Arrange the following ions in increasing $N a^{+}, M g^{2 +}, A l^{3 +}$
standard reduction potential $(E_{r e d n}^{o}),$
$M^{n +} (a q) + n e^{-} \to M (s)$

N a^{+} < M g^{2 +} < A l^{3 +}

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A l^{3 +} < M g^{2 +} < N a^{+}

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N a^{+} < M g^{2 +} > A l^{3 +}

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None of these

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Solution

The correct option is C $A l^{3 +} < M g^{2 +} < N a^{+}$
From SRP table it can be seen that,
$A l^{3 +} < M g^{2 +} < N a^{+} (E_{r e d}^{o} = - v e)$
(higher the metallic character, higher will be the reduction potential)

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Similar questions

N a^{+}

M g^{2 +}

A l^{3 +}

S i^{4 +}

$N a^{+} . M g^{2 +}, A l^{3 +}$ and $S i^{4 +}$ are isoelectronic. The order of their ionic size is

{N a}^{+}, {C a}^{2 +}, {M g}^{2 +}, {A l}^{3 +}

{C a}^{2 +} < {N a}^{+} < {M g}^{2 +} < {A l}^{3 +}

i) N e i i) O^{2 -} i i i) M g^{2 +} i v) F

v) A l^{3 +} v i) C l^{-} v i i) K v i i i) N a

N a^{+}, M g^{2 +}

A l^{3 +}

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