To solve the problem step by step, we need to analyze the reactions happening during the titrations and the role of zinc metal in the solution.
### Step 1: Understanding the First Titration
In the first part of the experiment, we have a 25.0 mL solution containing both `Fe^(2+)` and `Fe^(3+)` ions. This solution is titrated with 25.0 mL of 0.0200 M `KMnO4` in dilute `H2SO4`. The balanced reaction is:
\[ \text{MnO}_4^{-} + 5 \text{Fe}^{2+} + 8 \text{H}^{+} \rightarrow \text{Mn}^{2+} + 5 \text{Fe}^{3+} + 4 \text{H}_2\text{O} \]
From this reaction, we can see that 1 mole of `MnO4^-` reacts with 5 moles of `Fe^(2+)` to produce `Fe^(3+)`.
### Step 2: Calculate Moles of KMnO4 Used
First, we calculate the moles of `KMnO4` used in the first titration:
\[
\text{Moles of KMnO4} = \text{Concentration} \times \text{Volume} = 0.0200 \, \text{mol/L} \times 0.0250 \, \text{L} = 0.0005 \, \text{mol}
\]
### Step 3: Calculate Moles of Fe2+ Oxidized
Using the stoichiometry of the reaction, we can find the moles of `Fe^(2+)` that were oxidized:
\[
\text{Moles of Fe}^{2+} = 5 \times \text{Moles of KMnO4} = 5 \times 0.0005 \, \text{mol} = 0.0025 \, \text{mol}
\]
### Step 4: Understanding the Second Titration
In the second part, 25 mL of the original solution is treated with zinc metal. Zinc is a reducing agent and will reduce `Fe^(3+)` back to `Fe^(2+)`. After the addition of zinc, the solution requires 40.0 mL of the same `KMnO4` solution for oxidation to `Fe^(3+)`.
### Step 5: Calculate Moles of KMnO4 Used in the Second Titration
Now, we calculate the moles of `KMnO4` used in the second titration:
\[
\text{Moles of KMnO4} = 0.0200 \, \text{mol/L} \times 0.0400 \, \text{L} = 0.0008 \, \text{mol}
\]
### Step 6: Calculate Moles of Fe2+ Oxidized in the Second Titration
Using the stoichiometry again, we find the moles of `Fe^(2+)` that were oxidized in the second titration:
\[
\text{Moles of Fe}^{2+} = 5 \times \text{Moles of KMnO4} = 5 \times 0.0008 \, \text{mol} = 0.0040 \, \text{mol}
\]
### Step 7: Determine the Change in Iron Species
In the second titration, the total amount of `Fe^(2+)` that was oxidized is 0.0040 mol. Since we had 0.0025 mol of `Fe^(2+)` initially oxidized in the first titration, the increase in `Fe^(2+)` must have come from the reduction of `Fe^(3+)` by zinc.
### Step 8: Conclusion
The zinc metal reduces `Fe^(3+)` to `Fe^(2+)`. Therefore, the correct conclusion is that zinc added in the second titration will reduce `Fe^(3+)` to `Fe^(2+)`.
### Final Answer
The zinc added in the second titration will reduce `Fe^(3+)` to `Fe^(2+)`.
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