JEE Main 20268 Apr, Evening shiftChemistryEasyRedox Reactions and Electrochemistry138× since 2002Given at 298 K : EFe2+/Fe⊖=XE^\ominus_{\mathrm{Fe^{2+}/Fe}}=XEFe2+/Fe⊖=X Volt, EFe3+/Fe⊖=YE^\ominus_{\mathrm{Fe^{3+}/Fe}}=YEFe3+/Fe⊖=Y Volt. The EFe3+/Fe2+⊖E^\ominus_{\mathrm{Fe^{3+}/Fe^{2+}}}EFe3+/Fe2+⊖ in Volt at 298 K is given by :AOption A. 2X − 3YBOption B. 3Y − 2XCOption C. 3Y + 2XDOption D. Y + XCheckSkip · show solution