A solution is 1 molar in each of NaCl, `CdCl_(2), ZnCl_(2) and PbCl_(2)`. To this Sn metal is added, which of the following is true?
Given `E^(@) (Pb^(2+)//Pb = -0.126V)`
`E_(Sn^(2+)//Sn)^(@) = -0.136V, E_(Cd^(2+)//Cd)^(@) = -0.40V`
`E_(Zn^(2+)//Zn)^(@) = -0.763V, E_(Na^(+)//Na)^(@)= -2.71V`

Options

A. Sn can reduce `Na^(+)` to Na
B. Sn can reduce `Zn^(2+)` to Zn
C. Sn can reduce `Cd^(2+)` to Cd
D. Sn can reduce `Pb^(2+)` to Pb

Answer

To solve the problem, we need to analyze the ability of tin (Sn) to reduce the metal ions present in the solutions of NaCl, CdCl₂, ZnCl₂, and PbCl₂ based on their standard reduction potentials.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials:**
- For Pb²⁺/Pb: \( E^\circ = -0.126 \, \text{V} \)
- For Sn²⁺/Sn: \( E^\circ = -0.136 \, \text{V} \)
- For Cd²⁺/Cd: \( E^\circ = -0.40 \, \text{V} \)
- For Zn²⁺/Zn: \( E^\circ = -0.763 \, \text{V} \)
- For Na⁺/Na: \( E^\circ = -2.71 \, \text{V} \)

2. **Determine the Reduction Potential Comparison:**
- The higher the reduction potential, the greater the tendency of the species to gain electrons (be reduced).
- Compare the reduction potential of Sn with each of the metal ions:
- **Na⁺:** \( -2.71 \, \text{V} \) (lower than Sn)
- **Zn²⁺:** \( -0.763 \, \text{V} \) (lower than Sn)
- **Cd²⁺:** \( -0.40 \, \text{V} \) (lower than Sn)
- **Pb²⁺:** \( -0.126 \, \text{V} \) (higher than Sn)

3. **Evaluate Each Option:**
- **Option 1:** Can Sn reduce Na⁺ to Na?
- No, because Sn has a higher reduction potential than Na⁺.
- **Option 2:** Can Sn reduce Zn²⁺ to Zn?
- No, because Sn has a higher reduction potential than Zn²⁺.
- **Option 3:** Can Sn reduce Cd²⁺ to Cd?
- No, because Sn has a higher reduction potential than Cd²⁺.
- **Option 4:** Can Sn reduce Pb²⁺ to Pb?
- Yes, because Sn has a lower reduction potential than Pb²⁺, meaning Sn can reduce Pb²⁺ to Pb.

4. **Conclusion:**
- The only true statement is that Sn can reduce Pb²⁺ to Pb. Therefore, the correct answer is **Option 4**.