Comparing the standard electrode potentials \(\left(E^{\circ}\right)\) of the Group \(1 \mathrm{~A}(1)\) metals \(\mathrm{Li}, \mathrm{Na},\) and \(\mathrm{K}\) with the negative of their first ionization energies reveals a discrepancy: Ionization process reversed: \(\mathrm{M}^{+}(g)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(g)\) Electrode reaction: $$\mathrm{M}^{+}(a q)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(s)$$ $$\begin{array}{lcc}\text { Metal } & -\text { IE (kJ/mol) } & E^{\circ} \text { (V) } \\\\\hline \text { Li } & -520 & -3.05 \\\\\text { Na } & -496 & -2.71 \\\ \text { K } & -419 & -2.93\end{array}$$ Note that the electrode potentials do not decrease smoothly down the group, whereas 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 IE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smaller \(E^{\circ}\) ), yet \(\mathrm{Li}^{+}(a q)\) is more difficult to reduce than \(\mathrm{Na}^{+}(a q) .\) Applying Hess's law, use an approach similar to 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

01

Identify the problem

There is a discrepancy between the ionization energies and the electrode potentials of Group 1A metals. We need to use Hess's law to break down the electrode process into steps and analyze the energy involved in each step to understand this discrepancy.

02

Write down the ionization process

For a metal M transitioning from a gaseous ion to a gas: \text{\text{Ionization process reversed:}}\text{\(\text{M}^{+}(g) + \text{e}^{-} \rightleftharpoons \text{M}(g)\)}

03

Write down the electrode reaction

\text{This is the electrode reaction given:} \text{\(\text{M}^{+}(aq) + \text{e}^{-} \rightleftharpoons \text{M}(s)\)}

04

Break down the electrode process

We can decompose the electrode reaction into three steps using an approach similar to a Born-Haber cycle:1. \text{Sublimation of the metal:} \text{\(\text{M}(s) \rightarrow \text{M}(g)\) \text{with sublimation energy} \text{\(\text{ΔH}_{sub}\)}}2. \text{Ionization energy:} \text{\(\text{M}(g) \rightarrow \text{M}^{+}(g) + \text{e}^{-}\) \text{with ionization energy} \text{\(\text{IE}\)}}3. \text{Hydration of the ion:} \text{\(\text{M}^{+}(g) + \text{e}^{-} \rightarrow \text{M}^{+}(aq)\) \text{with hydration energy} \text{\(\text{ΔH}_{hyd}\)}}

05

Label the energies involved

The total energy change involved in the electrode process can be expressed as a combination of these steps:\text{Electrode potential \(\text{E}^{\text{°}}\) = - ΔΗ}_{sub} - IE + ΔH_{hyd}}.

06

Adjust the perspective on the discrepancy

While ionization energy decreases down the group, hydration energy (for ions in aqueous solution) typically becomes less negative (less exothermic) down the group. Since this deconvolution shows that specific terms can vary, it explains why the electrode potentials do not decrease smoothly like the ionization energies.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Group 1A metals

Group 1A metals, also known as alkali metals, include lithium (Li), sodium (Na), and potassium (K). These elements are characterized by having a single electron in their outermost shell, which makes them highly reactive. They tend to lose this electron easily to form positive ions. The ionization energy of these metals decreases as you move down the group. This means it requires less energy to remove the outermost electron in potassium compared to sodium, and even less for lithium.

Hess's law

Hess's law states that the total enthalpy change of a reaction is the same regardless of the path taken. This is highly useful when dealing with complex reactions, as it allows us to break down a reaction into simpler steps with known energy changes. By summing the enthalpy changes of these steps, we can determine the overall energy change of the reaction. In our case, we use Hess's law to decompose the electrode reaction into three steps: sublimation, ionization, and hydration. This helps us understand the overall energy involved in the formation of a metal from its ion.

Born-Haber cycle

The Born-Haber cycle is a thermodynamic cycle used to analyze the steps involved in the formation of ionic compounds. It includes multiple steps like sublimation, ionization, and hydration energy. In our context, it helps to break down the electrode process involving alkali metals into individual steps:

• Sublimation: Conversion of the metal from solid to gas.
• Ionization: Loss of an electron from the gaseous metal atom to form a positive ion.
• Hydration: Interaction of the gaseous ion with water molecules.

This cycle helps us understand the discrepancy between the ionization energy and electrode potential by separating the contributing factors.

Hydration energy

Hydration energy is the energy change that occurs when a gaseous ion dissolves in water to form an aqueous ion. It's usually exothermic, meaning energy is released in the process. The magnitude of hydration energy depends on the charge and size of the ion. Smaller ions with higher charge densities generally have larger (more negative) hydration energies. In Group 1A metals, hydration energy decreases as you go down the group, meaning lithium ions release more energy upon hydration compared to sodium and potassium ions. This plays a role in our electrode potential analysis.

Ionization energy

Ionization energy is the amount of energy needed to remove an electron from a gaseous atom to form a positive ion. In Group 1A elements, ionization energies decrease as you move down the group because the outermost electron is further from the nucleus and is more easily removed. This explains why lithium has a higher ionization energy than sodium and potassium. In our analysis, while we might expect a straightforward relationship between ionization energies and electrode potentials, other factors like hydration energy complicate the picture, leading to the observed discrepancies.