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: Mar 24, 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

Solution and Explanation

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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