Number of moles of `K_(2)Cr_(2)O_(7)` reduced by one mole of `Sn^(2+)` ions is

Options

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

Answer

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

### Step 1: Write the Ionic Reaction
The ionic reaction for the reduction of dichromate ions \( Cr_2O_7^{2-} \) by \( Sn^{2+} \) ions is:
\[ Cr_2O_7^{2-} + Sn^{2+} + H^+ \rightarrow Sn^{4+} + Cr^{3+} + H_2O \]

### Step 2: Assign Oxidation States
- In \( Cr_2O_7^{2-} \), the oxidation state of chromium (Cr) is +6.
- In \( Sn^{2+} \), the oxidation state is +2.
- In \( Sn^{4+} \), the oxidation state is +4.
- In \( Cr^{3+} \), the oxidation state is +3.
- Oxygen (O) has an oxidation state of -2.

### Step 3: Identify Changes in Oxidation States
- Chromium is reduced from +6 to +3, which means it gains 6 electrons (2 Cr atoms, each gaining 3 electrons).
- Tin is oxidized from +2 to +4, which means it loses 2 electrons.

### Step 4: Balance the Electrons
To balance the number of electrons transferred:
- 2 moles of \( Sn^{2+} \) will provide 4 electrons (2 electrons per \( Sn^{2+} \)).
- 1 mole of \( Cr_2O_7^{2-} \) requires 6 electrons for the reduction of 2 moles of Cr.

To equalize the number of electrons, we need 3 moles of \( Sn^{2+} \) to provide 6 electrons:
\[ 3 Sn^{2+} \rightarrow 3 Sn^{4+} \]

### Step 5: Write the Balanced Overall Reaction
The balanced overall reaction becomes:
\[ Cr_2O_7^{2-} + 3 Sn^{2+} + 14 H^+ \rightarrow 2 Cr^{3+} + 3 Sn^{4+} + 7 H_2O \]

### Step 6: Determine the Moles of \( K_2Cr_2O_7 \) Reduced
From the balanced equation, we see that:
- 3 moles of \( Sn^{2+} \) reduce 1 mole of \( K_2Cr_2O_7 \).
- Therefore, 1 mole of \( Sn^{2+} \) will reduce \( \frac{1}{3} \) moles of \( K_2Cr_2O_7 \).

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