The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions is

Options

A. `1//3`
B. `1//6`
C. `2//3`
D. `3//4`

Explanation

To determine the number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of \( Sn^{2+} \) ions, we need to analyze the redox reaction between \( K_2Cr_2O_7 \) and \( Sn^{2+} \).

### Step-by-Step Solution:

1. **Identify the Redox Reaction:**
The reaction involves the reduction of \( K_2Cr_2O_7 \) (potassium dichromate) and the oxidation of \( Sn^{2+} \) (tin ions).
The reduction half-reaction is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
Here, \( Cr \) is reduced from an oxidation state of +6 to +3, which means each chromium atom gains 3 electrons. Since there are 2 chromium atoms, a total of 6 electrons are required.

2. **Oxidation Half-Reaction:**
The oxidation half-reaction for \( Sn^{2+} \) is:
\[ Sn^{2+} \rightarrow Sn^{4+} + 2e^- \]
In this reaction, each \( Sn^{2+} \) ion loses 2 electrons to become \( Sn^{4+} \).

3. **Balancing the Overall Reaction:**
To balance the overall reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. Since the reduction half-reaction requires 6 electrons, we need to find out how many \( Sn^{2+} \) ions are required to provide these 6 electrons.
\[ 3 \, Sn^{2+} \rightarrow 3 \, Sn^{4+} + 6e^- \]
This means that 3 moles of \( Sn^{2+} \) will provide 6 electrons.

4. **Writing the Balanced Equation:**
The balanced overall reaction can be written as:
\[ K_2Cr_2O_7 + 3Sn^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O \]

5. **Calculating Moles of \( K_2Cr_2O_7 \) Reduced:**
From the balanced equation, we can see that 1 mole of \( K_2Cr_2O_7 \) is reduced by 3 moles of \( Sn^{2+} \). Therefore, if we have 1 mole of \( Sn^{2+} \):
\[ \text{Moles of } K_2Cr_2O_7 \text{ reduced} = \frac{1 \text{ mole of } Sn^{2+}}{3} = \frac{1}{3} \text{ moles of } K_2Cr_2O_7 \]

### Final Answer:
The number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of \( Sn^{2+} \) ions is \( \frac{1}{3} \) moles.