Analysing the oxidation enthalpies for $\ce{Li}$, $\ce{Na}$ and $\ce{K}$, we have their respective values:
Sodium has the largest positive oxidation enthalpy, which means that its oxidation will be less likely to happen according to its Gibbs free energy. But, why does this happens to sodium?
The general trend in the reactivity of the group 1 elements is that it increases down the group. This can be attributed to the decrease in 1st ionisation energy as the atomic radius increases.
Reducing power can be measured by the standard electrode potential E0. This refers to the half - cell:
$$M^+(aq)+e⇌M(s)$$
If you look at the values for group 1 we see a different pattern:
Electrode Potential (V) Li. -3.02 Na. -2.71 K. -2.92 Rb. -2.99 Cs. -3.02
Lithium has an anomalously large and negative electrode potential. After Li the values become more large and negative, indicating, as you rightly say, that sodium has the lowest reducing power.
The reason for the unusual value for lithium is that E0 values are measured in solution whereas ionisation energies are measured in the gaseous phase.
To turn a metal atom in the solid state into an aqueous ion we can think of a 3 step process:
The metal is atomized. $\ce{Li(s)→Li(g)}$
The metal is ionised. $\ce{Li(g)→Li+(g)+e}$
The ion is hydrated. $\ce{Li+(g) +(aq)→Li+(aq)}$
Steps 1 and 2 are both endothermic, i.e. they require energy. Step 3 is exothermic as it is a bond forming process. It is referred to as the enthalpy of hydration.
The small size of the lithium ion gives it a large charge density and so water molecules are strongly attracted giving a large, negative enthalpy of hydration.
This is the main factor responsible for lithium's large, negative electrode potential.
In aqueous conditions it is as reducing as cesium though much less reactive in anhydrous conditions.
This anomaly accounts for sodium having the lowest reducing power, but only in aqueous conditions.
