Consider the following standard electrode potentials ($$E°$$ in volts) in aqueous solution:

Element	$$M^{3+}/M$$	$$M^+/M$$
Al	$$-1.66$$	$$+0.55$$
Tl	$$+1.26$$	$$-0.34$$

Based on these data, which of the following statements is correct?

First, we recall the definition of the standard reduction potential. For a general half-reaction of the form $$\text{Ox}^{n+}+ne^- \rightarrow \text{Red},$$ the symbol $$E^\circ$$ tells us how readily the oxidised species $$\text{Ox}^{n+}$$ gains the electrons.

More positive $$E^\circ$$  ⇒  reduction is easy, so the oxidised species is less stable and prefers to be reduced.
More negative $$E^\circ$$  ⇒  reduction is difficult, so the oxidised species is more stable and prefers to remain as it is.

Now we write the relevant half-reactions given in the table.

For aluminium:

$$Al^{3+}+3e^- \rightarrow Al,\qquad E^\circ=-1.66\ \text{V}$$ $$Al^{+}+e^- \rightarrow Al,\qquad E^\circ=+0.55\ \text{V}$$

For thallium:

$$Tl^{3+}+3e^- \rightarrow Tl,\qquad E^\circ=+1.26\ \text{V}$$ $$Tl^{+}+e^- \rightarrow Tl,\qquad E^\circ=-0.34\ \text{V}$$

Let us discuss the stability of each oxidation state.

For aluminium, the potential $$-1.66\text{ V}$$ is quite negative, so the reduction of $$Al^{3+}$$ is hard; therefore $$Al^{3+}$$ is comparatively stable. The potential $$+0.55\text{ V}$$ for $$Al^{+}$$ is positive, so $$Al^{+}$$ is easily reduced to metallic aluminium and is consequently less stable than $$Al^{3+}$$.

For thallium, the potential $$+1.26\text{ V}$$ for $$Tl^{3+}$$ is strongly positive, so $$Tl^{3+}$$ is easily reduced and is not very stable. The potential $$-0.34\text{ V}$$ for $$Tl^{+}$$ is negative, so the reduction of $$Tl^{+}$$ is relatively difficult and $$Tl^{+}$$ is the more stable oxidation state for thallium.

We now compare the species mentioned in the options.

$$Tl^{+}: E^\circ=-0.34\text{ V}$$ is less positive (more negative) than $$Al^{+}: E^\circ=+0.55\text{ V}$$. Because a more negative potential means a lower tendency to get reduced, $$Tl^{+}$$ remains as it is more readily than $$Al^{+}$$; thus $$Tl^{+}$$ is more stable than $$Al^{+}$$.

The other options conflict with the sign logic we have just applied: $$Al^{+}$$ is not more stable than $$Al^{3+}$$, $$Tl^{3+}$$ is not more stable than $$Al^{3+}$$, and $$Tl^{+}$$ is not more stable than $$Al^{3+}$$.

Therefore, only the statement “$$Tl^{+}$$ is more stable than $$Al^{+}$$” is correct.

Hence, the correct answer is Option 4.