PbO2- PbO - G298 0Most probable oxidation state of Pb and Sn will be -

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Oxidising and Reducing Nature

PbO2→ PbO Δ...

Question

$P b O_{2} \to P b O$ $Δ G_{298} < 0$
$S n O_{2} \to S n O$ $Δ G_{298} > 0$
Most probable oxidation state of $P b$ and $S n$ will be :

P b^{4 +}, S n^{4 +}

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P b^{4 +}, S n^{2 +}

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P b^{2 +}, S n^{2 +}

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P b^{2 +}, S n^{4 +}

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Solution

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Similar questions

[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 b O_{2}

P b O Δ G_{298} < 0

S n O_{2}

S n O

Δ G_{298}

Consider the reactions;

P b O_{2} \to P b O, Δ G_{298} < 0;

$S n O_{2} \to S n O, Δ G_{298} < 0$

The most probable oxidation state of $P b$ and $S n$ will be :

Stability of ions of $G e, S n, a n d P b$ will be in the order:

P b^{4 +} + S n^{2 +} \to S n^{4 +} + P b^{2 +}

S n^{4 +} + P b^{2 +} \to P b^{4 +} + S n^{2 +}

E_{P b^{4 +} / P b^{2 +}}^{o} = 1.69 V, E_{S n^{4 +} / S n^{2 +}}^{o} = 0.15 V

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