Consider the reaction fo extraction of gold from its ore
`Au +2CN^(-) (aq)+(1)/(4)O_(2) (g)+(1)/(2)H_(2)O rarr Au(CN)_(2)^(-) +OH^(-)`
Use the following data to calculate `DeltaG^(@)` for the reaction
`K_(f) [Au(CN)_(2)^(-)] = X`
`{:(O_(2)+2H_(2)O +4e^(-)rarr4OH^(-),:,E^(@) =+0.41 "volt"),(Au^(3+)+3e^(-)rarr Au,:,E^(@) = +1.5 "volt"),(Au^(3+)+2e^(-)rarr Au^(+),:,E^(@) =+1.4 "volt"):}`

Options

A. `-RT In X +1.29 F`
B. `-RT In X - 2.11 F`
C. `-RT In(1)/(X) +2.11F`
D. `-RT In X -1.29 F`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the extraction of gold from its ore, we will follow these steps:

### Step 1: Identify the reactions and their standard reduction potentials
We have three half-reactions with their standard reduction potentials (E°):

1. \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \) (E° = +1.5 V)
2. \( \text{Au}^{3+} + 2e^- \rightarrow \text{Au}^+ \) (E° = +1.4 V)
3. \( \text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^- \) (E° = +0.41 V)

### Step 2: Determine the overall cell reaction
The overall reaction for the extraction of gold is:
\[ \text{Au} + 2\text{CN}^- + \frac{1}{4}\text{O}_2 + \frac{1}{2}\text{H}_2\text{O} \rightarrow \text{Au(CN)}_2^- + \text{OH}^- \]

### Step 3: Calculate E° for the overall reaction
To find the E° for the overall reaction, we need to combine the half-reactions. The oxidation of Au to Au+ can be derived from the first two half-reactions.

1. From reaction 1, we can write:
\[ \text{Au} \rightarrow \text{Au}^+ + e^- \]

2. From reaction 2, we can write:
\[ \text{Au}^+ + 2e^- \rightarrow \text{Au}^{3+} \]
(We reverse this reaction for oxidation)

Now, we can combine these reactions:
- The overall reaction will be:
\[ \text{Au} \rightarrow \text{Au}^{3+} + 3e^- \]
\[ \text{Au}^{3+} + 2e^- \rightarrow \text{Au}^+ \]
\[ \text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^- \]

### Step 4: Calculate E°cell
Using the Nernst equation, we can find the E°cell for the overall reaction:
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \]
Where:
- \( E°_{\text{cathode}} = +0.41 \, \text{V} \) (for the reduction of O2)
- \( E°_{\text{anode}} = 1.7 \, \text{V} \) (calculated from the difference of the first two reactions)

Thus,
\[ E°_{\text{cell}} = 0.41 - 1.7 = -1.29 \, \text{V} \]

### Step 5: Calculate ΔG°
Using the formula:
\[ \Delta G° = -nFE°_{\text{cell}} \]
Where:
- \( n = 1 \) (number of moles of electrons transferred)
- \( F = 96485 \, \text{C/mol} \)

Now substituting the values:
\[ \Delta G° = -1 \times 96485 \times (-1.29) \]
\[ \Delta G° = 124,000 \, \text{J/mol} \]

### Final Answer
Thus, the standard Gibbs free energy change for the reaction is:
\[ \Delta G° \approx 124 \, \text{kJ/mol} \]