Question

The reducing power of a metal depends on various factor. Suggest the factor which makes Li, the strongest reducing agent in aquous solution.
(a) Sublimation enthalpy (b) Lonisation enthalpy
(c) Hydration enthalpy (d) Electron - gain enthalpy

Answer 1

(c)

Standard reduction potential $(E_{RP}^{\circ }$ Is ameasure f tendency fo an element to lose electron in aqueous solution. Higher the negative $E_{RP}^{\circ }$ greater is the ability to lose electrons.
$E_{RP}^{\circ }$
(i) enthalpy of sublimation
(ii) ionisation enthalpy
(iii) enthalpy of hydration
Thus, in aqueous medium, order of reactivity of alkali metals is Na < K Rb < Cs < Li,
$E_{RP}^{\circ }$ value of Li is least (-3, 04V) among all alkali imetals
The formation's of Li^{+} (aq) from Li involves following steps
(i) $Li\left(s\right)\stackrel{Sublimation}{\to }Li\left(g\right)△H_5=$ Enthalpy of sublimation
$Li\left(g\right)L{i}^{+}\left(s\right)I{E}_1=$ Ionisation enthalpy
(iii) $L{i}^{+}\left(g\right)⟶L{i}^{+}\left(aq\right)△H_n$ = Enthalpy of hydration
For alkali metals, enthalpies of sublimation are almost same, $I{E}_1$ value of Li is endohermic and highest and hydration is exothermic and maximum for $L{i}^{+}$ Endothermic and highest and hydration is exotherrnic and maximum for $L{i}^{+}$ The highly exothermic step (iii) for smallest $L{i}^{+}$ makes it strongest reducing agent.