Consider two half cells based on the reaction `Ag_((aq))^(+)etoAg_((s))`.t he left half cell contain `Ag^(+)` ions at concentration of `Ag^(+)` ions, but just enough `NaCl_((aq))` has been added to completely precipitate the `Ag_((aq))^(+)` as `AgCl`. If the emf of the cell is 0.29V, then `log_(10)K_(sp)` would have been

Options

A. 9.804
B. `-9.804`
C. `-4.902`
D. `10.004`

Explanation

To solve the problem, we need to find the value of \( \log_{10} K_{sp} \) given the emf of the cell and the reaction involving silver ions. Let's break this down step by step.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The half-reaction for silver is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag (s)} \]
This reaction occurs at both the anode and cathode since they involve the same species.

2. **Understand the Cell Setup**:
In this cell, the left half-cell contains \( \text{Ag}^+ \) ions that are completely precipitated as \( \text{AgCl} \). Therefore, the concentration of \( \text{Ag}^+ \) ions is effectively zero in the presence of solid \( \text{AgCl} \).

3. **Use the Nernst Equation**:
The Nernst equation relates the cell potential (emf) to the standard potential and the concentrations of the reactants/products:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log_{10} Q \]
Here, \( E^0_{cell} \) is the standard cell potential, \( n \) is the number of electrons transferred, and \( Q \) is the reaction quotient.

4. **Determine \( E^0_{cell} \)**:
Since both half-reactions involve the same species (silver), the standard potential \( E^0_{cell} \) is zero:
\[ E^0_{cell} = 0 \, \text{V} \]

5. **Calculate the Reaction Quotient \( Q \)**:
For the precipitation of \( \text{AgCl} \), the solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Since \( \text{Ag}^+ \) is precipitated, we can assume \( [\text{Ag}^+] = 0 \) and \( [\text{Cl}^-] \) is constant. However, for the purpose of calculating \( \log_{10} K_{sp} \), we can set \( Q = K_{sp} \).

6. **Substituting into the Nernst Equation**:
Substitute the known values into the Nernst equation:
\[ 0.29 = 0 - \frac{0.0591}{1} \log_{10} K_{sp} \]
Here, \( n = 1 \) because one electron is involved in the half-reaction.

7. **Rearranging the Equation**:
Rearranging gives:
\[ \log_{10} K_{sp} = -\frac{0.29}{0.0591} \]

8. **Calculating \( \log_{10} K_{sp} \)**:
Now, calculate the value:
\[ \log_{10} K_{sp} = -4.907 \]

9. **Final Result**:
Therefore, the value of \( \log_{10} K_{sp} \) is approximately:
\[ \log_{10} K_{sp} \approx -4.907 \]

### Summary:
The value of \( \log_{10} K_{sp} \) is approximately -4.907.