Question

The reducing power of metal depends on various factors. Suggest the factors which make Li, the strongest reducing agent in aqueous solution.
                 (A) Sublimation enthalpy
                 (B) Ionisation enthalpy
                 (C) Hydration enthalpy
                 (D) Electron-gain enthalpy

Answer 1

Hint: Reducing power is associated with the ability of a metal to lose electrons itself and oxidise other atoms. While undergoing a reduction reaction the metal converts from ${M_{(s)}}$ state to ${M^ + }_{(aq)}$ state which includes it going from solid to gaseous state, then losing an electron and then going into aqueous state.

Complete step by step solution:
-High reducing power means more negative value of ${E^ \circ }$ (reduction potential). All metals are good reducing agents and so have highly negative reduction potentials. But among all metals Li (Lithium) is the strongest reducing agent.
-The reduction potential of any metal depends on various factors like:

     -Sublimation enthalpy: Energy required in converting one mole of solid substance into gaseous state.

    -Ionisation enthalpy: Energy required in removing an electron from the outermost shell of an isolated gaseous atom.

     -Hydration enthalpy: energy released when new bonds are made between the ions and water molecules.

-${E^ \circ }$ (reduction potential) of Li is the most negative = ( -3.04 V ). So, Li atom itself gets oxidised and reduces other atoms.
-When lithium acts as a reducing agent in aqueous medium, its conversion is as follows:
                                                    $L{i_{(s)}} \to L{i^ + }_{(aq)}$

There are 3 steps involved in this. They are:
1)Sublimation process to convert solid Li to gaseous Li:
                                       $L{i_{(s)}}\xrightarrow{{subli.}}L{i_{(g)}}$
           $\Delta {H_s}$= Enthalpy of sublimation
      -Sublimation energy involved in this process is the same for all alkali metals. So, it does not make Li as the strongest reducing agent in aqueous medium.

2)Ionisation of Li atom and the energy involved is ionisation enthalpy:
                                                    $L{i_{(g)}} \to L{i^ + }_{(g)}$
           $I{E_1}$ = First ionisation enthalpy (+ve)
      -For Li atom this is an endothermic process. It is endothermic because of the small size of the Li atom and the high value of ${Z_{eff}}$ (charge). So, this also does not contribute in making Li the strongest reducing agent in aqueous medium.

3)Hydration of Li ion: $L{i_{(g)}} \to L{i^ + }_{(aq)}$
        $\Delta {H_n}$ = Enthalpy of hydration (-ve)
       - For the Li atom this is a highly exothermic process. The amount of energy released during this process is so high that it compensates the ionisation enthalpy (energy required during the ionisation process). We can also say that the hydration enthalpy of Li is very high (highly exothermic). So, due to this highest hydration enthalpy of Li atom it has become the strongest reducing agent in aqueous medium.
-Hence, the high hydration enthalpy of Li atom makes it the strongest reducing agent in aqueous medium.

So, the correct option is: (C) Hydration enthalpy

Note: Metals are the best reducing agents because they have the ability to lose electrons easily and form ions. More negative is the reducing potential; more is its reducing power.