Q.Standard electrode potential values, E°V for Al3+/Al is –1.66 V and that of Tl3+/Tl is +1.26 V. Predict about the formation of M3+ ion in solution and compare the electropositive character of the two metals.
Step 1 - Read the sign and magnitude of
A large negative for means the reverse process, , is thermodynamically favourable - the metal oxidises easily and is the stable species in solution. A positive means the forward reduction is favoured - is a strong oxidising agent and is not the stable form.
Step 2 - Apply to Al
The strongly negative value shows Al loses its three valence electrons () easily; Al(aq) is thermodynamically stable and Al is a strong reducing agent.
Step 3 - Apply to Tl
The positive value shows Tl readily gains electrons, i.e. it is a strong oxidising agent that is reduced back to the far more stable (the pair resists bonding - the inert pair effect - so it stays as a lone pair rather than being used to form the third bond).
Step 4 - Compare electropositive character
Electropositive character is measured by how readily a metal loses electrons to form its cation. Al does this far more readily (very negative ) than Tl does for the +3 state (positive ), so Al is more electropositive than Tl as far as the state is concerned; for Tl the electropositive character instead shows up in the state.
Al3+ forms readily and is stable in solution (Al is a strong reducing agent, highly electropositive). Tl3+ is unstable and strongly oxidising - it is reduced to the far more stable Tl+ because of the inert pair effect. So, in the +3 state, Al is markedly more electropositive than Tl.
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