Q21.125 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Comparing the standard electrode potentials ${\text{(E}}^{\text{o}}\text{)}$ of the Group $\text{1A(1)}$  metals $\text{Li, Na, and K}$with the negative of their first ionization energies reveals a discrepancy:

Ionization process reversed: ${\text{M}}^{\text{ + }}{\text{(g) + e}}^{\text{ - }}\rightleftharpoons \text{M(g) ( - IE)}$

Electrode reaction: ${\text{M}}^{\text{ + }}{\text{(aq) + e}}^{\text{ - }}\rightleftharpoons {\text{M(s) (E}}^{\text{o}}\text{)}$

Note that the electrode potentials do not decrease smoothly down the group, as the ionization energies do. You might expect that if it is more difficult to remove an electron from an atom to form a gaseous ion (larger $\text{IE}$), then it would be less difficult to add an electron to an aqueous ion to form an atom (smaller ${\text{E}}^{\text{o}}$ ), yet  ${\text{Li}}^{\text{ + }}\text{(aq)}$is more difficult to reduce than ${\text{Na}}^{\text{ + }}\text{(aq)}$. Applying Hess’s law, use an approach similar to that for a Born-Haber cycle to break down the process occurring at the electrode into three steps and label the energy involved in each step. How can you account for the discrepancy?

Step-by-Step Solution

Answer

The three steps occurring at electrode is –

${\text{(1) M}}_{\text{(aq)}}^{\text{ + }}\to {\text{M}}_{\text{(g)}}^{\text{ + }}$

${\text{(2) M}}_{\text{(g)}}^{\text{ + }}{\text{ + e}}^{\text{ - }}\to {\text{M}}_{\text{(g)}}$

${\text{(3) M}}_{\text{(g)}}^{\text{ + }}\to {\text{M}}_{\text{(s)}}$

$\text{Li}$ has a higher potential than $\text{Na and K}$ because of higher hydration energy in changing phases from aqueous to gas.

Step 1: Concept Introduction

The Born Haber cycle is a series of enthalpy changes in a process that results in the synthesis of a solid crystalline ionic compound from elemental atoms in their standard state, with the net enthalpy of formation being zero. The standard electrode potential is a measurement of the equilibrium potential. The potential of the electrode is the difference in potential between the electrode and the electrolyte. The electrode potential is known as the standard electrode potential when the concentrations of all the species involved in a semi-cell are equal.

Step 2: The Reaction Breakdown

Given the initial equation, ${\text{M}}^{\text{ + }}{\text{(aq) + e}}^{\text{ - }}\rightleftharpoons \text{M(s)}$ equations can be –

$$\begin{gathered} {\text{(1) M}}_{\text{(aq)}}^{\text{ + }}\to {\text{M}}_{\text{(g)}}^{\text{ + }} \\ {\text{(2) M}}_{\text{(g)}}^{\text{ + }}{\text{ + e}}^{\text{ - }}\to {\text{M}}_{\text{(g)}} \\ {\text{(3) M}}_{\text{(g)}}^{\text{ + }}\to {\text{M}}_{\text{(s)}} \end{gathered}$$

In equation 1 , the energy involved is $△{\text{H}}_{\text{hydration}}$ converted from aqueous to gas phases. In equation 2, lionization Energy is involved. Lastly, the last equation involves atomization energy, converting from a gas to solid phase.

The trend for the cell potential does not follow the trend for ionization energy because there can be major difference in the hydration energy of $\text{Li, Na, K}$. $\text{Li}$is smaller and holds water molecules more, therefore giving it a higher cell potential.

 

Therefore, lithium has higher potential than Sodium and Potassium.