JEE Advanced2021ChemistryElectrochemistryActual
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 const
Options
- ACd and Ni
- BCd and Fe
- CNi and Pb
- DNi and Fe
Correct answer
A. Cd and Ni
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 &#