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 reduction potentials of the metals involved and determine which metal can reduce the others in the solution. Let's break down the steps:

### Step 1: Identify the Reduction Potentials
We have the following standard reduction potentials given:
- \( E^\circ (Pb^{2+}/Pb) = -0.126 \, V \)
- \( E^\circ (Sn^{2+}/Sn) = -0.136 \, V \)
- \( E^\circ (Cd^{2+}/Cd) = -0.40 \, V \)
- \( E^\circ (Zn^{2+}/Zn) = -0.763 \, V \)
- \( E^\circ (Na^+/Na) = -2.71 \, V \)

### Step 2: Determine the Reducing Strength
The more negative the standard reduction potential, the stronger the reducing agent. Therefore, we can arrange the metals based on their reduction potentials from least negative (strongest oxidizing agent) to most negative (strongest reducing agent):
1. \( Pb^{2+}/Pb \) (-0.126 V)
2. \( Sn^{2+}/Sn \) (-0.136 V)
3. \( Cd^{2+}/Cd \) (-0.40 V)
4. \( Zn^{2+}/Zn \) (-0.763 V)
5. \( Na^+/Na \) (-2.71 V)

### Step 3: Analyze the Reaction
When Sn metal is added to the solution, it can potentially reduce any metal cation that has a higher reduction potential than itself. We compare the reduction potentials of Sn with those of the other metals:
- \( Sn \) can reduce \( Pb^{2+} \) because \( E^\circ (Pb^{2+}/Pb) > E^\circ (Sn^{2+}/Sn) \).
- \( Sn \) cannot reduce \( Cd^{2+} \) because \( E^\circ (Cd^{2+}/Cd) < E^\circ (Sn^{2+}/Sn) \).
- \( Sn \) cannot reduce \( Zn^{2+} \) or \( Na^+ \) for the same reason.

### Step 4: Conclusion
The only metal that can be reduced by Sn in the solution is \( Pb^{2+} \). Therefore, the correct statement is:
- Sn can reduce \( Pb^{2+} \) to \( Pb \).

### Final Answer
**Sn can reduce \( Pb^{2+} \) to \( Pb \) only.**

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