PbO2- PbO - G2980Most probable oxidation state of Pb Sn will be -

The correct option is C Pb^+2, Sn^+4 For Pb it will be +2 as Δ G_298 of PbO lt;0 . (Spontaneous formation or more stable).For ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Reactivity with Halogens and Hydrogen (Group 2)

PbO2—— PbO ...

Question

$P b O_{2}$ ------ $P b O Δ G_{298} < 0$
$S n O_{2}$ ------- $S n O$ $Δ G_{298}$ >0
Most probable oxidation state of Pb & Sn will be :

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

P b O_{2} \to P b O

Δ G_{298} < 0

S n O_{2} \to S n O

Δ G_{298} > 0

P b

S n

[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 +}

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 :

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

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

Join BYJU'S Learning Program