Chapter 14, Problem 14.97QP

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is ${\text{Na}}^{+}+e^{-}\stackrel{}{\to }\text{Na}$. We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at 892°C:

$\text{Na(}g\text{)}+\text{KCl(}l\text{)}\stackrel{}{\rightleftharpoons }\text{NaCl(}l\text{)}+\text{K(}g\text{)}$

In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are 892°C and 770°C, respectively.)