inorganic chemistry - Why is lithium the most reducing alkali metal, and not caesium?

Caesium has a larger size, and the effective nuclear charge that the valence electron experiences will be far less compared to that of lithium's, right? But lithium is still considered the strongest reducing agent among all the alkali metals, and this is evidenced by its large and negative reduction potential. Why is this so?

Answer

The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated.

$$\begin{array}{cc} \hline \ce{M} & E^\circ(\ce{M+}/\ce{M}) \\ \hline \ce{Li} & -3.045 \\ \ce{Na} & -2.714 \\ \ce{K} & -2.925 \\ \ce{Rb} & -2.925 \\ \ce{Cs} & -2.923 \\ \hline \end{array}$$ ^{Source: Chemistry of the Elements 2nd ed., Greenwood & Earnshaw, p 75}

However, a full description of the middle three metals is beyond the scope of this question. I just thought it was worth pointing out that the trend is not really straightforward.

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The $\ce{M+}/\ce{M}$ standard reduction potential is related to $\Delta_\mathrm{r}G^\circ$ for the reaction

$$\ce{M(s) -> M+(aq) + e-}$$

by the equation

$$E^\circ = \frac{\Delta_\mathrm{r}G^\circ + K}{F}$$

where $K$ is the absolute standard Gibbs free energy for the reaction

$$\ce{H+ + e- -> 1/2 H2}$$

and is a constant (which means we do not need to care about its actual value). Assuming that $\Delta_\mathrm{r} S^\circ$ is approximately independent of the identity of the metal $\ce{M}$, then the variations in $\Delta_\mathrm{r}H^\circ$ will determine the variations in $\Delta_\mathrm{r}G^\circ$ and hence $E^\circ$. We can construct an energy cycle to assess how $\Delta_\mathrm{r}H^\circ$ will vary with the identity of $\ce{M}$. The standard state symbol will be dropped from now on.

$$\require{AMScd} \begin{CD} \ce{M (s)} @>{\large \Delta_\mathrm{r}H}>> \ce{M+(aq) + e-} \\ @V{\large\Delta_\mathrm{atom}H(\ce{M})}VV @AA{\large\Delta_\mathrm{hyd}H(\ce{M+})}A \\ \ce{M (g)} @>>{\large IE_1(\ce{M})}> \ce{M+ (g) + e-} \end{CD}$$

We can see, as described in Prajjawal's answer, that there are three factors that contribute to $\Delta_\mathrm{r}H$:

$$\Delta_\mathrm{r}H = \Delta_\mathrm{atom}H + IE_1 + \Delta_\mathrm{hyd}H$$

(the atomisation enthalpy being the same as the sublimation enthalpy). You are right in saying that there is a decrease in $IE_1$ going from $\ce{Li}$ to $\ce{Cs}$. If taken alone, this would mean that $E(\ce{M+}/\ce{M})$ would decrease going from $\ce{Li}$ to $\ce{Cs}$, which would mean that $\ce{Cs}$ is a better reducing agent than $\ce{Li}$.

However, looking at the very first table, this is clearly not true. So, some numbers will be needed. All values are in $\mathrm{kJ~mol^{-1}}$.

$$\begin{array}{ccccc} \hline \ce{M} & \Delta_\mathrm{atom}H & IE_1 & \Delta_\mathrm{hyd}H & \text{Sum} \\ \hline \ce{Li} & 161 & 520 & \mathbf{-520} & 161 \\ \ce{Cs} & 79 & 376 & \mathbf{-264} & 211 \\ \hline \end{array}$$ ^{Source: Inorganic Chemistry 6th ed., Shriver et al., p 160}

This is, in fact, an extremely crude analysis. However, it hopefully does serve to show in a more quantitative way why $E(\ce{Cs+}/\ce{Cs}) > E(\ce{Li+}/\ce{Li})$: it's because of the extremely exothermic hydration enthalpy of the small $\ce{Li+}$ ion.

Just as a comparison, the ionic radii of $\ce{Li+}$ and $\ce{Cs+}$ ($\mathrm{CN} = 6$) are $76$ and $167~\mathrm{pm}$ respectively (Greenwood & Earnshaw, p 75).