A half cell is prepared by `K_(2)Cr_(2)O_(7)` in a buffer solution of `pH =1`. Concentration of `K_(2)Cr_(2)O_(7)` is `1M`. To 3 litre of this solution `570 gm` of `SnCI_(2)` is added which is oxidised completely to `SnCI_(4)`.
Given: `E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06`,
Atomic of mass `Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15`
Number of moles of `Cr^(+3)` formed are

Options

A. 2
B. 6
C. 4
D. 3

Explanation

To solve the problem step by step, we will follow the reactions and calculations systematically.

### Step 1: Identify the half-reactions
At the anode, the oxidation of Sn²⁺ to Sn⁴⁺ occurs:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]

At the cathode, the reduction of Cr₂O₇²⁻ to Cr³⁺ occurs:
\[ \text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7H_2O \]

### Step 2: Combine the half-reactions
To combine the half-reactions, we need to balance the number of electrons. Since the anode reaction involves 2 electrons and the cathode reaction involves 6 electrons, we multiply the anode reaction by 3:
\[ 3\text{Sn}^{2+} \rightarrow 3\text{Sn}^{4+} + 6e^- \]

Now, we can combine the two half-reactions:
\[ 3\text{Sn}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14H^+ \rightarrow 3\text{Sn}^{4+} + 2\text{Cr}^{3+} + 7H_2O \]

### Step 3: Calculate the moles of SnCl₂
Given that 570 g of SnCl₂ is added, we first need to calculate the molar mass of SnCl₂:
- Atomic mass of Sn = 119 g/mol
- Atomic mass of Cl = 35.5 g/mol
- Molar mass of SnCl₂ = 119 + 2(35.5) = 190 g/mol

Now, we calculate the number of moles of SnCl₂:
\[ \text{Moles of SnCl}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{570 \text{ g}}{190 \text{ g/mol}} = 3 \text{ moles} \]

### Step 4: Determine the concentration of Sn²⁺
Since the volume of the solution is 3 liters, the concentration of Sn²⁺ is:
\[ \text{Concentration of Sn}^{2+} = \frac{\text{moles}}{\text{volume}} = \frac{3 \text{ moles}}{3 \text{ L}} = 1 \text{ M} \]

### Step 5: Calculate the concentration of H⁺
Given that the pH of the solution is 1:
\[ \text{pH} = -\log[H^+] \]
Thus,
\[ [H^+] = 10^{-1} = 0.1 \text{ M} \]

### Step 6: Calculate the moles of Cr³⁺ formed
From the balanced equation, for every 1 mole of Cr₂O₇²⁻ reduced, 2 moles of Cr³⁺ are produced. The initial concentration of Cr₂O₇²⁻ is 1 M in 3 L, so:
\[ \text{Moles of Cr}_2\text{O}_7^{2-} = 1 \text{ M} \times 3 \text{ L} = 3 \text{ moles} \]

Now, we need to determine how much of Cr₂O₇²⁻ will react. The limiting reagent will be determined by the stoichiometry of the reaction.

From the balanced equation:
- 3 moles of Sn²⁺ react with 1 mole of Cr₂O₇²⁻.

Since we have 3 moles of Sn²⁺ and 3 moles of Cr₂O₇²⁻, all of the Sn²⁺ will react, consuming:
\[ \frac{3 \text{ moles of Sn}^{2+}}{3} = 1 \text{ mole of Cr}_2\text{O}_7^{2-} \]

Thus, 1 mole of Cr₂O₇²⁻ will produce:
\[ 2 \text{ moles of Cr}^{3+} \]

### Step 7: Final calculation of moles of Cr³⁺
Thus, the total moles of Cr³⁺ formed are:
\[ \text{Moles of Cr}^{3+} = 2 \text{ moles} \]

### Conclusion
The number of moles of Cr³⁺ formed is **2 moles**.

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