Ncert Exercises & Solutions

The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.

1. Sublimation enthalpy

2. Ionisation enthalpy

3. Hydration enthalpy

4. Electron-gain enthalpy

Hint: ${Li}^{+}$ ion is the most stable in aqueous medium

Standard reduction potential

(E_{RP}^{o})

is a measure of tendency of an element to lose electron in aqueous solution. Higher the negative

E_{RP}^{o}

greater is the ability to lose electrons.

E_{RP}^{o}

depends on

(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

E_{RP}^{o}

value of Li is least (-3.04V) among all alkali metals.

The formation of Li⁺(aq) from Li involves following steps are as follows:

(i)

Li (s) \overset{Sublimation}{\to} Li (g) ∆ H_{s} = Enthalpy of sublimation

(ii)

Li (g) \to {Li}^{+} (s) {IE}_{1} = ionisation enthalpy

(iii)

{Li}^{+} (g) \to {Li}^{+} (aq) ∆ H_{n} = Enthalpy of hydration

For alkali metals, enthalpies of sublimation are almost same. IE₁ value of Li is endothermic and highest and hydration is exothermic and maximum for Li⁺.

The highly exothermic step (iii) for smallest Li⁺ makes it strongest reducing agent.

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