Question

Elemental Li and $\mathbf{Na}$ are prepared by electrolysis of a molten salt, whereas$\mathbf{\text{K,Rb}}$ , and$\mathbf{\text{Cs}}$ are prepared by chemical reduction.

(a) In general terms, explain why the alkali metals cannot be prepared by electrolysis of their aqueous salt solutions.

(b) Use ionization energies (see the Family Portraits, pp. 572 and 576 ) to explain why Calcium should not be able to isolate$\mathbf{\text{Rb}}$ from molten$\mathbf{\text{RbX(X=}}$ halide).

(c) Use physical properties to explain why Calcium is used to isolate Rb from molten RbX.

(d) Can Ca be used to isolate Cs from molten CsX? Explain.

Short answer

1. In all of the cases, it is energetically more favorable to reduce ${\text{H}}_{\text{2}}\text{O}$ to ${\text{H}}_{\text{2}}$ than to reduce ${\text{M}}^{\text{+}}$to ${\text{M}}_{\text{.}}$where M is the alkali metal.
2. The energy required to remove an electron is the ionization energy. It requires more energy to ionize Calcium's electrons, so it seems unlikely that Calcium could reduce ${\text{Rb}}^{\text{+}}$ . Based on values of IE and a positive change in enthalpy for the forward reaction, it seems more reasonable that ${\text{Rb}}^{\text{0}}$ would reduce ${\text{Ca}}^{\text{2+}}$.
3. Calcium remains in the molten phase and remains separated from gaseous rubidium.
4. Calcium can be used to separate gaseous Caesium from molten CsX.

Step-by-step solution

Le Chatlier's principle

If a system in equilibrium is disturbed by changes in concentration, temperature, volume, or pressure, Le Chatelier's principle states that it will reset to counteract the effect of the disturbance.

Subpart (a)

Aqueous salt solutions are ions and water mixtures. When two half-reactions are possible at the same electrode, the one with the higher electrode potential wins. The two half-reactions in this case are:

${\text{M}}^{\text{+}}{\text{+e}}^{\text{-}}\to {\text{M}}^{\text{°}}{\text{E}}_{\text{red}}^{\text{o}}\text{=-3.05V,-2.93,and-2.71V}$

For ${\text{Li}}^{\text{+}}{\text{,K}}^{\text{+}}$ , and ${\text{Na}}^{\text{+}}$, respectively.

In all of these cases, it is energetically more favorable to reduce ${\text{H}}_{\text{2}}\text{O}$ to ${\text{H}}_{\text{2}}$ than to reduce ${\text{M}}^{\text{+}}$ to ${\text{M}}^{}$.

Subpart (b) 

The element Ca chemically reduce $\text{RbX}$, i.e., convert ${\text{Rb}}^{\text{+}}$ to ${\text{Rb}}^{\text{0}}$. In order for this to occur, Calcium loses electrons $\left({\text{Ca}}^{\text{0}}\to {\text{Ca}}^{\text{2+}}{\text{+2e}}^{\text{-}}\right)$ and each ${\text{Rb}}^{\text{+}}$ gains an electron$\left({\text{2Rbb}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{2Rb}}^{\text{0}}\right)$ . The reaction is written as follows:$\text{2RbX+Ca}\to {\text{CaX}}_{\text{2}}\text{+2Rb}$

Where

$\begin{array}{c}{\text{ΔH=IE}}_{\text{2}}{\text{(Ca)-2IE}}_{\text{l}}\text{(Rb)}\\ \text{=590+1145-2(403)}\\ \text{=+929kJ/mol}\end{array}$

Recall that Calcium acts as a reducing agent for the ${\text{Rb}}^{\text{+}}$ ion because it oxidizes. The energy required to remove an electron is the ionization energy. It requires more energy to ionize Calcium's electrons, so it seems unlikely that Calcium could reduce ${\text{Rb}}^{\text{+}}$ . Based on values of IE and a positive change in enthalpy for the forward reaction, it seems more reasonable that ${\text{Rb}}^{\text{0}}$ would reduce ${\text{Ca}}^{\text{2+}}$.

Subpart (c)

If the reaction is carried out at a temperature greater than ${\text{688}}^{\text{°}}\text{C}$ (the boiling point of rubidium), the product mixture will contain gaseous rubidium. This can be removed from the reaction vessel, causing a shift in equilibrium to form more rubidium products. If the reaction is carried out between data-custom-editor="chemistry" ${\text{688}}^{\text{°}}\text{C}$ and data-custom-editor="chemistry" ${\text{1484}}^{\text{°}}\text{C}$ (bp for Calcium), then Calcium remains in the molten phase and is separated from gaseous rubidium.

Subpart (d)

The reaction of Calcium with molten CsX is written as follows:

$\begin{array}{c}{\text{ΔH=IE}}_{\text{2}}\text{(Ca)-2}\times {\text{IE}}_{\text{1}}\text{(Cs)}\\ \text{=590+1145-2(376)}\\ \text{=+983kJ/mol}\end{array}$

This reaction is more unfavorable than for rubidium, but Caesium has a lower boiling point of ${\text{671}}^{\text{°}}\text{C}$ . If the reaction is carried out between ${\text{671}}^{\text{°}}\text{C}$ and ${\text{1484}}^{\text{°}}\text{C}$, then Calcium can be used to separate gaseous Caesium from molten CsX.