Calculate the emf of the following cell at 298K:
`Mg(s)|Mg^(2+)(0.1M)||Cu^(2+)(1.0xx10^(-3)M)|Cu(s)` [Given=`E_(Cell)^(@)=2.71V`]

Options

A. 1.426 V
B. 2.503V
C. 2.651 V
D. 1.8V

Explanation

To calculate the EMF of the given galvanic cell at 298 K, we will follow these steps:

### Step 1: Write the overall cell reaction
The cell reaction can be derived from the oxidation and reduction half-reactions.

- **Oxidation half-reaction**:
\[ \text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- \]

- **Reduction half-reaction**:
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \]

Combining these half-reactions gives us the overall cell reaction:
\[ \text{Mg}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu}(s) \]

### Step 2: Identify the standard cell potential (E°cell)
From the problem statement, we are given:
\[ E°_{cell} = 2.71 \, \text{V} \]

### Step 3: Determine the concentrations of the ions
The concentrations provided in the problem are:
- \([\text{Mg}^{2+}] = 0.1 \, \text{M}\)
- \([\text{Cu}^{2+}] = 1.0 \times 10^{-3} \, \text{M}\)

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E°_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
where \(n\) is the number of moles of electrons transferred in the reaction. Here, \(n = 2\).

### Step 5: Substitute values into the Nernst equation
Substituting the known values into the Nernst equation:
\[ E_{cell} = 2.71 \, \text{V} - \frac{0.0591}{2} \log \left( \frac{0.1}{1.0 \times 10^{-3}} \right) \]

### Step 6: Calculate the logarithm
Calculating the logarithm:
\[ \log \left( \frac{0.1}{1.0 \times 10^{-3}} \right) = \log(100) = 2 \]

### Step 7: Substitute the logarithm back into the equation
Now substituting back:
\[ E_{cell} = 2.71 \, \text{V} - \frac{0.0591}{2} \times 2 \]
\[ E_{cell} = 2.71 \, \text{V} - 0.0591 \]
\[ E_{cell} = 2.71 \, \text{V} - 0.0591 = 2.6509 \, \text{V} \]

### Step 8: Round the result
Rounding the result gives:
\[ E_{cell} \approx 2.651 \, \text{V} \]

### Final Answer
Thus, the EMF of the cell at 298 K is:
\[ \boxed{2.651 \, \text{V}} \]