The electrode potential of the following half cell at 298 ~K X|X²⁺(0.001 M) \| Y²⁺(0.01 M)| Y is _ _ _ _ × 10⁻² ~V (Nearest integer) Given: E⁰ X²⁺ X =-2.36 ~V E Y²⁺ Y ⁰=+0.36 ~V (2.303 RT)/F=0.06 ~V

Redox Reactions and Electrochemistry: JEE Main 2023 Chemistry question

JEE Main 2023 · 30 Jan, Evening shift · Q9

JEE Main 202330 Jan, Evening shiftChemistryHardRedox Reactions and Electrochemistry138× since 2002

The electrode potential of the following half cell at

298 \mathrm{~K}

\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}

is _______

\times 10^{-2} \mathrm{~V}

(Nearest integer) Given:

\mathrm{E}^{0} _ {\mathrm{X}^{2+} \mid \mathrm{X}}=-2.36 \mathrm{~V}

\mathrm{E}_{\mathrm{Y}^{2+} \mid \mathrm{Y}}^{0}=+0.36 \mathrm{~V}

\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}

Your answerType your answer. Integers and decimals are accepted.