Consider the cell reaction `:`
`Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq)`
If `E^(c-)._(Mg^(2+)|Mg(s))` and `E^(c-)._(Cu^(2+)|Cu(s))` are `-2.37` and `0.34V`, respectively. `E^(c-)._(cell)` is

Options

A. `2.03V`
B. `-2.03V`
C. `-2.17 V`
D. `2.71 V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials
The cell reaction is:
\[ \text{Mg(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Cu(s)} + \text{Mg}^{2+}(aq) \]

From the problem, we have the following standard reduction potentials:
- For the reduction of magnesium:
\[ \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg(s)} \quad E^\circ = -2.37 \, \text{V} \]
- For the reduction of copper:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \quad E^\circ = 0.34 \, \text{V} \]

### Step 2: Determine the anode and cathode
In this reaction:
- Magnesium (Mg) is oxidized to magnesium ions (Mg²⁺), so it acts as the anode.
- Copper ions (Cu²⁺) are reduced to copper metal (Cu), so it acts as the cathode.

### Step 3: Write the formula for the cell potential
The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

### Step 4: Substitute the values into the formula
Here, the cathode is Cu²⁺ and the anode is Mg:
- \( E^\circ_{\text{cathode}} = 0.34 \, \text{V} \)
- \( E^\circ_{\text{anode}} = -2.37 \, \text{V} \)

Now substituting these values:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-2.37 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} + 2.37 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 2.71 \, \text{V} \]

### Step 5: Conclusion
The standard cell potential \( E^\circ_{\text{cell}} \) is:
\[ \boxed{2.71 \, \text{V}} \]

---