The alkali metal with the lowest E^{circ}_{M^ | vedclass

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Question

(A) The standard reduction potential $E^{\circ}_{M^{+}/M}$ depends on the sublimation enthalpy,ionization enthalpy,and hydration enthalpy of the metal

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$Li, Na$

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(A) The standard reduction potential $E^{\circ}_{M^{+}/M}$ depends on the sublimation enthalpy,ionization enthalpy,and hydration enthalpy of the metal ion.For alkali metals,$Li$ has the most negative standard reduction potential (lowest value) due to its very high hydration enthalpy,which compensates for its high ionization energy.Therefore,$X = Li$.Among the alkali metals,$Li$ has the most negative value,while $Na$ has the least negative (highest) standard reduction potential value compared to other alkali metals like $K, Rb, Cs$.Thus,$Y = Na$.Hence,$X$ and $Y$ are $Li$ and $Na$ respectively.

The alkali metal with the lowest $E^{\circ}_{M^{+}/M} (V)$ is $X$ and the alkali metal with the highest $E^{\circ}_{M^{+}/M} (V)$ is $Y$. Then $X$ and $Y$ are respectively:

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