For a cell reaction `DeltaG^(@)=-1930(kJ)/(mol)` and the cell e.m.f `(E^(@))` at standard condition is 2.5 V. What is value of `(DeltaG^(@))/(2FE^(@))`?

Answer

To solve the problem, we need to find the value of \(\frac{\Delta G^\circ}{2FE^\circ}\) given:

- \(\Delta G^\circ = -1930 \, \text{kJ/mol}\)
- \(E^\circ = 2.5 \, \text{V}\)
- \(F = 96500 \, \text{C/mol}\) (Faraday's constant)

### Step 1: Convert \(\Delta G^\circ\) from kJ to J
Since \(\Delta G^\circ\) is given in kilojoules, we need to convert it to joules for consistency with the other units.

\[ \Delta G^\circ = -1930 \, \text{kJ/mol} = -1930 \times 10^3 \, \text{J/mol} = -1930000 \, \text{J/mol} \]

### Step 2: Calculate \(2FE^\circ\)
Next, we need to calculate \(2FE^\circ\).

\[ 2F = 2 \times 96500 \, \text{C/mol} = 193000 \, \text{C/mol} \]

Now, multiply this by \(E^\circ\):

\[ 2FE^\circ = 193000 \, \text{C/mol} \times 2.5 \, \text{V} = 482500 \, \text{J/mol} \]

### Step 3: Calculate \(\frac{\Delta G^\circ}{2FE^\circ}\)
Now we can substitute the values we have calculated into the formula:

\[ \frac{\Delta G^\circ}{2FE^\circ} = \frac{-1930000 \, \text{J/mol}}{482500 \, \text{J/mol}} \]

### Step 4: Perform the division
Now, we perform the division:

\[ \frac{\Delta G^\circ}{2FE^\circ} = -4 \]

Thus, the final answer is:

\[ \frac{\Delta G^\circ}{2FE^\circ} = -4 \]

### Summary of the Solution
The value of \(\frac{\Delta G^\circ}{2FE^\circ}\) is \(-4\).

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