Using the `DeltaG^(@)` for the reactions
`{:(C+O_(2)rarrCO_(2),DeltaG^(@)=-395kJ//"mole"),(2Al(l)+3//2O_(2)rarrAl_(2)O_(3)(s),DeltaG^(@)=-1269kJ //"mole"),(Al_(2)O_(3)(s)rarrAl_(2)O_(3)("melt"),DeltaG^(@)=16kJ //"mole"):}`
What is the EMF for the cell reaction
`2Al_(2)O_(3)("melt")+3Crarr4Al(l)+3CO_(2)(g)`
The number of electrons involved in the reaction is 12.

Options

A. `-1.14 `volt
B. `-2.28` volt
C. `+1.14` volt
D. `+2.28`volt

Explanation

To find the EMF for the cell reaction given, we will follow these steps:

### Step 1: Write the reactions and their ΔG values
We have the following reactions with their respective ΔG values:
1. \( C + O_2 \rightarrow CO_2, \Delta G^\circ = -395 \, \text{kJ/mol} \)
2. \( 2Al(l) + \frac{3}{2}O_2 \rightarrow Al_2O_3(s), \Delta G^\circ = -1269 \, \text{kJ/mol} \)
3. \( Al_2O_3(s) \rightarrow Al_2O_3(\text{melt}), \Delta G^\circ = 16 \, \text{kJ/mol} \)

### Step 2: Write the target reaction
The target reaction is:
\[ 2Al_2O_3(\text{melt}) + 3C \rightarrow 4Al(l) + 3CO_2(g) \]

### Step 3: Balance the reactions
To derive the target reaction from the given reactions, we need to manipulate them:
- For the melting of \( Al_2O_3 \):
\[ 2Al_2O_3(s) \rightarrow 2Al_2O_3(\text{melt}), \Delta G^\circ = 2 \times 16 \, \text{kJ/mol} = 32 \, \text{kJ/mol} \]
- For the formation of \( Al_2O_3 \):
\[ 2Al(l) + \frac{3}{2}O_2 \rightarrow Al_2O_3(s), \Delta G^\circ = 2 \times (-1269) \, \text{kJ/mol} = -2538 \, \text{kJ/mol} \]
- For the combustion of carbon:
\[ 3C + 3O_2 \rightarrow 3CO_2, \Delta G^\circ = 3 \times (-395) \, \text{kJ/mol} = -1185 \, \text{kJ/mol} \]

### Step 4: Calculate the total ΔG for the target reaction
Now, we can calculate the total ΔG for the overall reaction:
\[ \Delta G^\circ_{\text{total}} = \Delta G^\circ_{Al_2O_3(\text{melt})} + \Delta G^\circ_{Al_2O_3} + \Delta G^\circ_{C} \]
\[ = 32 \, \text{kJ/mol} + (-2538 \, \text{kJ/mol}) + (-1185 \, \text{kJ/mol}) \]
\[ = 32 - 2538 - 1185 = -3691 \, \text{kJ/mol} \]

### Step 5: Convert ΔG to Joules
Convert ΔG from kJ to J:
\[ \Delta G^\circ_{\text{total}} = -3691 \times 10^3 \, \text{J/mol} = -3691000 \, \text{J/mol} \]

### Step 6: Use the relationship between ΔG, n, F, and E
The relationship is given by:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( n = 12 \) (number of electrons)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)

### Step 7: Solve for E
Rearranging the equation to find E:
\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]
Substituting the values:
\[ E^\circ = -\frac{-3691000 \, \text{J/mol}}{12 \times 96500 \, \text{C/mol}} \]
Calculating:
\[ E^\circ = \frac{3691000}{1158000} \approx 3.18 \, \text{V} \]

### Final Answer
The EMF for the cell reaction is approximately \( 3.18 \, \text{V} \).

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