Amongst the following ions, which is considered as most stable in $M^{2 +}$ state?

{T i}^{2 +} / T i (- 1.63 V)

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V^{2 +} / V (- 1.11 V)

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{C r}^{2 +} / C r (- 0.90 V)

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{M n}^{2 +} / M n (1.18 V)

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Solution

The correct option is A ${T i}^{2 +} / T i (- 1.63 V)$
Half reactions with lowest standard reduction have stable reactants (ions).
Here, $T i^{2 +} + 2 e^{-} ⟶ T i [R e d u c t i o n]$
The standard reduction potential of this reaction is least. So, the reactant $T i^{2 +}$ is more stable.

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Similar questions

M^{2 +}

Z n^{2 +} > S c^{2 +} = C u^{2 +} > T i^{2 +} = N i^{2 +} > V^{2 +} = C o^{2 +} > C r^{2 +} = F e^{2 +} > M n^{2 +}

Which one of the following ions is the most stable in aqueous solution?
(Atomic number. Ti = 22, V = 23, Cr = 24, Mn = 25)

T i = 22, V = 23, C r = 24, M n = 25

{T i}^{2 +} (a q) + M g (s) \to T i (s) + {M g}^{2 +} (a q)

{T i}^{2 +} (a q) + 2 e^{-} \to T i (s);

E^{o} = - 1.63 V

{M g}^{2 +} (a q) + 2 e^{-} \to M g (s);

E^{o} = - 2.38 V

Δ G^{o}

3 d

T i^{4 +}, V^{2 +}, M n^{3 +}, C r^{3 +}

T i = 22, V = 23, M n = 25, C r = 24)

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