# 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?

• I think it has to do with solvation. The more solvated the resulting ions, the more favourable electron loss is. Even though its ionization energy is higher, lithium creates a very small cation which is strongly solvated in water, which drives the oxidation of metallic lithium thermodynamically towards the products. – Nicolau Saker Neto Apr 12 '15 at 16:01

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).

• I did not understand the formulas you provided. I hope to be able one day. My Russian textbook says: "lithium ions are much smaller than Na and K ions. That's why they will have stronger electrical fields around them. Hence, more active hydration. This eases the transition of electrons from Li into the solution and explains Li's position in the electrochemical series". – CowperKettle May 5 '16 at 8:24
• That's pretty much the same as what I wrote, except it's more descriptive and less quantitative. OP's argument is that it is difficult to remove the electron off lithium, therefore lithium should be hardest to oxidise. However, in reality, lithium is coerced to give up its electron by the surrounding water molecules, which are very happy to form ion-dipole interactions with the small ion $\ce{Li+}$. – orthocresol Dec 8 '16 at 9:58

To decide which is the best reducing agent we only not consider that who has less ionisation energy yet it follows 3 steps:

1. Metal (solid) to Metal (gaseous state) sublimation energy
2. Metal from gaseous state to M+ ionisation energy
3. 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.

Well there might be more reasons than these two:

1. Lithium has a higher reduction potential.
2. If you also look at the electronegativities of just Lithium and Cesium then you would notice that the shielding effect is more prevalent in Cesium, thereby reducing the electronegativity and affecting the reduction potential. So Lithium however, just compared to Cesium, has a higher electronegativity.

I think these are the two main reasons, please correct me If I am wrong.

• (-1) Saying that "lithium is a stronger reducing agent because it has a higher reduction potential" is just like saying "fluorine is reactive because it reacts with a lot of species readily". It's a tautology. Secondly, the greater shielding effect in caesium works to make caesium a stronger reducing agent because it reduces the first ionisation energy. It's exactly the same thing that OP states in the question. – orthocresol Jan 9 '16 at 7:35