standard electrode potential values, $E^{θ}$ for $\frac{A l^{+ 3}}{A l}$ is -1.66V and that of $\frac{T i^{+ 3}}{T i}$ is +1.26V. Predict about the formation of $M^{3 +}$ ion in solution and compare the electropositive character of the two metals.

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Solution

$A l^{+ 3} + 3 e^{-} ⇌ A l E^{θ} = - 1.66 v$
$T i^{+ 3} + 3 e^{-} ⇌ T i E^{θ} = + 1.26 v$
In solution $A l$ would be readily ionised while $T i$ would not ionise.
Also Aluminium $(A l)$ is more electropositive than titanium $(T i)$

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E^{o}

A l^{3 +} / A l

T l^{3 +} / T l

- 1.66 V

+ 1.26 V

A l^{+ 3} + 3 e^{-} \to A l

[A l^{+ 3}]

A l^{3 +} (a q .) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V

C u^{2 +} (a q .) + 2 e^{-} \to C u (s); E^{\circ} = + 0.34 V

M g < A l > Z n > C u > A g

E^{0}

Z n / Z n + 2, E^{0} = - 0.76 V

C u^{+ 2} / C u, E^{0} = + 0.34 V

A l / A l^{+ 3}, E^{0} = - 1.66 V

H g^{2 +} / H g, E^{0} = + 0.885 V

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