Chemistry - Why is lithium the most reducing alkali metal, and not caesium?
Solution 1:
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.
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).
Solution 2:
To decide which is the best reducing agent we only not consider that who has less ionisation energy yet it follows 3 steps:
- Metal (solid) to Metal (gaseous state) sublimation energy
- Metal from gaseous state to M+ ionisation energy
- M+ to M+ (aqueous state) hydration energy
Lithium having more charge density has more sublimation energy and ionisation energy than caesium but hydration energy is released in such a big amount that it compensates the S.E and I.E. and caesium's hydration is less than lithium. That's why lithium is good reducing agent.