For galvanic cell
`M^(2+)(aq) + Zn(s) rarr M(s) +Zn^(2+)(aq)` `triangleG^@` = -386KJ/mole
the value of `E^@_(cell)` is

Answer

To find the standard cell potential \( E^\circ_{\text{cell}} \) for the given galvanic cell reaction:

\[ M^{2+}(aq) + Zn(s) \rightarrow M(s) + Zn^{2+}(aq) \]

with the given standard Gibbs free energy change \( \Delta G^\circ = -386 \, \text{kJ/mol} \), we can use the relationship between Gibbs free energy and cell potential:

\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

where:
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( F \) is Faraday's constant (\( F = 96500 \, \text{C/mol} \)),
- \( E^\circ_{\text{cell}} \) is the standard cell potential.

### Step-by-Step Solution:

1. **Identify the number of electrons transferred (\( n \))**:
In the given reaction, zinc (Zn) is oxidized from \( Zn \) to \( Zn^{2+} \), losing 2 electrons. The metal \( M^{2+} \) is reduced to \( M \) by gaining these 2 electrons. Therefore, \( n = 2 \).

2. **Convert \( \Delta G^\circ \) from kJ to J**:
Since Faraday's constant is in coulombs per mole, we need to convert \( \Delta G^\circ \) to joules:
\[ \Delta G^\circ = -386 \, \text{kJ/mol} = -386 \times 1000 \, \text{J/mol} = -386000 \, \text{J/mol} \]

3. **Substitute the known values into the equation**:
Now, we can substitute \( \Delta G^\circ \), \( n \), and \( F \) into the equation:
\[ -386000 \, \text{J/mol} = -2 \times 96500 \, \text{C/mol} \times E^\circ_{\text{cell}} \]

4. **Rearrange the equation to solve for \( E^\circ_{\text{cell}} \)**:
\[ E^\circ_{\text{cell}} = \frac{-386000 \, \text{J/mol}}{-2 \times 96500 \, \text{C/mol}} \]

5. **Calculate \( E^\circ_{\text{cell}} \)**:
\[ E^\circ_{\text{cell}} = \frac{386000}{193000} = 2 \, \text{V} \]

### Final Answer:
The value of \( E^\circ_{\text{cell}} \) is \( 2 \, \text{V} \).