Let's analyze the data step by step.

We are given the following standard electrode reactions and their potentials:

For aluminum (Al):

$$\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}, \quad E^\circ = -1.66 \text{ V}$$

$$\mathrm{Al}^{+} + e^- \to \mathrm{Al}, \quad E^\circ = +0.55 \text{ V}$$

For thallium (Tl):

$$\mathrm{Tl}^{3+} + 3e^- \to \mathrm{Tl}, \quad E^\circ = +1.26 \text{ V}$$

$$\mathrm{Tl}^{+} + e^- \to \mathrm{Tl}, \quad E^\circ = -0.34 \text{ V}$$

A key idea in interpreting these potentials is that a more positive reduction potential means the ion is a stronger oxidizing agent and is more eager to gain electrons (i.e., it is less “stable” in its oxidized form). In contrast, a more negative reduction potential indicates that the species is less likely to gain electrons and is more stable in that oxidation state.

Let’s apply this reasoning:

• For aluminum:

The reaction $$\mathrm{Al}^{3+} \to \mathrm{Al}$$ has a very negative potential (–1.66 V), meaning that once aluminum is in the +3 state, it does not easily pick up electrons to revert to the metal. This makes $$\mathrm{Al}^{3+}$$ a very stable oxidation state.

The reaction involving $$\mathrm{Al}^{+}$$ has a positive potential (+0.55 V), indicating that $$\mathrm{Al}^{+}$$ is readily reduced to aluminum, and thus is less stable in the +1 state.

• For thallium:

The reaction $$\mathrm{Tl}^{3+} \to \mathrm{Tl}$$ has a highly positive potential (+1.26 V), so $$\mathrm{Tl}^{3+}$$ is not very stable; it tends strongly to gain electrons.

In contrast, the reaction involving $$\mathrm{Tl}^{+}$$ has a negative potential (–0.34 V), which indicates that $$\mathrm{Tl}^{+}$$ does not easily accept an electron, making it a relatively stable oxidation state.

Now, let’s look at the options:

Option A: “$$\mathrm{Tl}^{3+}$$ is more stable than $$\mathrm{Al}^{3+}$$.”

• Because $$\mathrm{Tl}^{3+}$$ (with +1.26 V) is eager to gain electrons (a strong oxidizing agent), it is less stable compared to $$\mathrm{Al}^{3+}$$ (with –1.66 V). Thus, Option A is incorrect.

Option B and Option C: “$$\mathrm{Al}^{+}$$ is more stable than $$\mathrm{Al}^{3+}$$.”

• Our data indicates that $$\mathrm{Al}^{+}$$ (with +0.55 V) is less stable (more readily reduced) than $$\mathrm{Al}^{3+}$$ (with –1.66 V). So, these statements are not correct.

Option D: “$$\mathrm{Tl}^{+}$$ is more stable than $$\mathrm{Al}^{+}$$.”

• Comparing the two, the reduction reaction for $$\mathrm{Tl}^{+}$$ has $$E^\circ = -0.34\text{ V}$$ while that for $$\mathrm{Al}^{+}$$ has $$E^\circ = +0.55\text{ V}.$$ The more negative potential for $$\mathrm{Tl}^{+}$$ shows it is less likely to be reduced to elemental Tl, meaning $$\mathrm{Tl}^{+}$$ is more stable than $$\mathrm{Al}^{+}$$.

Thus, Option D is the correct statement.

In summary, the correct answer is:

Option D: $$\mathrm{Tl}^{+}$$ is more stable than $$\mathrm{Al}^{+}$$.

Electrochemistry

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