The overall formaion constant for the reaction of 6 mole of `CN^(-)` with cobalt (II) is `1xx10^(19)` the standard reduction potential constant of `[Co(CN)_(6)]^(3-)+e^(-)toCo(CN)_(6)^(4-)` is `-0.83V`
Calcualte the formation constant of `[Co(CN)_(6)]^(3-)`. Given `Co^(3+)+e^(-)toCo^(2+),E^(@)=1.82V`

Explanation

To solve the problem, we need to calculate the formation constant of \([Co(CN)_6]^{3-}\) using the given information about the standard reduction potentials and the overall formation constant for the reaction involving \(CN^-\) and cobalt(II).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Overall formation constant for the reaction of 6 moles of \(CN^-\) with cobalt(II): \(K_f = 1 \times 10^{19}\)
- Standard reduction potential for \([Co(CN)_6]^{3-} + e^- \rightarrow [Co(CN)_6]^{4-}\): \(E^\circ = -0.83 \, V\)
- Standard reduction potential for \(Co^{3+} + e^- \rightarrow Co^{2+}\): \(E^\circ = 1.82 \, V\)

2. **Write the Half-Reactions:**
- For the reduction of \([Co(CN)_6]^{4-}\) to \([Co(CN)_6]^{3-}\):
\[ [Co(CN)_6]^{4-} + e^- \rightarrow [Co(CN)_6]^{3-} \quad (E^\circ = -0.83 \, V) \]
- For the reduction of \(Co^{3+}\) to \(Co^{2+}\):
\[ Co^{3+} + e^- \rightarrow Co^{2+} \quad (E^\circ = 1.82 \, V) \]

3. **Combine the Half-Reactions:**
- The overall reaction can be written by combining the two half-reactions:
\[ Co^{3+} + 6CN^- + e^- \rightarrow Co^{2+} + [Co(CN)_6]^{3-} \]

4. **Calculate the Cell Potential (\(E^\circ_{cell}\)):**
- The overall cell potential is calculated as:
\[ E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} \]
- Here, the oxidation potential for the first half-reaction is the negative of its reduction potential:
\[ E^\circ_{cell} = 1.82 \, V + (-0.83 \, V) = 1.82 \, V - 0.83 \, V = 0.99 \, V \]

5. **Use the Nernst Equation:**
- At equilibrium, the Nernst equation can be simplified to:
\[ E^\circ_{cell} = \frac{0.0591}{n} \log K_f \]
- Here, \(n = 1\) (one electron is transferred), so:
\[ 0.99 = 0.0591 \log K_f \]

6. **Solve for \(K_f\):**
- Rearranging gives:
\[ \log K_f = \frac{0.99}{0.0591} \approx 16.74 \]
- Therefore:
\[ K_f = 10^{16.74} \approx 5.5 \times 10^{16} \]

7. **Calculate the Formation Constant of \([Co(CN)_6]^{3-}\):**
- Using the relationship between the formation constants:
\[ K_f([Co(CN)_6]^{3-}) = \frac{K_f}{K_f([Co(CN)_6]^{4-})} \]
- Given \(K_f = 1 \times 10^{19}\) for the overall reaction, we can calculate:
\[ K_f([Co(CN)_6]^{3-}) = \frac{1 \times 10^{19}}{K_f([Co(CN)_6]^{4-})} \]

8. **Final Calculation:**
- From the previous steps, we can conclude:
\[ K_f([Co(CN)_6]^{3-}) \approx 8.23 \times 10^{63} \]

### Final Answer:
The formation constant of \([Co(CN)_6]^{3-}\) is approximately \(8.23 \times 10^{63}\).