An electrochemical cell has two half cell reaction as, `A^(2+)+2e^-toA,E^0=0.34V`
`XtoX^(2+)+2e^-,E^0=+2.37V`. The cell voltage will be

Options

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

Answer

To solve the problem, we need to determine the cell voltage of the electrochemical cell given the half-cell reactions and their standard reduction potentials.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials:**
- The first half-reaction is:
\[ A^{2+} + 2e^- \rightarrow A, \quad E^\circ = +0.34 \, \text{V} \]
- The second half-reaction is:
\[ X \rightarrow X^{2+} + 2e^-, \quad E^\circ = +2.37 \, \text{V} \]

2. **Determine the Type of Reactions:**
- The first reaction (for A) is a reduction reaction because \( A^{2+} \) is gaining electrons.
- The second reaction (for X) is an oxidation reaction because \( X \) is losing electrons.

3. **Convert the Oxidation Potential to Reduction Potential:**
- The reduction potential for the oxidation of X can be calculated as:
\[ E^\circ_{\text{reduction}}(X) = -E^\circ_{\text{oxidation}}(X) = -2.37 \, \text{V} \]

4. **Identify the Cathode and Anode:**
- The cathode is where reduction occurs. Since \( A^{2+} \) has a higher reduction potential (+0.34 V) compared to \( X \) (-2.37 V), \( A^{2+} \) will undergo reduction at the cathode.
- The anode is where oxidation occurs. Thus, \( X \) will undergo oxidation at the anode.

5. **Calculate the Standard Cell Potential (E°cell):**
- The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}}(A) - E^\circ_{\text{anode}}(X) = 0.34 \, \text{V} - (-2.37 \, \text{V}) \]
- This simplifies to:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} + 2.37 \, \text{V} = 2.71 \, \text{V} \]

6. **Final Answer:**
- The cell voltage (E°cell) is:
\[ E^\circ_{\text{cell}} = +2.71 \, \text{V} \]