple; lowest E strongest.

Cl- (1.36 V), Mn2+ (1.51 V), Cr (-0.74 V); MnO4- is an oxidant.

Cr is the strongest reducing agent.

Answer: (3)

Keep practising — Electrochemistry

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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
Ques. 7JEE Main 2024, 9 April (Morning Shift)

The molar conductivity for electrolytes A and B are plotted against C1/2 as shown below. Electrolytes A and B respectively are: [figure]

  1. A - weak electrolyte B - weak electrolyte
  2. A - weak electrolyte B - strong electrolyte
  3. A - strong electrolyte B - strong electrolyte
  4. A - strong electrolyte B - weak electrolyte
Ques. 8JEE Main 2024, 9 April (Morning Shift)

The standard reduction potentials at 298 K for the following half cells are given below : Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O, E = 1.33 V Fe3+(aq) + 3e- Fe E = -0.04 V Ni2+(aq) + 2e- Ni E = -0.25 V Ag+(aq) + e- Ag E = 0.80 V Au3+(aq) + 3e- Au E = 1.40 V Consider the given electrochemical reactions, The number of metal(s) which will be oxidized by Cr2O72-, in aqueous solution is .

Ques. 9JEE Main 2024, 9 April (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-I, B-II, C-III, D-IV
  3. A-IV, B-III, C-I, D-II
  4. A-II, B-III, C-IV, D-I
Ques. 10JEE Main 2024, 9 April (Evening Shift)

Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning :

  1. m - m + AC 12 = 0
  2. m2C + Ka m 2 - Ka m m = 0
  3. m - m - AC 12 = 0
  4. m2C - Ka m 2 + Ka m m = 0
Ques. 11JEE Main 2024, 8 April (Evening Shift)

The emf of cell Tl | Tl+ (0.001 M) | | Cu2+ (0.01 M) | Cu is 0.83 V at 298 K. It could be increased by :

  1. increasing concentration of Tl+ ions
  2. increasing concentration of Cu2+ ions
  3. increasing concentration of both Tl+ and Cu2+ ions
  4. decreasing concentration of both Tl+ and Cu2+ ions

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