PbO2+Pb xrightarrow2PbO ; Δ rG° 0 Which oxidation states are more characteristic for lead and tin?

Question

PbO2+Pb xrightarrow2PbO ; Δ rG° <0 SnO2+Sn xrightarrow2SnO; Δ rG° >0 Which oxidation states are more characteristic for lead and tin? (A) For lead +4, for tin +2 (B) for lead +2 , for tin +4 (C) For lead +2, for tin +2 (D) For lead +4, for tin +4

Tardigrade Admin · Accepted Answer

Since Δ rG° is negative for first reaction, therefore, +2 oxidation state of Pb is more stable. Further, since Δ rG° is positive for second reaction, therefore, +4 oxidation state of Sn is more stable. Thus option is correct.

Q. $P b O_{2} + P b 2 P b O; Δ_{r} G^{\circ} < 0$ $S n O_{2} + S n 2 S n O; Δ_{r} G^{\circ} > 0$ Which oxidation states are more characteristic for lead and tin?

For lead +4, for tin +2

for lead +2 , for tin +4

For lead +2, for tin +2

For lead +4, for tin +4

Since $Δ_{r} G^{\circ}$ is negative for first reaction, therefore, +2 oxidation state of Pb is more stable. Further, since $Δ_{r} G^{\circ}$ is positive for second reaction, therefore, +4 oxidation state of Sn is more stable. Thus option is correct.

Q. PbO2​+Pb​2PbO;Δr​G∘<0 SnO2​+Sn​2SnO;Δr​G∘>0 Which oxidation states are more characteristic for lead and tin?