JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Redox Reactions and Electrochemistry

Some standard electrode potentials at 298 K are given below:

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}

)

Answer: (A), (B), (C)

Solution

For X to dissolve (act as anode), the cell potential

E_{cell}

must be positive. Using the Nernst equation:

E_{cell} = E_{cell}^{\circ} - \frac{0.06}{2} lo g \frac{[ X ^{2 +} ]}{[ Y ^{2 +} ]} = E_{cell}^{\circ} - \frac{0.06}{2} lo g \frac{0.001}{0.1} = E_{cell}^{\circ} + 0.06

. For (A) Cd and Ni:

E^{\circ} = - 0.24 - (- 0.40) = + 0.16

, so

E_{cell} = 0.22

V (positive). For (B) Cd and Fe:

E^{\circ} = - 0.44 - (- 0.40) = - 0.04

, so

E_{cell} = 0.02

V (positive). For (C) Ni and Pb:

E^{\circ} = - 0.13 - (- 0.24) = + 0.11

, so

E_{cell} = 0.17

V (positive). For (D) Ni and Fe:

E^{\circ} = - 0.44 - (- 0.24) = - 0.20

, so

E_{cell} = - 0.14

V (negative). Options (A), (B), and (C) have positive cell potentials.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · One or more correct options