Given the cell reactions
`MX_((s))+e^(-)rarrM_((s))+X_((aq))^(-),E^(@)=0.207V`
and `M_((aq))^(+)+e^(-)rarrM_((s)),E^(@)=0.799V`
The solubility of `MX_((s))` at 298K is

Options

A. `1.0xx10L^(-1)`
B. `1.0x10^(-9)"mole L"^(-1)`
C. `1.0xx10^(-4)"mole L"^(-1)`
D. `1.0xx10^(-5)"mole L"^(-1)`

Explanation

To find the solubility of \( MX_{(s)} \) at 298 K, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials (E°)
We are given two half-reactions:
1. \( MX_{(s)} + e^- \rightarrow M_{(s)} + X_{(aq)}^- \), with \( E° = 0.207 \, V \)
2. \( M_{(aq)}^+ + e^- \rightarrow M_{(s)} \), with \( E° = 0.799 \, V \)

### Step 2: Determine the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
In our case, the cathode is the reduction of \( M_{(aq)}^+ \) and the anode is the oxidation of \( MX_{(s)} \):
\[ E°_{cell} = E°_{M_{(aq)}^+ / M_{(s)}} - E°_{MX_{(s)} / M_{(aq)}^-} \]
Substituting the values:
\[ E°_{cell} = 0.799 \, V - 0.207 \, V = 0.592 \, V \]

### Step 3: Relate E°cell to Ksp using the Nernst equation
The relationship between the standard cell potential and the solubility product (Ksp) is given by:
\[ E°_{cell} = \frac{-RT}{nF} \ln K_{sp} \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \)
- \( n = 1 \) (number of electrons transferred)
- \( F = 96485 \, C/mol \)

### Step 4: Calculate Ksp
Rearranging the equation gives:
\[ \ln K_{sp} = -\frac{nFE°_{cell}}{RT} \]
Substituting the known values:
\[ \ln K_{sp} = -\frac{(1)(96485)(0.592)}{(8.314)(298)} \]
Calculating the right side:
\[ \ln K_{sp} \approx -19.2 \]
Now, exponentiating to find \( K_{sp} \):
\[ K_{sp} = e^{-19.2} \approx 1.03 \times 10^{-9} \]

### Step 5: Determine the solubility from Ksp
For the dissolution of \( MX_{(s)} \):
\[ MX_{(s)} \rightleftharpoons M_{(aq)}^+ + X_{(aq)}^- \]
Let the solubility of \( MX_{(s)} \) be \( s \). Then:
\[ K_{sp} = [M_{(aq)}^+][X_{(aq)}^-] = s \cdot s = s^2 \]
Thus,
\[ s^2 = K_{sp} \implies s = \sqrt{K_{sp}} = \sqrt{1.03 \times 10^{-9}} \approx 1.015 \times 10^{-5} \, mol/L \]

### Final Answer
The solubility of \( MX_{(s)} \) at 298 K is approximately \( 1.015 \times 10^{-5} \, mol/L \).

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