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Electrolytic Dissociation

The element i...

Question

The element (ion) which is easiest to reduce is:

F e^{2 +}

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F e^{3 +}

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A g^{+}

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S n^{4 +}

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Solution

The correct option is D $A g^{+}$
SRP of following reactions are as follows:
$F e^{2 +}$ $+$ $2 e^{-}$ $\to$ $F e;$ ----- $E^{o} = - 0.44 V$
$F e^{3 +}$ $+$ $e^{-}$ $\to$ $F e^{2 +};$ ----- $E^{o} = 0.77 V$
$A g^{+}$ $+$ $e^{-}$ $\to$ $A g;$ ------- $E^{o} = 0.80 V$
$S n^{4 +}$ $+$ $2 e^{-}$ $\to$ $S n^{2 +}$ ---- $E^{o} = 0.15 V$
Now, we know that electrode with high SRP acts as a powerful oxidising agent, i.e., can be reduced easily.

So, among the above given options, $A g^{+}$ can be reduced most easily as SRP is the highest.

Option C is correct.

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