The electrode potential of the following half cell at 298 K. $\mathrm{X}|\mathrm{X}^{2+}(0.001~\mathrm{M}) || \mathrm{Y}^{2+}(0.01~\mathrm{M})|\mathrm{Y}$ is ______ $\times 10^{-2}~\mathrm{V}$ (Nearest integer). Given: $\mathrm{E}_{\mathrm{X}^{2+}|\mathrm{X}}^{0}=-2.36~\mathrm{V}$ $\mathrm{E}_{\mathrm{Y}^{2+}|\mathrm{Y}}^{0}=+0.36~\mathrm{V}$ $\frac{2.303RT}{F}=0.06~\mathrm{V}$

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Redox Reactions and Electrochemistry

The electrode potential of the following half cell at 298 K.

X ∣ X^{2 +} (0.001 M) ∣∣ Y^{2 +} (0.01 M) ∣ Y

is ______

\times 1 0^{- 2} V

(Nearest integer).

Given:

E_{X^{2 +} ∣ X}^{0} = - 2.36 V

E_{Y^{2 +} ∣ Y}^{0} = + 0.36 V

\frac{2.303 R T}{F} = 0.06 V

Answer: 275

The cell runs on the two half-reactions, the more positive electrode serving as cathode, so the standard cell potential is

E_{cell}^{\circ} = 0.36 - (- 2.36) = 2.72 V .

The spontaneous change is

X + Y^{2 +} \to X^{2 +} + Y

, a two-electron reaction, so the Nernst equation reads

E_{cell} = 2.72 - \frac{0.06}{2} lo g \frac{[ X ^{2 +} ]}{[ Y ^{2 +} ]} = 2.72 - 0.03 lo g \frac{0.001}{0.01} .

The logarithm is

- 1

, so the correction adds

0.03 V

and the cell potential is

2.75 V

, that is

275 \times 1 0^{- 2} V

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer