The efficiency of a hypothetical cell is about `84%` which involves the following reactions:
`A(s) + B^(2+)(aq) rightarrow A^(2+)(aq) _ B(s)` `DeltaH = -285kJ`
Then, the standard electrode potential of the cell will be: (Asume `DeltaS = 0)

Options

A. 1.2
B. 2.40 V
C. 1.10 V
D. 1.24 V

Answer

To find the standard electrode potential of the cell given the efficiency and the reaction, we will follow these steps:

### Step 1: Identify the given data
- The reaction is:
\[ A(s) + B^{2+}(aq) \rightarrow A^{2+}(aq) + B(s) \]
- The enthalpy change (\(\Delta H\)) is given as \(-285 \, \text{kJ}\).
- The efficiency of the cell is \(84\%\).
- We assume \(\Delta S = 0\).

### Step 2: Calculate \(\Delta G\)
Since \(\Delta S = 0\), we can simplify the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
This simplifies to:
\[ \Delta G = \Delta H \]
Thus,
\[ \Delta G = -285 \, \text{kJ} \]

### Step 3: Relate \(\Delta G\) to the electrode potential
The relationship between Gibbs free energy change and electrode potential is given by:
\[ \Delta G = -nFE \]
where:
- \(n\) = number of moles of electrons transferred (in this case, \(n = 2\))
- \(F\) = Faraday's constant (\(96485 \, \text{C/mol}\))
- \(E\) = standard electrode potential

### Step 4: Substitute \(\Delta G\) into the equation
Substituting the values we have:
\[ -285 \times 10^3 \, \text{J} = -2 \times 96485 \, \text{C/mol} \times E \]
The negative signs cancel out:
\[ 285 \times 10^3 = 2 \times 96485 \times E \]

### Step 5: Solve for \(E\)
Rearranging the equation to solve for \(E\):
\[ E = \frac{285 \times 10^3}{2 \times 96485} \]
Calculating this gives:
\[ E = \frac{285000}{192970} \approx 1.477 \, \text{V} \]

### Step 6: Adjust for efficiency
Since the efficiency of the cell is \(84\%\), we adjust the standard electrode potential:
\[ E_{\text{actual}} = E \times \text{Efficiency} \]
\[ E_{\text{actual}} = 1.477 \, \text{V} \times 0.84 \]
Calculating this gives:
\[ E_{\text{actual}} \approx 1.24 \, \text{V} \]

### Final Answer
The standard electrode potential of the cell is approximately \(1.24 \, \text{V}\).

---