JEE Advanced 2021 · Paper 2 · Chemistry Q4

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ChemistryMultiple correct+4 / -2Question 23 of 57

Some standard electrode potentials at 298 K are given below:

Couple	$E^\circ$
$\ce{Pb^{2+}/Pb}$	−0.13 V
$\ce{Ni^{2+}/Ni}$	−0.24 V
$\ce{Cd^{2+}/Cd}$	−0.40 V
$\ce{Fe^{2+}/Fe}$	−0.44 V

To a solution containing 0.001 M of $\mathbf X^{2+}$ and 0.1 M of $\mathbf Y^{2+}$ , the metal rods $\mathbf X$ and $\mathbf Y$ are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of $\mathbf X$ . The correct combination(s) of $\mathbf X$ and $\mathbf Y$ , respectively, is(are)

(Given: Gas constant, R = 8.314 J K $^{-1}$ mol $^{-1}$ , Faraday constant, F = 96500 C mol $^{-1}$ )

Cd and Ni

Cd and Fe

Ni and Pb

Ni and Fe

Answer and worked solution

Official answer: A, B, C

Condition for X to dissolve. X dissolves if it is the anode, i.e. if the cell $\ce{X|X^{2+}(0.001)||Y^{2+}(0.1)|Y}$ has $E>0$ :
$E=E^\circ_Y-E^\circ_X-\frac{0.059}2\log\frac{0.001}{0.1}=E^\circ_Y-E^\circ_X+0.059.$
So we need $E^\circ_Y-E^\circ_X>-0.059$ V.

Check each option.

(A) Cd, Ni: $+0.16$ . Yes.
(B) Cd, Fe: $-0.04>-0.059$ . Yes.
(C) Ni, Pb: $+0.11$ . Yes.
(D) Ni, Fe: $-0.20$ . No.

(A), (B), (C)

Topics:Nernst equation Galvanic cells and EMF

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JEE Advanced 2021 · Paper 2, Chemistry question 4