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JEE Advanced Chemistry Electrochemistry 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Chemistry Question (2021) — Solution

Question

Some standard electrode potentials at 298   K are given below: Pb 2 + / Pb - 0 . 13   V Ni 2 + / Ni - 0 . 24   V Cd 2 + / Cd - 0 . 40   V Fe 2 + / Fe - 0 . 44   V To a solution containing 0 . 001   M of X 2 + and 0 . 1   M of Y 2 + , the metal rods X and Y are inserted ( at 298   K ) and connected by a conducting wire. This resulted in dissolution of X . The correct combination(s) of X and Y , respectively, is(are) (Given: Gas constant, R = 8 . 314   J   K - 1   mol - 1 , Faraday constant, F = 96500   C   mol - 1 )

Options

  1. A. Cd and  Ni
  2. B. Cd and  Fe
  3. C. Ni and  Pb
  4. D. Ni and  Fe

Answer

C. Ni and  Pb

Step-by-step solution

Since  X  is getting dissolved, so reaction is X → X 2 + + 2 e -     Anode Y 2 + + 2 e - → Y     cathode Y 2 + + X → X 2 + + Y E cell = E cell ° - 0 . 06 2 log 10 - 3 10 - 1 E cell = E cell ° + 0 . 06 (A)  Cd   &   Ni \(=0.16+0.05912 2\) \(=.16+0.0591\) \(=0.21(+ ve )\) E cell < 0  (Non-spontaneous) (B)  Cd   &   Fe E cell = − 0 .044 − − 0 .40 + 0 .06 = - 0 . 04 + 0 . 06 = 0 . 02   V E cell > 0  (Spontaneous) (C)  Ni   &   Pb E cell = − 0 .13 − − 0 .24 + 0 .06 = 0 . 11 + 0 . 06 = 0 . 17   V E cell > 0  (spontaneous) (D)  Ni   &   Fe E cell = − 0 .44 − − 0 .24 + 0 .06 = - 0 . 20 + 0 . 06 = - 0 . 14   V E cell < 0  (Non-Spontaneous)

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