In view of the signs of `Delta_rG^@` for the following reactions
`PbO_2 + Pb to 2PbO, Delta_r G^@ lt 0`
`SnO_2 +Sn to 2SnO, Delta_rG^@ gt 0`
Which oxidation states are more characteristic for lead and tin?

Options

A. For lead + 4, for tin + 2
B. For lead + 2, for tin + 2
C. For lead + 4, for tin + 4
D. For lead + 2, for tin + 4.

Explanation

To determine the more characteristic oxidation states for lead (Pb) and tin (Sn) based on the given reactions and the signs of their Gibbs free energy changes (Δ_rG^@), we can follow these steps:

### Step 1: Analyze the first reaction
The first reaction is:
\[ \text{PbO}_2 + \text{Pb} \rightarrow 2\text{PbO} \]
with Δ_rG^@ < 0.

### Step 2: Determine oxidation states in the first reaction
- In PbO2, the oxidation state of Pb can be calculated as follows:
- Let the oxidation state of Pb be \( x \).
- The equation becomes: \( x + 2(-2) = 0 \) (since O is -2).
- Thus, \( x - 4 = 0 \) → \( x = +4 \).

- In PbO, the oxidation state of Pb is:
- Let the oxidation state of Pb be \( y \).
- The equation becomes: \( y + (-2) = 0 \).
- Thus, \( y - 2 = 0 \) → \( y = +2 \).

### Step 3: Identify the change in oxidation state for lead
- In this reaction, lead goes from +4 in PbO2 to +2 in PbO.
- Since the reaction is favored (Δ_rG^@ < 0), the +2 oxidation state is more stable and characteristic for lead.

### Step 4: Analyze the second reaction
The second reaction is:
\[ \text{SnO}_2 + \text{Sn} \rightarrow 2\text{SnO} \]
with Δ_rG^@ > 0.

### Step 5: Determine oxidation states in the second reaction
- In SnO2, the oxidation state of Sn can be calculated as follows:
- Let the oxidation state of Sn be \( a \).
- The equation becomes: \( a + 2(-2) = 0 \).
- Thus, \( a - 4 = 0 \) → \( a = +4 \).

- In SnO, the oxidation state of Sn is:
- Let the oxidation state of Sn be \( b \).
- The equation becomes: \( b + (-2) = 0 \).
- Thus, \( b - 2 = 0 \) → \( b = +2 \).

### Step 6: Identify the change in oxidation state for tin
- In this reaction, tin goes from +4 in SnO2 to +2 in SnO.
- However, since the reaction is not favored (Δ_rG^@ > 0), it indicates that the +4 oxidation state is more stable and characteristic for tin.

### Conclusion
- The more characteristic oxidation state for lead (Pb) is +2.
- The more characteristic oxidation state for tin (Sn) is +4.

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