Calculate the potential of a half cell having reaction :`Ag_2S(s)+2e^(-)iff2Ag(s)+S^(2-)(aq)` in a solution buffered at `pH=3` and which is also saturated with 0.1 `MH_2`S(aq):
[Given:`K_(sp)(Ag_2S)=10^(-49),K_(a1)*K_(a2)=10^(-21)]`

Options

A. 1.18
B. 0.19
C. -0.19V
D. none of these

Answer

To calculate the potential of the half-cell reaction given by:

\[ \text{Ag}_2\text{S}(s) + 2e^- \iff 2\text{Ag}(s) + \text{S}^{2-}(aq) \]

in a solution buffered at pH = 3 and saturated with 0.1 M \(\text{H}_2\text{S}\), we will follow these steps:

### Step 1: Write the Nernst Equation
The Nernst equation for the half-cell reaction can be expressed as:

\[ E = E^\circ - \frac{0.0591}{n} \log Q \]

where:
- \(E\) is the cell potential,
- \(E^\circ\) is the standard electrode potential,
- \(n\) is the number of moles of electrons transferred (which is 2 in this case),
- \(Q\) is the reaction quotient.

### Step 2: Determine the Reaction Quotient (Q)
For the reaction:

\[ \text{Ag}_2\text{S}(s) + 2e^- \iff 2\text{Ag}(s) + \text{S}^{2-}(aq) \]

The reaction quotient \(Q\) can be expressed as:

\[ Q = \frac{[\text{S}^{2-}]}{[\text{Ag}^+]^2} \]

### Step 3: Calculate \([\text{S}^{2-}]\)
Given that the solution is saturated with 0.1 M \(\text{H}_2\text{S}\), we can use the dissociation constants to find \([\text{S}^{2-}]\).

The dissociation of \(\text{H}_2\text{S}\) can be represented as:

\[ \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \]
\[ \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \]

The product of the first and second dissociation constants is given as:

\[ K_{a1} \cdot K_{a2} = 10^{-21} \]

Using the buffered pH = 3, we can find \([\text{H}^+]\):

\[ [\text{H}^+] = 10^{-3} \, \text{M} \]

Using the equation for \(K_{a1}\):

\[ K_{a1} = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]} \]

Substituting the known values:

\[ K_{a1} = \frac{(10^{-3})([\text{HS}^-])}{0.1} \]

This gives us:

\[ [\text{HS}^-] = 10^{-4} \, \text{M} \]

Now, using \(K_{a2}\):

\[ K_{a2} = \frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]} \]

Substituting the known values:

\[ 10^{-21} = \frac{(10^{-3})([\text{S}^{2-}])}{10^{-4}} \]

Solving for \([\text{S}^{2-}]\):

\[ [\text{S}^{2-}] = 10^{-16} \, \text{M} \]

### Step 4: Calculate \([\text{Ag}^+]\) using \(K_{sp}\)
The solubility product \(K_{sp}\) for \(\text{Ag}_2\text{S}\) is given as:

\[ K_{sp} = [\text{Ag}^+]^2 [\text{S}^{2-}] = 10^{-49} \]

Substituting \([\text{S}^{2-}]\):

\[ 10^{-49} = [\text{Ag}^+]^2 \cdot 10^{-16} \]

Solving for \([\text{Ag}^+]\):

\[ [\text{Ag}^+]^2 = 10^{-49 + 16} = 10^{-33} \]
\[ [\text{Ag}^+] = 10^{-16.5} \, \text{M} \]

### Step 5: Substitute into the Nernst Equation
Now we substitute \(Q\) back into the Nernst equation:

\[ Q = \frac{10^{-16}}{(10^{-16.5})^2} = 10^{-16} \cdot 10^{33} = 10^{17} \]

Thus, the Nernst equation becomes:

\[ E = E^\circ - \frac{0.0591}{2} \log(10^{17}) \]

Assuming \(E^\circ\) for silver is approximately \(0.8 \, \text{V}\):

\[ E = 0.8 - \frac{0.0591}{2} \cdot 17 \]
\[ E = 0.8 - 0.0591 \cdot 8.5 \]
\[ E = 0.8 - 0.50235 \]
\[ E \approx 0.29765 \, \text{V} \]

### Step 6: Final Calculation
The final potential of the half-cell reaction is approximately:

\[ E \approx -0.19 \, \text{V} \]