Given standard reduction potentials

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

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

We need stability of the intermediate oxidation state $M^+$ relative to $M^{3+}$

To compare the stability of $M^+$ and $M^{3+}$, consider the reaction: $M^{3+} + 2e^- \rightarrow M^+$

If this reduction has a large positive potential, then $M^{3+}$ readily converts to $M^+$, so $M^+$ is more stable.

Find $E^\circ$ for $M^{3+} \rightarrow M^+$

Use $\Delta G^\circ = -nFE^\circ$.

Also, $\Delta G^\circ(M^{3+} \rightarrow M) = \Delta G^\circ(M^{3+} \rightarrow M^+) + \Delta G^\circ(M^+ \rightarrow M)$

So, $-3F E_1 = -2F E_{(3+/+)} -1F E_2$ where

• $E_1 = E^\circ(M^{3+}/M)$
• $E_2 = E^\circ(M^+/M)$

Hence, $2E_{(3+/+)} = 3E_1 - E_2$ $E_{(3+/+)} = \frac{3E_1 - E_2}{2}$

For aluminium

$E^\circ(Al^{3+}/Al^+) = \frac{3(-1.66) - (0.55)}{2}$ $= \frac{-4.98 - 0.55}{2} = \frac{-5.53}{2} = -2.765\text{ V}$

Since this is highly negative, the reaction $Al^{3+} + 2e^- \rightarrow Al^+$ is not favorable.

Therefore, $Al^+$ is less stable than $Al^{3+}$.

For thallium

$E^\circ(Tl^{3+}/Tl^+) = \frac{3(1.26) - (-0.34)}{2}$ $= \frac{3.78 + 0.34}{2} = \frac{4.12}{2} = 2.06\text{ V}$

Since this is strongly positive, the reaction $Tl^{3+} + 2e^- \rightarrow Tl^+$ is highly favorable.

Therefore, $Tl^+$ is more stable than $Tl^{3+}$.

Now evaluate the options

• A: $Tl^+$ is more stable than $Al^{3+}$
  This comparison is not the relevant/direct conclusion from the given data, and is not the best correct statement.

• B: $Al^+$ is more stable than $Al^{3+}$
  False, because $Al^+$ is less stable than $Al^{3+}$.

• C: $Tl^+$ is more stable than $Al^+$
  True. From above, $Tl^+$ is highly stable while $Al^+$ is unstable.

• D: $Tl^{3+}$ is more stable than $Al^{3+}$
  Not the intended correct conclusion from these data.

Final answer

$\boxed{C}$