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(D) The stability of an oxidation state can be inferred from the standard electrode potential $(E^0)$. $A$ more negative $E^0$ value for the reduction

Vedclass · Accepted Answer

$Tl^{+}$ is more stable than $Al^{+}$

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(D) The stability of an oxidation state can be inferred from the standard electrode potential $(E^0)$. $A$ more negative $E^0$ value for the reduction half-reaction $(M^n+ + ne^- \rightarrow M)$ indicates that the metal is more easily oxidized,meaning the higher oxidation state is more stable relative to the metal.For $Al$: $E^0(Al^{3+}/Al) = -1.66 \ V$ and $E^0(Al^{+}/Al) = +0.55 \ V$. Since $E^0(Al^{3+}/Al) For $Tl$: $E^0(Tl^{3+}/Tl) = +1.26 \ V$ and $E^0(Tl^{+}/Tl) = -0.34 \ V$. Since $E^0(Tl^{+}/Tl) Comparing $Tl^{+}$ and $Al^{+}$: $Tl^{+}$ has a negative reduction potential $(-0.34 \ V)$,making it relatively stable,whereas $Al^{+}$ has a positive reduction potential $(+0.55 \ V)$,making it highly unstable and a strong oxidizing agent.Therefore,$Tl^{+}$ is more stable than $Al^{+}$.

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

Consider the following standard electrode potentials ($E^0$ in volts) in aqueous solution. Based on this data,which of the following statements is correct?
Element $M^{3+}/M$ $M^{+}/M$
$Al$ $-1.66$ $+0.55$
$Tl$ $+1.26$ $-0.34$

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Consider the following standard electrode potentials ($E^0$ in volts) in aqueous solution. Based on this data,which of the following statements is correct?Element$M^{3+}/M$$M^{+}/M$$Al$$-1.66$$+0.55$$Tl$$+1.26$$-0.34$