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Electrochemistry · Question 1

JEE Advanced 2026 Paper 2MCQMedium📈 Topic in 17% papers+4−1
1

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: array|c|c|c| Salt & Concentration (M) & Molar conductivity ( S cm2 mol-1) \ NaNO3 & 0.01 & 111 \ & 0.04 & 101 \ NaCl & 0.01 & 117 \ & 0.04 & 107 \ AgNO3 & 0.01 & 125 \ & 0.04 & 116 \ array The conductivity of a saturated aqueous solution of AgCl is 1.40 10-6 S cm-1 at 300 K. If the solubility of AgCl in water at 300 K is X mol L-1, then 10(X-1) is (Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

  1. 3
  2. 4
  3. 5
  4. 6

Answer

(3) 5

Solution

Strong electrolytes obey Kohlrausch's law: m= m -A c; c=0.1 and 0.2.

NaNO3: slope = 111-1010.1=100, so =111+10=121; similarly NaCl: 117+10=127; AgNO3: 125+9=134 S cm2 mol-1.

(AgCl)= (AgNO3)+ (NaCl)- (NaNO3)=134+127-121=140 S cm2 mol-1

X= 1000 = 1000 1.40 10-6140=1.0 10-5 mol L-1

10(X-1)=5

Answer: (C)

JEE Advanced 2026 Paper 2

Keep practising — Electrochemistry

2

In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to Cr3+. The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of Cr3+ is . Use: 1 Faraday = 96500 C mol-1

JEE Advanced 2025 Paper 1
3

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is XF 103 volts, where F is the Faraday constant. The value of X is . Use: Standard Gibbs energies of formation at 298 K are: f GoCO2 = -394 kJ mol-1; f Gowater = -237 kJ mol-1; f Gobutane = -18 kJ mol-1

JEE Advanced 2025 Paper 2
4

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
JEE Advanced 2024 Paper 1
5

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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