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 value of \( E^\circ_{\text{cell}} \) for the given galvanic cell reaction, we can use the relationship between Gibbs free energy change (\( \Delta G^\circ \)) and the cell potential (\( E^\circ_{\text{cell}} \)). The equation is:

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

Where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change.
- \( n \) is the number of moles of electrons transferred in the reaction.
- \( F \) is the Faraday constant (\( 96500 \, \text{C/mol} \)).
- \( E^\circ_{\text{cell}} \) is the standard cell potential.

### Step 1: Convert \( \Delta G^\circ \) to Joules
Given \( \Delta G^\circ = -386 \, \text{kJ/mol} \), we need to convert this to Joules:

\[ \Delta G^\circ = -386 \times 1000 \, \text{J/mol} = -386000 \, \text{J/mol} \]

### Step 2: Determine the number of electrons transferred (\( n \))
From the balanced cell reaction:

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

We can see that 2 electrons are transferred (since \( \text{M}^{2+} \) gains 2 electrons to become \( \text{M} \)). Therefore, \( n = 2 \).

### Step 3: Substitute values into the equation
Now we can substitute the values into the equation:

\[ -386000 = -nFE^\circ_{\text{cell}} \]

Substituting \( n = 2 \) and \( F = 96500 \):

\[ -386000 = -2 \times 96500 \times E^\circ_{\text{cell}} \]

### Step 4: Solve for \( E^\circ_{\text{cell}} \)
Rearranging the equation gives:

\[ E^\circ_{\text{cell}} = \frac{386000}{2 \times 96500} \]

Calculating the denominator:

\[ 2 \times 96500 = 193000 \]

Now substituting back:

\[ E^\circ_{\text{cell}} = \frac{386000}{193000} \approx 2 \, \text{V} \]

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

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