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`
Half cell potential `E_(Cr_(2)O_(7)^(-2)//Cr^(+3))^(@)` after teh reaction of `SnCI_(2)` is:

Options

A. `1.187`
B. `1.191`
C. `1.285`
D. None of these

Explanation

To solve the problem step by step, we will calculate the half-cell potential \( E_{Cr_2O_7^{2-}/Cr^{3+}} \) after the reaction of \( SnCl_2 \) with \( K_2Cr_2O_7 \) in a buffer solution of pH 1.

### Step 1: Determine the moles of \( SnCl_2 \) added
First, we need to calculate the number of moles of \( SnCl_2 \) added to the solution. The molecular weight of \( Sn \) is 119 g/mol, and since \( SnCl_2 \) contains one \( Sn \) atom, we can calculate the moles of \( SnCl_2 \):

\[ \text{Molar mass of } SnCl_2 = 119 + (2 \times 35.5) = 119 + 71 = 190 \text{ g/mol} \]

Now, calculate the moles of \( SnCl_2 \):

\[ \text{Moles of } SnCl_2 = \frac{570 \text{ g}}{190 \text{ g/mol}} = 3 \text{ moles} \]

### Step 2: Determine the concentration of \( Cr^{3+} \)
The reaction between \( SnCl_2 \) and \( K_2Cr_2O_7 \) can be represented as follows:

\[ Cr_2O_7^{2-} + 6e^- + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O \]

From the stoichiometry of the reaction, 1 mole of \( Cr_2O_7^{2-} \) reacts with 6 moles of electrons to produce 2 moles of \( Cr^{3+} \).

Given that the concentration of \( K_2Cr_2O_7 \) is 1 M in a 3 L solution:

\[ \text{Moles of } K_2Cr_2O_7 = 1 \text{ M} \times 3 \text{ L} = 3 \text{ moles} \]

Since \( K_2Cr_2O_7 \) dissociates to give \( Cr_2O_7^{2-} \), we have 3 moles of \( Cr_2O_7^{2-} \).

### Step 3: Calculate the concentration of \( Cr^{3+} \)
From the reaction, 1 mole of \( Cr_2O_7^{2-} \) produces 2 moles of \( Cr^{3+} \). Therefore, 3 moles of \( Cr_2O_7^{2-} \) will produce:

\[ \text{Moles of } Cr^{3+} = 2 \times 3 = 6 \text{ moles} \]

Now, the concentration of \( Cr^{3+} \) in the 3 L solution is:

\[ \text{Concentration of } Cr^{3+} = \frac{6 \text{ moles}}{3 \text{ L}} = 2 \text{ M} \]

### Step 4: Calculate the concentration of \( H^+ \)
Given that the solution is buffered at pH 1, the concentration of \( H^+ \) ions is:

\[ [H^+] = 10^{-1} = 0.1 \text{ M} \]

### Step 5: Calculate the half-cell potential using the Nernst equation
The Nernst equation is given by:

\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]

Where:
- \( E^0_{cell} = 1.33 \text{ V} \)
- \( n = 6 \) (number of electrons transferred)
- \( Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} \)

First, we need to find \( [Cr_2O_7^{2-}] \):

Since 3 moles of \( Cr_2O_7^{2-} \) were present initially and none was consumed (as the reaction is with \( SnCl_2 \)), the concentration remains:

\[ [Cr_2O_7^{2-}] = \frac{3 \text{ moles}}{3 \text{ L}} = 1 \text{ M} \]

Now we can calculate \( Q \):

\[ Q = \frac{(2)^2}{(1)(0.1)^{14}} = \frac{4}{(0.1)^{14}} = 4 \times 10^{14} \]

Now substituting the values into the Nernst equation:

\[ E_{cell} = 1.33 - 0.06 \cdot \log(4 \times 10^{14}) \]

Calculating \( \log(4 \times 10^{14}) \):

\[ \log(4 \times 10^{14}) = \log(4) + \log(10^{14}) = 0.602 + 14 = 14.602 \]

Now substituting back:

\[ E_{cell} = 1.33 - 0.06 \cdot 14.602 \]
\[ E_{cell} = 1.33 - 0.87612 \]
\[ E_{cell} \approx 1.19188 \text{ V} \approx 1.19 \text{ V} \]

### Final Answer
The half-cell potential \( E_{Cr_2O_7^{2-}/Cr^{3+}} \) after the reaction of \( SnCl_2 \) is approximately **1.19 V**.

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