For the electrochemical cell,
`Mg(s)|Mg^(2+) (aq. 1M)||Cu^(2+) (aq. 1M)|Cu(s)`
the standard emf of the cell is 2.70 V at 300 K. When the concentration of `Mg^(2+)` is chaged to x M, the cell potential changes to 2.67 V at 300 K. The value of x is ________ .
(Given `F/R=11500 kV^(-1)`. where F is the Faraday constant and R is the gas constant, ln (10) = 2.30)

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write down the given data
- Standard EMF of the cell (E°) = 2.70 V
- Cell potential when concentration of Mg²⁺ is changed (E) = 2.67 V
- Temperature (T) = 300 K
- F/R = 11500 kV⁻¹
- ln(10) = 2.30

### Step 2: Write the half-reactions
For the electrochemical cell:
1. Oxidation half-reaction:
\[ \text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- \]
2. Reduction half-reaction:
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \]

### Step 3: Combine the half-reactions
The overall cell reaction can be written as:
\[ \text{Mg}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu}(s) \]

### Step 4: Write the Nernst equation
The Nernst equation for the cell can be expressed as:
\[ E = E° - \frac{RT}{nF} \ln \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \]

### Step 5: Simplify the equation
Since the concentration of Cu²⁺ is 1 M and Mg is a solid (concentration = 1), the equation simplifies to:
\[ E = E° - \frac{RT}{nF} \ln [\text{Mg}^{2+}] \]
Where:
- n = 2 (number of electrons transferred)

### Step 6: Substitute the known values
Substituting the known values into the Nernst equation:
\[ 2.67 = 2.70 - \frac{RT}{2F} \ln x \]
Where \( x \) is the concentration of Mg²⁺.

### Step 7: Rearrange the equation
Rearranging gives:
\[ 2.67 - 2.70 = -\frac{RT}{2F} \ln x \]
\[ -0.03 = -\frac{RT}{2F} \ln x \]
\[ 0.03 = \frac{RT}{2F} \ln x \]

### Step 8: Substitute R, T, and F/R
Using \( F/R = 11500 \):
\[ 0.03 = \frac{(8.314)(300)}{2F} \ln x \]
Now substituting \( F \) using \( F = R \times 11500 \):
\[ 0.03 = \frac{(8.314)(300)}{2 \times (8.314 \times 11500)} \ln x \]

### Step 9: Simplify the equation
This simplifies to:
\[ 0.03 = \frac{300}{2 \times 11500} \ln x \]
\[ 0.03 = \frac{300}{23000} \ln x \]
\[ 0.03 = \frac{3}{230} \ln x \]

### Step 10: Solve for ln x
Now, multiplying both sides by 230:
\[ 230 \times 0.03 = 3 \ln x \]
\[ 6.9 = 3 \ln x \]
\[ \ln x = \frac{6.9}{3} = 2.3 \]

### Step 11: Find x
Now, exponentiating both sides to solve for \( x \):
\[ x = e^{2.3} \]
Using the fact that \( e^{\ln 10} = 10 \):
\[ x = 10 \]

### Final Answer
Thus, the value of \( x \) is 10.

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