Adding powdered Pb and Fe to a solution containing `1.0 `M each of `Pb^(+2) and Fe^(+2)` ions would result in the formation of :

Options

A. More of Pb and `Fe^(+2)` ions
B. More of Fe and `Pb^(2+)` ions
C. More of Fe and Pb both
D. More of `Fe^(+2 )and Pb^(+2)` ions

Answer

To solve the problem of what happens when powdered Pb and Fe are added to a solution containing 1.0 M each of Pb²⁺ and Fe²⁺ ions, we can analyze the reduction potentials of the two metals and their respective ions.

### Step-by-Step Solution:

1. **Identify the Species Involved**:
We have two metal ions in solution: Pb²⁺ and Fe²⁺. We are adding solid Pb and Fe to this solution.

2. **Write the Reduction Reactions**:
- For lead (Pb):
\[ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \quad (E^\circ = -0.126 \, \text{V}) \]
- For iron (Fe):
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad (E^\circ = -0.440 \, \text{V}) \]

3. **Compare the Standard Reduction Potentials**:
The standard reduction potential for Pb²⁺ is -0.126 V and for Fe²⁺ is -0.440 V. A more positive (or less negative) reduction potential indicates a greater tendency to be reduced.

4. **Determine Which Ion Will Be Reduced**:
Since Pb²⁺ has a higher reduction potential than Fe²⁺, Pb²⁺ will be reduced preferentially over Fe²⁺. This means that Pb²⁺ ions will gain electrons and form more solid Pb.

5. **Analyze the Effect on Iron**:
The addition of solid Fe will not significantly reduce the concentration of Fe²⁺ ions because the reduction potential for Fe²⁺ is lower. Therefore, Fe²⁺ will remain in the solution.

6. **Conclusion**:
After adding powdered Pb and Fe to the solution, we will have an increase in the amount of solid Pb formed from Pb²⁺ ions, while the concentration of Fe²⁺ ions will remain relatively unchanged.

### Final Answer:
The addition of powdered Pb and Fe to the solution will result in the formation of more Pb and Fe²⁺ ions. Thus, the correct answer is: **More Pb and Fe²⁺ ions.**