The electrochemical cell shown below is a concentrstion cell.
`M|M^(2+)` (saturated solution of a sparingly soluble salt,`Mx_2`)||`M^(2+)(0.001 mol dm^3)|M` The emf of the cell depends on the difference in concentrations of `M^(2+)` ions at the two electrods. The emf of the cell at 298K is 0.059V.
The value of `/_\G(kJ mol^-1)` for the given cell is (take 1F = `96500Cmol^-1`)

Options

A. `-5.7`
B. 5.7
C. 11.4
D. `-11.4`

Answer

To solve the problem, we need to calculate the Gibbs free energy change (ΔG) for the given electrochemical concentration cell. The formula we will use is:

\[ \Delta G = -nFE \]

where:
- \( n \) = number of moles of electrons transferred in the reaction,
- \( F \) = Faraday's constant (96500 C/mol),
- \( E \) = emf of the cell (0.059 V).

### Step 1: Determine the number of moles of electrons (n)
In the given concentration cell, the half-reaction can be represented as:

\[ M^{2+} + 2e^- \rightarrow M \]

From this half-reaction, we can see that 2 moles of electrons are transferred. Therefore,

\[ n = 2 \]

### Step 2: Use the given values
We have:
- \( n = 2 \)
- \( F = 96500 \, \text{C/mol} \)
- \( E = 0.059 \, \text{V} \)

### Step 3: Substitute the values into the Gibbs free energy formula
Now we can substitute the values into the formula:

\[ \Delta G = -nFE = -(2)(96500 \, \text{C/mol})(0.059 \, \text{V}) \]

### Step 4: Calculate ΔG
Calculating this gives:

\[ \Delta G = -2 \times 96500 \times 0.059 \]

Calculating the multiplication:

\[ \Delta G = -2 \times 96500 \times 0.059 = -11387 \, \text{J/mol} \]

### Step 5: Convert to kJ/mol
To convert joules to kilojoules, we divide by 1000:

\[ \Delta G = -11.387 \, \text{kJ/mol} \approx -11.4 \, \text{kJ/mol} \]

### Final Answer
Thus, the value of \( \Delta G \) for the given cell is:

\[ \Delta G \approx -11.4 \, \text{kJ/mol} \]