State the following is oxidation reaction or reduction reaction: $S n^{4 +} + 2 e^{-} \to S n^{2 +}$

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Solution

Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion. Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.
Here,
$S n^{4 +} + 2 e^{-} \to S n^{2 +}$
$S n^{+ 4}$ accepts two electrons and forms $S n^{+ 2}$ so its a reduction process.

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[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

P t / F e^{2 -}, F e^{3 +}

P t / S n^{4 +}, S n^{2 +}

25^{o}

S n^{4 +} + 2 F e^{2 +} \to S n^{2 +} + 2 F e^{3 +}

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