Calculate the equilibrium constant for the reaction:
`3Sn(s) +2Cr_(2)O_(7)^(2-) +28 H^(+) rarr 3Sn^(+4) +4Cr^(3+) +14H_(2)O`
`E^(@)` for `Sn//Sn^(+2) = 0.136 V E^(@)` for4 `Sn^(2+)//Sn^(4+) =- 0.154V`
`E^(@)` for `Cr_(2)O_(7)^(2-)//Cr^(3+) = 1.33V`

Explanation

To calculate the equilibrium constant for the reaction:

\[ 3 \text{Sn}(s) + 2 \text{Cr}_2\text{O}_7^{2-} + 28 \text{H}^+ \rightarrow 3 \text{Sn}^{+4} + 4 \text{Cr}^{3+} + 14 \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.

1. **Oxidation of Sn:**
- \( \text{Sn}(s) \rightarrow \text{Sn}^{2+} + 2e^- \) (E° = 0.136 V)
- \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \) (E° = -0.154 V)

2. **Reduction of \( \text{Cr}_2\text{O}_7^{2-} \):**
- \( \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \) (E° = 1.33 V)

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

To find the overall standard cell potential, we first need to combine the oxidation and reduction half-reactions.

#### Calculate E° for the oxidation of Sn to Sn4+:

- The overall oxidation of Sn involves 4 electrons:
\[ 3 \text{Sn}(s) \rightarrow 3 \text{Sn}^{4+} + 12e^- \]

- The combined standard reduction potential (E°) for Sn is calculated using the two half-reactions:
\[ E_3 = \frac{n_1 E_1 + n_2 E_2}{n_1 + n_2} \]
where \( n_1 = 2 \) (for Sn to Sn2+) and \( n_2 = 2 \) (for Sn2+ to Sn4+).

Plugging in the values:
\[ E_3 = \frac{2(0.136) + 2(-0.154)}{2 + 2} = \frac{0.272 - 0.308}{4} = \frac{-0.036}{4} = -0.009 \, \text{V} \]

#### Calculate E°cell:

- The overall cell potential is given by:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
Here, the cathode is the reduction of \( \text{Cr}_2\text{O}_7^{2-} \) (E° = 1.33 V) and the anode is the oxidation of Sn (E° = -0.009 V).

Therefore:
\[ E°_{cell} = 1.33 - (-0.009) = 1.33 + 0.009 = 1.339 \, \text{V} \]

### Step 3: Calculate the equilibrium constant (K).

Using the Nernst equation at equilibrium:
\[ E°_{cell} = -\frac{0.0591}{n} \log K \]
where \( n \) is the number of moles of electrons transferred in the balanced equation. Here, \( n = 12 \) (6 electrons per \( \text{Cr}_2\text{O}_7^{2-} \) and 2 moles of \( \text{Cr}_2\text{O}_7^{2-} \)).

Substituting the values:
\[ 1.339 = -\frac{0.0591}{12} \log K \]

Rearranging gives:
\[ \log K = -\frac{1.339 \times 12}{0.0591} \]
Calculating:
\[ \log K = -\frac{16.068}{0.0591} \approx -271.5 \]

Thus,
\[ K \approx 10^{-271.5} \approx 10^{268} \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is approximately \( 10^{268} \).

---