Strong reducing power of lithium in aqueous solution can be understood in terms of electrode potential. Electrode potential is a measure of the tendency of an element to lose electrons in the aqueous solution. It mainly depends upon the following three factors i.e.,

(i) $\mathrm{Li}(\mathrm{s}) \xrightarrow[\text { Enthalpy }]{\text { Sublimation }} \mathrm{Li}(g)$

(ii) $\mathrm{Li}(g) \xrightarrow[\text { Enthalpy }]{\text { Ionisation }} \mathrm{Li}^{+}(g)+\mathrm{e}^{-}$

(iii) $\mathrm{Li}^{+}(g)+a q \longrightarrow \mathrm{Li}^{+}(a q)+$ enthalpy of hydration

With the small size of its ion, lithium has the highest hydration enthalpy. However, ionisation enthalpy of Li is highest among alkali metals but hydration enthalpy predominates over ionisation enthalpy.

Therefore, lithium is the strongest reducing agent in aqueous solution mainly because of its high enthalpy of hydration.

The reactivity of alkali metals towards oxygen increases on moving down the group with the increase in atomic size. Thus, Li forms only lithium oxide $\left(\mathrm{Li}_2 \mathrm{O}\right)$, sodium forms mainly sodium peroxide $\mathrm{Na}_2 \mathrm{O}_2$ along with a small amount of sodium oxide while potassium forms only potassium superoxide $\left(\mathrm{KO}_2\right)$.

$4 \mathrm{Li}+\mathrm{O}_2 \xrightarrow{\Delta} 2 \mathrm{Li}_2 \mathrm{O}$

$6 \mathrm{Na}+2 \mathrm{O}_2 \xrightarrow{\Delta} \underset{\substack{\text { Sodium peroxide } \\ \text { (major) }}}{\mathrm{Na}_2 \mathrm{O}_2}+\underset{\substack{\text { Monooxide } \\ \text { (minor) }}}{2 \mathrm{Na}_2 \mathrm{O}}$

$\mathrm{K}+\mathrm{O}_2 \xrightarrow{\Delta} \underset{\begin{array}{c}\text { Potassium } \\ \text { super oxides }\end{array}}{\mathrm{KO}_2}+\underset{\text { Peroxide }}{\mathrm{K}_2 \mathrm{O}_2}+\underset{\text { Monoxide }}{\mathrm{K}_2(\mathrm{O})}$

The superoxide, $\mathrm{O}_2^{-}$ion is stable only in presence of large cations such as $\mathrm{K}, \mathrm{Rb}$ etc.

$\mathrm{O}_2{ }^{2-}$ represents a peroxide ion

$\mathrm{O}_2^-$ represents a superoxide ion

(i) Peroxide ion react with water to form $\mathrm{H}_2 \mathrm{O}_2$

$$\mathrm{O_2^{2 - } + 2{H_2}O\buildrel {} \over \longrightarrow 2O{H^ - } + \mathop {{H_2}{O_2}}\limits_{Hydrogen\,peroxide}}$$

(ii) Superoxide ion react with water to form $\mathrm{H}_2 \mathrm{O}_2$ and $\mathrm{O}_2$

$2 \mathrm{O}_2^{-}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{OH}^{-}+\underset{\substack{\text { Hydrogen } \\ \text { peroxide }}}{\mathrm{H}_2 \mathrm{O}_2}+\mathrm{O}_2$