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Electrochemistry · Question 4

JEE Advanced 2024 Paper 1MatchingMedium📈 Topic in 17% papers+4−1
1

In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition. The limiting ionic conductivity ( 0) values (in mS m2 mol-1) for different ions in aqueous solutions are given below: array|c|c|c|c|c|c|c|c|c|c| Ions & Ag+ & K+ & Na+ & H+ & NO3- & Cl- & SO42- & OH- & CH3COO- \ 0 & 6.2 & 7.4 & 5.0 & 35.0 & 7.2 & 7.6 & 16.0 & 19.9 & 4.1 \ array For different combinations of titrates and titrants given in List-I, the graphs of ‘conductance’ versus ‘volume of titrant’ are given in List-II. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

  1. P-4, Q-3, R-2, S-5
  2. P-2, Q-4, R-3, S-1
  3. P-3, Q-4, R-2, S-5
  4. P-4, Q-3, R-2, S-1

Answer

(3) P-3, Q-4, R-2, S-5

Solution

Before the end point the titrate ion is replaced by the titrant ion; after it, the excess titrant adds ions and conductance rises.

P (KCl + AgNO3): Cl- (7.6) replaced by NO3- (7.2) → slight fall, then rise → graph 3.

Q (AgNO3 + KCl): Ag+ (6.2) replaced by K+ (7.4) → slow rise, then steeper rise → graph 4.

R (NaOH + HCl): OH- (19.9) replaced by Cl- (7.6) → sharp fall; excess H+ (35.0) → sharp rise → graph 2.

S (NaOH + CH3COOH): OH- replaced by CH3COO- (4.1) → fall; excess weak acid hardly ionises (common ion) → almost constant → graph 5.

P-3, Q-4, R-2, S-5

Answer: (C)

JEE Advanced 2024 Paper 1

Keep practising — Electrochemistry

2

An aqueous solution of hydrazine (N2H4) is electrochemically oxidized by O2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2(g). Choose the correct statement(s) about the above process

  1. OH- ions react with N2H4 at the anode to form N2(g) and water, releasing 4 electrons to the anode.
  2. At the cathode, N2H4 breaks to N2(g) and nascent hydrogen released at the electrode reacts with oxygen to form water.
  3. At the cathode, molecular oxygen gets converted to OH-.
  4. Oxides of nitrogen are major by-products of the electrochemical process.
JEE Advanced 2024 Paper 2

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