Chemistry Question from JEE Main 2023 · 30 January, Shift 2

The electrode potential of the following half cell at $298 \, \text{K}$.

$$X \mid X^{2+} (0.001 \, \text{M}) \parallel Y^{2+} (0.01 \, \text{M}) \mid Y$$

is ..... $\times 10^{-2} \, \text{V}$ (Nearest integer).

Given: $E^{0}_{X^{2+} \mid X} = -2.36 \, \text{V}$ $E^{0}_{Y^{2+} \mid Y} = +0.36 \, \text{V}$

$$\frac{2.303RT}{F} = 0.06 \, \text{V}$$