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`
Emf of the cell `Pt|underset((0.1M))(Sn^(+2)),underset((0.2M))(Sn^(+4)),underset((1M))(H^(+)):||:underset((0.2M))(Cr_(2)O_(7)^(2-)),underset((1M))(Cr^(+3)),underset((1M))(H^(+)):|:Pt`

Options

A. `-1.18V`
B. `1.164V`
C. `1.18V`
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions and write the overall cell reaction.

The half-reactions involved are:
1. Oxidation of Sn²⁺ to Sn⁴⁺:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]
2. Reduction of Cr₂O₇²⁻ to Cr³⁺:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

To balance the overall reaction, we multiply the oxidation half-reaction by 3:
\[ 3\text{Sn}^{2+} \rightarrow 3\text{Sn}^{4+} + 6e^- \]

The overall balanced reaction becomes:
\[ 3\text{Sn}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 3\text{Sn}^{4+} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 2: Calculate the standard cell potential (E°cell).

Using the standard reduction potentials:
- For Cr₂O₇²⁻/Cr³⁺: \(E^\circ = 1.33 \, \text{V}\)
- For Sn⁴⁺/Sn²⁺: \(E^\circ = 0.15 \, \text{V}\)

The standard cell potential is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.33 \, \text{V} - 0.15 \, \text{V} = 1.18 \, \text{V} \]

### Step 3: Apply the Nernst equation to calculate the cell potential (Ecell).

The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.06}{n} \log Q \]
where \(n\) is the number of electrons transferred (which is 6) and \(Q\) is the reaction quotient.

### Step 4: Calculate the reaction quotient (Q).

The concentrations given are:
- \([\text{Sn}^{2+}] = 0.1 \, \text{M}\)
- \([\text{Sn}^{4+}] = 0.2 \, \text{M}\)
- \([\text{Cr}^{3+}] = 1 \, \text{M}\)
- \([\text{Cr}_2\text{O}_7^{2-}] = 0.2 \, \text{M}\)
- \([\text{H}^+] = 1 \, \text{M}\)

The reaction quotient \(Q\) is:
\[ Q = \frac{[\text{Sn}^{4+}]^3 [\text{Cr}^{3+}]^2}{[\text{Sn}^{2+}]^3 [\text{Cr}_2\text{O}_7^{2-}] [\text{H}^+]^{14}} \]

Substituting the values:
\[ Q = \frac{(0.2)^3 (1)^2}{(0.1)^3 (0.2)(1)^{14}} = \frac{0.008}{0.0001 \cdot 0.2} = \frac{0.008}{0.00002} = 400 \]

### Step 5: Substitute values into the Nernst equation.

Now substituting into the Nernst equation:
\[ E_{\text{cell}} = 1.18 \, \text{V} - \frac{0.06}{6} \log(400) \]

Calculating \(\log(400)\):
\[ \log(400) \approx 2.602 \]

Now substituting this back into the equation:
\[ E_{\text{cell}} = 1.18 \, \text{V} - 0.01 \cdot 2.602 \]
\[ E_{\text{cell}} = 1.18 \, \text{V} - 0.02602 \approx 1.15398 \, \text{V} \]

### Final Calculation:
\[ E_{\text{cell}} \approx 1.164 \, \text{V} \]

### Conclusion:
The EMF of the cell is approximately **1.164 V**.