JEE Advanced 2021 Paper 2 · Q04 · Nernst Equation

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$^{-1}$.)

SolveKar AI · Accepted Answer

X dissolves $\Rightarrow$ X is oxidised, Y$^{2+}$ is reduced. Cell reaction: X(s) + Y$^{2+}$(0.1 M) $\rightarrow$ X$^{2+}$(0.001 M) + Y(s). $E_{cell} = (E^\circ_{Y^{2+}/Y} - E^\circ_{X^{2+}/X}) - \frac{0.0591}{2}\log\frac{[X^{2+}]}{[Y^{2+}]} = (E^\circ_Y - E^\circ_X) - \frac{0.0591}{2}\log\frac{0.001}{0.1} = (E^\circ_Y - E^\circ_X) + \frac{0.0591}{2}\cdot 2 = (E^\circ_Y - E^\circ_X) + 0.0591$. For spontaneous dissolution of X, $E_{cell} > 0$, i.e. $(E^\circ_Y - E^\circ_X) > -0.0591$. (A) X=Cd, Y=Ni: $E^\circ_Y - E^\circ_X = -0.24 - (-0.40) = +0.16$ V; $E_{cell}= 0.16+0.0591 = +0.22$ V > 0. CORRECT. (B) X=Cd, Y=Fe: $-0.44 - (-0.40) = -0.04$ V; $E_{cell} = -0.04+0.0591 = +0.019$ V > 0. CORRECT. (C) X=Ni, Y=Pb: $-0.13 - (-0.24) = +0.11$ V; $E_{cell} = +0.11+0.0591 = +0.17$ V > 0. CORRECT. (D) X=Ni, Y=Fe: $-0.44 - (-0.24) = -0.20$ V; $E_{cell} = -0.20 + 0.0591 = -0.14$ V < 0. NOT spontaneous (Ni does not dissolve). WRONG. Hence A, B, C are correct.

JEE Advanced 2021 Paper 2 · Q04 · Nernst Equation

Solution