A cell constituted by two electrodes A
`(E^(@)A//A+0.35V)` and `B(E^(@)B//B=-0.42V)`
Has value of `E_("Cell")^(@)` equal to

Options

A. 0.07V
B. 0.77V
C. `0.77V`
D. `-0.07V`.

Explanation

To find the standard cell potential \( E_{\text{Cell}}^{\circ} \) for the given electrochemical cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Oxidation Potentials:**
- Electrode A has an oxidation potential of \( E^{\circ}_A = +0.35 \, \text{V} \).
- Electrode B has an oxidation potential of \( E^{\circ}_B = -0.42 \, \text{V} \).

2. **Convert Oxidation Potentials to Reduction Potentials:**
- The reduction potential for electrode A is the negative of its oxidation potential:
\[ E^{\circ}_{\text{red}, A} = -E^{\circ}_A = -0.35 \, \text{V} \]
- The reduction potential for electrode B is the negative of its oxidation potential:
\[ E^{\circ}_{\text{red}, B} = -E^{\circ}_B = +0.42 \, \text{V} \]

3. **Determine the Anode and Cathode:**
- The anode is where oxidation occurs, and the cathode is where reduction occurs.
- Since B has a higher reduction potential than A, B will act as the cathode (where reduction occurs), and A will act as the anode (where oxidation occurs).

4. **Calculate the Standard Cell Potential:**
- The standard cell potential \( E_{\text{Cell}}^{\circ} \) is given by the formula:
\[ E_{\text{Cell}}^{\circ} = E^{\circ}_{\text{red}, \text{cathode}} - E^{\circ}_{\text{red}, \text{anode}} \]
- Substituting the values:
\[ E_{\text{Cell}}^{\circ} = E^{\circ}_{\text{red}, B} - E^{\circ}_{\text{red}, A} = 0.42 \, \text{V} - (-0.35 \, \text{V}) \]
- This simplifies to:
\[ E_{\text{Cell}}^{\circ} = 0.42 \, \text{V} + 0.35 \, \text{V} = 0.77 \, \text{V} \]

5. **Final Answer:**
- Therefore, the standard cell potential \( E_{\text{Cell}}^{\circ} \) is \( 0.77 \, \text{V} \).