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PbO(2)rarrPbO,DeltaG(298)lt0 SnO(2)rar...

`PbO_(2)rarrPbO,DeltaG_(298)lt0`
`SnO_(2)rarrSnO,DeltaG_(298)gt0`
Most proble oxidation states of `Pb` and `Sn` will be

A

`Pb^(+4),Sn^(+2)`

B

`Pb^(+4),Sn^(+2)`

C

`Pb^(+2),Sn^(+2)`

D

`Pb^(+2),Sn^(+4)`

Text Solution

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The correct Answer is:
To determine the most probable oxidation states of lead (Pb) and tin (Sn) based on the given reactions and their Gibbs free energy changes (ΔG), we can follow these steps: ### Step-by-Step Solution: 1. **Analyze the First Reaction:** - The first reaction is: \[ \text{PbO}_2 \rightarrow \text{PbO} \] - In this reaction, lead (Pb) is going from an oxidation state of +4 in PbO2 to +2 in PbO. - The ΔG for this reaction at 298 K is less than 0 (ΔG < 0), indicating that this reaction is spontaneous. 2. **Determine the Most Probable Oxidation State for Pb:** - Since the reaction is spontaneous (ΔG < 0), the most stable and probable oxidation state of lead (Pb) in this context is +2. - Therefore, the most probable oxidation state of Pb is: \[ \text{Pb: +2} \] 3. **Analyze the Second Reaction:** - The second reaction is: \[ \text{SnO}_2 \rightarrow \text{SnO} \] - Here, tin (Sn) is going from an oxidation state of +4 in SnO2 to +2 in SnO. - The ΔG for this reaction at 298 K is greater than 0 (ΔG > 0), indicating that this reaction is non-spontaneous. 4. **Determine the Most Probable Oxidation State for Sn:** - Since the reaction is non-spontaneous (ΔG > 0), the most stable and probable oxidation state of tin (Sn) in this context is +4. - Therefore, the most probable oxidation state of Sn is: \[ \text{Sn: +4} \] ### Final Answer: - The most probable oxidation states are: - **Pb: +2** - **Sn: +4**
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