Calculate the cell potential of a cell having reaction `Ag_(2)S +2e^(-) hArr 2Ag +S^(2-)` in a solution buffered at `pH =3` and which is also saturated with `0.1M H_(2)S.`
For `H_(2)S: K_(1) = 10^(-8)` and `K_(2) = 1.1 xx 10^(-13), K_(sp) (Ag_(2)S) = 2xx 10^(-49), E_(Ag^(+)//Ag)^(@) =0.8`

Explanation

To calculate the cell potential of the given reaction, we will follow these steps:

### Step 1: Write the half-reaction and determine standard reduction potential
The half-reaction given is:
\[ \text{Ag}_2\text{S} + 2e^- \rightleftharpoons 2\text{Ag} + \text{S}^{2-} \]

Let \( E^\circ \) for this reaction be \( x \) volts. We also know the standard reduction potential for the silver electrode:
\[ \text{Ag}^+ + e^- \rightleftharpoons \text{Ag} \quad E^\circ = 0.8 \, \text{V} \]

The oxidation reaction for silver is:
\[ 2\text{Ag} \rightleftharpoons 2\text{Ag}^+ + 2e^- \quad E^\circ = -0.8 \, \text{V} \]

### Step 2: Combine the half-reactions
Combining the half-reactions gives:
\[ \text{Ag}_2\text{S} + 2e^- \rightleftharpoons 2\text{Ag}^+ + \text{S}^{2-} \]

The overall standard potential \( E^\circ \) for the reaction is:
\[ E^\circ = x - 0.8 \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]

For our reaction, \( n = 2 \), and the products are \( [\text{Ag}^+]^2 \) and \( [\text{S}^{2-}] \). Thus, we can write:
\[ E = x - 0.8 - \frac{0.0591}{2} \log \left( [\text{Ag}^+]^2 [\text{S}^{2-}] \right) \]

### Step 4: Calculate the concentration of \( \text{S}^{2-} \)
Given that the solution is saturated with \( 0.1 \, \text{M} \, \text{H}_2\text{S} \) and buffered at \( \text{pH} = 3 \):
- The concentration of \( \text{H}^+ \) is \( 10^{-3} \, \text{M} \).
- We can find the concentration of \( \text{S}^{2-} \) using the ionization constants \( K_a1 \) and \( K_a2 \).

The overall ionization constant for \( \text{H}_2\text{S} \) can be calculated as:
\[ K = K_{a1} \times K_{a2} = (10^{-8}) \times (1.1 \times 10^{-13}) = 1.1 \times 10^{-21} \]

Using the equilibrium expression:
\[ K = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]} \]
Substituting the known values:
\[ 1.1 \times 10^{-21} = \frac{(10^{-3})^2 [\text{S}^{2-}]}{0.1} \]

Solving for \( [\text{S}^{2-}] \):
\[ [\text{S}^{2-}] = \frac{1.1 \times 10^{-21} \times 0.1}{10^{-6}} = 1.1 \times 10^{-16} \, \text{M} \]

### Step 5: Substitute concentrations into the Nernst equation
Now substituting \( [\text{S}^{2-}] \) into the Nernst equation:
\[ E = x - 0.8 - \frac{0.0591}{2} \log \left( [\text{Ag}^+]^2 \times 1.1 \times 10^{-16} \right) \]

At equilibrium, we can use the solubility product \( K_{sp} \) for \( \text{Ag}_2\text{S} \):
\[ K_{sp} = [\text{Ag}^+]^2 [\text{S}^{2-}] = 2 \times 10^{-49} \]

Thus,
\[ [\text{Ag}^+]^2 = \frac{K_{sp}}{[\text{S}^{2-}]} = \frac{2 \times 10^{-49}}{1.1 \times 10^{-16}} \approx 1.818 \times 10^{-33} \]

Now substituting back into the Nernst equation:
\[ E = x - 0.8 - \frac{0.0591}{2} \log(1.818 \times 10^{-33} \times 1.1 \times 10^{-16}) \]

### Step 6: Calculate the final cell potential
Calculating the logarithm:
\[ \log(1.818 \times 10^{-33} \times 1.1 \times 10^{-16}) \approx \log(2 \times 10^{-49}) \approx -49 \]

Thus,
\[ E = x - 0.8 - \frac{0.0591}{2} \times (-49) \]
\[ E = x - 0.8 + 1.45 \]

Substituting \( x = -0.639 \):
\[ E = -0.639 - 0.8 + 1.45 \]
\[ E \approx -0.167 \, \text{V} \]

### Final Answer
The cell potential \( E \) for the given reaction is approximately:
\[ E \approx -0.167 \, \text{V} \]