Q.Explain the following:
Standard reduction potential of alkali metals have high negative values.

The reactivity and reducing strength of alkali-metal species can be read off their standard electrode (reduction) potentials, $E^{\ominus}$: the more negative the $E^{\ominus}$ of a couple $M^{n+}/M$, the more strongly the reduced form $M$ acts as a reducing agent, because it releases electrons more readily. Given $E^{\ominus}(Cl_2/Cl^-)=+1.36V$, $E^{\ominus}(I_2/I^-)=+0.53V$, $E^{\ominus}(Ag^+/Ag)=+0.79V$, $E^{\ominus}(Na^+/Na)=-2.71V$ and $E^{\ominus}(Li^+/Li)=-3.04V$, reversing each to the oxidation direction gives the reducing power of $I^-$, $Ag$, $Cl^-$, $Na$ and $Li$ as roughly $-0.53V$, $-0.79V$, $-1.36V$, $+2.71V$ and $+3.04V$ respectively, so reducing strength increases in the order $I^- < Ag < Cl^- < Na < Li$. Down Group 1, atomic size grows and ionisation enthalpy falls faster than hydration enthalpy declines, so the net ease of losing an electron in solution rises from Li to Cs; hence sodium, with a higher ionisation enthalpy than potassium, is less reactive than potassium. The same electrochemical theme explains why the aqueous ionic mobility of the alkali-metal ions rises as $Li^+ < Na^+ < K^+ < Rb^+ < Cs^+$ — the smallest bare ion, $Li^+$, binds the thickest shell of water molecules and so migrates most slowly despite being the smallest ion; why lithium alone forms a nitride directly with $N_2$ (its high lattice/enthalpy terms favour nitride formation, unlike the heavier, less strongly bonding alkali metals); and why $E^{\ominus}$ for $M^{2+}(aq)+2e^- \rightarrow M(s)$ is almost the same (about $-2.8$ to $-2.9V$) for $Ca$, $Sr$ and $Ba$ — the sublimation, ionisation and hydration enthalpy terms that make up this potential change down the group in a way that largely cancels out, so the three metals are almost equally strong reducing agents.