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. `3`
C. `1//6`
D. `6`

Explanation

To determine the number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of \( Sn^{2+} \) ions, we can follow these steps:

### Step 1: Identify the reduction of \( K_2Cr_2O_7 \)
The dichromate ion \( Cr_2O_7^{2-} \) is reduced in the reaction. The reduction half-reaction can be written as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
This shows that 1 mole of \( K_2Cr_2O_7 \) (or \( Cr_2O_7^{2-} \)) requires 6 moles of electrons for reduction.

### Step 2: Identify the oxidation of \( Sn^{2+} \)
The oxidation half-reaction for \( Sn^{2+} \) is:
\[ Sn^{2+} \rightarrow Sn^{4+} + 2e^- \]
This indicates that 1 mole of \( Sn^{2+} \) can donate 2 moles of electrons.

### Step 3: Calculate the number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of \( Sn^{2+} \)
Now, we need to find out how many moles of \( K_2Cr_2O_7 \) can be reduced by the electrons provided by 1 mole of \( Sn^{2+} \).

From the half-reactions, we know:
- 1 mole of \( K_2Cr_2O_7 \) requires 6 moles of electrons.
- 1 mole of \( Sn^{2+} \) provides 2 moles of electrons.

To find the number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of \( Sn^{2+} \), we set up the following proportion:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{\text{Electrons provided by } Sn^{2+}}{\text{Electrons required by } K_2Cr_2O_7} \]
Substituting the values:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{2 \text{ moles of electrons}}{6 \text{ moles of electrons}} = \frac{1}{3} \text{ moles of } K_2Cr_2O_7 \]

### Final Answer
Thus, the number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of \( Sn^{2+} \) ions is:
\[ \frac{1}{3} \text{ moles} \]

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