An acid solution of 0.2 " mol of "`KReO_(4)` was reduced with Zn and then titrated with 1.6 " Eq of "acidic `KMnO_(4)` solution for the reoxidation of the ehenium `(Re)` to the perrhenate ion `(ReO_(4)^(ɵ))`. Assuming that rhenium was the only elements reduced, what is the oxidation state to which rhenium was reduced by Zn?

Options

A. 1
B. 2
C. `-1`
D. `-2`

Explanation

To determine the oxidation state to which rhenium (Re) was reduced by zinc (Zn) in the given reaction, we can follow these steps:

### Step 1: Understand the Reaction
We start with potassium perrhenate, \( KReO_4 \), which contains rhenium in the +7 oxidation state. When it is reduced by zinc, rhenium will lose some of its oxidation state.

### Step 2: Write the Reduction Reaction
The reduction can be represented as:
\[ KReO_4 + Zn \rightarrow Re^x + Zn^{2+} + \text{other products} \]
Here, \( Re^x \) represents rhenium in its reduced state.

### Step 3: Determine the n-factor for the Reaction
We know that \( KReO_4 \) is reduced to \( Re^x \). The n-factor is defined as the change in oxidation state per mole of the substance.

For \( KReO_4 \):
- The oxidation state of rhenium in \( KReO_4 \) is +7.
- Let the oxidation state of rhenium after reduction be \( x \).

The change in oxidation state can be calculated as:
\[ \text{n-factor} = 7 - x \]

### Step 4: Relate n-factor to the Given Equivalents
We are given that 0.2 moles of \( KReO_4 \) were reduced and subsequently titrated with 1.6 equivalents of acidic \( KMnO_4 \).

Since 1 equivalent of \( ReO_4^- \) reacts with 1 equivalent of \( MnO_4^- \), we can set up the equation:
\[ n \times \text{number of moles of } KReO_4 = \text{equivalents of } KMnO_4 \]
Substituting the values:
\[ (7 - x) \times 0.2 = 1.6 \]

### Step 5: Solve for x
Now we can solve for \( x \):
1. Rearranging the equation gives:
\[ 7 - x = \frac{1.6}{0.2} \]
\[ 7 - x = 8 \]
2. Solving for \( x \):
\[ -x = 8 - 7 \]
\[ -x = 1 \]
\[ x = -1 \]

### Conclusion
Thus, the oxidation state to which rhenium was reduced by zinc is -1.