Question

Which one is the best reducing agent among alkali metal in aqueous solution ?
(A) $Li$
(B) $Be$
(C) B
(D) None of these

Answer 1

Alkali metals are the elements of group one. It includes lithium $\left( {Li} \right)$, Sodium $\left( {Na} \right)$, Potassium $\left( K \right)$, rubidium $\left( {Rb} \right)$, Cesium $\left( {Cs} \right)$, and Francium $\left( {Fr} \right)$. Alkali metals are good reducing agents as they have low value of ionization energy that decrease as we move down the group.

Complete step by step answer:
Reducing agent :- Element which loses their electrons to an electron recipient in a redox reaction are known as reducing agent
Alkali metals are good reducing agents as they readily lose their electrons due to low ionization enthalpy.
Among all the alkali metals, Lithium $\left( {Li} \right)$is the strongest reducing agent in aqueous solution.
There are two cases, one is a free gaseous state and other one is aqueous state.
I . In free gaseous state :-
In alkali metal, ionization enthalpy decreases down the group from $Li\;\operatorname{to} \;Cs$. Thus, lithium has the least reduction in nature in a free gaseous state.
II. aqueous state :-
In aqueous state, Oxidation potential $\left( {{E^0}} \right)$is the measure of reducing property of an element. Among all the alkali metals, $Li$ have the highest oxidation potential therefore, it has the highest reducing property.
Moreover, due to its small size $Li$ has very high hydration enthalpy. So, $Li$ has a high tendency to lose electrons in solution. Hence, $Li$ is the strongest reducing agent.

Hence, the correct option is (A) .

Note:
Ionization energy is the energy required to remove the loosely bound electron from an isolated neutral gaseous atom. But oxidation potential is taken into concentration when metal is present in the solution.