Question:

Consider the following standard electrode potentials (\(E^\circ\) in volts) in aqueous solution: \includegraphics[]{135.png} Based on this data, which of the following statements is correct?

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Lower oxidation states of thallium (Tl) are more stable due to the inert pair effect, while higher oxidation states of aluminum (Al) are more stable due to stronger metallic bonding.
Updated On: May 18, 2025
  • \( Tl^{3+} \) is more stable than \( Al^{3+} \)
  • \( Tl^{1+} \) is more stable than \( Al^{3+} \)
  • \( Al^{1+} \) is more stable than \( Al^{3+} \)
  • \( Tl^{1+} \) is more stable than \( Al^{1+} \)
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The Correct Option is D

Approach Solution - 1

Step 1: Understanding Electrode Potentials 
- Negative \(E^\circ\) means the ion is stable in its oxidized form. 
- Positive \(E^\circ\) means the ion is easily reduced, implying instability in oxidized form.
Step 2: Stability of \( Al^{3+} \) vs. \( Al^+ \) 
- \(E^\circ\) for \( Al^{3+}/Al \) is -1.66 V, so \( Al^{3+} \) is stable. 
- \(E^\circ\) for \( Al^+/Al \) is +0.55 V, so \( Al^+ \) is unstable. 
\( Al^{3+} \) is more stable than \( Al^+ \) → Eliminates Option (3).
Step 3: Stability of \( Tl^{3+} \) vs. \( Tl^+ \) 
- \(E^\circ\) for \( Tl^{3+}/Tl \) is +1.26 V, so \( Tl^{3+} \) is unstable. 
- \(E^\circ\) for \( Tl^+/Tl \) is -0.34 V, so \( Tl^+ \) is stable. 
\( Tl^+ \) is more stable than \( Tl^{3+} \) → Eliminates Option (1).
Step 4: Comparing \( Tl^+ \) and \( Al^+ \) 
- \( Al^+ \) is highly unstable. 
- \( Tl^+ \) is relatively stable. 
\( Tl^+ \) is more stable than \( Al^+ \) → Confirms Option (4) as the correct answer.

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Approach Solution -2

Step 1: Interpreting Electrode Potentials
- A negative standard electrode potential (\( E^\circ \)) suggests that the ion is stable in its oxidized form and less prone to reduction.
- A positive \( E^\circ \) indicates that the ion is more easily reduced and less stable as an oxidized species.

Step 2: Comparing Stability of \( Al^{3+} \) and \( Al^{+} \)
- The electrode potential for the \( Al^{3+}/Al \) couple is \( -1.66 \, \text{V} \), which means \( Al^{3+} \) is relatively stable in its oxidized state.
- For the \( Al^{+}/Al \) couple, \( E^\circ \) is \( +0.55 \, \text{V} \), indicating \( Al^{+} \) is unstable and readily reduced.
- Thus, \( Al^{3+} \) is more stable than \( Al^{+} \), ruling out Option (3).

Step 3: Comparing Stability of \( Tl^{3+} \) and \( Tl^{+} \)
- The electrode potential for \( Tl^{3+}/Tl \) is \( +1.26 \, \text{V} \), showing that \( Tl^{3+} \) is unstable and easily reduced.
- The potential for \( Tl^{+}/Tl \) is \( -0.34 \, \text{V} \), suggesting that \( Tl^{+} \) is stable.
- Therefore, \( Tl^{+} \) is more stable than \( Tl^{3+} \), which eliminates Option (1).

Step 4: Stability Comparison Between \( Tl^{+} \) and \( Al^{+} \)
- \( Al^{+} \) is very unstable and tends to gain electrons quickly.
- In contrast, \( Tl^{+} \) is comparatively more stable.
- This confirms that \( Tl^{+} \) is more stable than \( Al^{+} \), making Option (4) the correct choice.

Summary:
- The stability of oxidation states can be assessed using their standard electrode potentials.
- \( Al^{3+} \) and \( Tl^{+} \) are the favored stable oxidation states for aluminum and thallium, respectively.
- This understanding helps in correctly identifying the most stable ions and selecting the right answer.
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