The zinc/silver oxide cell is used in electric current reaction is as following:
`Zn^(2+) + 2e^(-) to Zn, E^(@) = - 0.760 V`
`Ag_(2)O + H_(2)O + 2e^(-) to 2Ag + 2OH, E^(@) = 0.344 V` If F is 96,500 C `mol^(-1)`, `DeltaG^(@)` of the cell will be:

Options

A. `413.21 kJ mol^(-1)`
B. `113.072 kJ mol^(-1)`
C. `213.072 kJ mol^(-1)`
D. `313.082 kJ mol^(-1)`

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the zinc/silver oxide cell using the given standard reduction potentials and the Faraday constant. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the half-reactions and their standard reduction potentials
The half-reactions provided are:
1. For zinc:
\[ Zn^{2+} + 2e^{-} \rightarrow Zn, \quad E^{\circ} = -0.760 \, V \]
2. For silver oxide:
\[ Ag_{2}O + H_{2}O + 2e^{-} \rightarrow 2Ag + 2OH^{-}, \quad E^{\circ} = 0.344 \, V \]

### Step 2: Determine the anode and cathode
- The anode is where oxidation occurs. In this case, zinc is oxidized:
\[ Zn \rightarrow Zn^{2+} + 2e^{-} \quad (E^{\circ} = +0.760 \, V) \]
- The cathode is where reduction occurs. Silver oxide is reduced:
\[ Ag_{2}O + H_{2}O + 2e^{-} \rightarrow 2Ag + 2OH^{-} \quad (E^{\circ} = 0.344 \, V) \]

### Step 3: Calculate the standard cell potential (E°cell)
Using the formula:
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Substituting the values:
\[ E^{\circ}_{cell} = 0.344 \, V - (-0.760 \, V) = 0.344 \, V + 0.760 \, V = 1.104 \, V \]

### Step 4: Calculate the number of electrons transferred (n)
From the half-reactions, we see that:
- 2 electrons are involved in the overall cell reaction.
Thus, \( n = 2 \).

### Step 5: Use the Gibbs free energy formula
The formula for Gibbs free energy change is:
\[ \Delta G^{\circ} = -n E^{\circ}_{cell} F \]
Where:
- \( F = 96,500 \, C \, mol^{-1} \)

### Step 6: Substitute the values into the Gibbs free energy formula
\[ \Delta G^{\circ} = -2 \times 1.104 \, V \times 96,500 \, C \, mol^{-1} \]

### Step 7: Calculate ΔG°
Calculating:
\[ \Delta G^{\circ} = -2 \times 1.104 \times 96,500 = -213,072 \, J/mol \]
To convert to kilojoules:
\[ \Delta G^{\circ} = -213.072 \, kJ/mol \]

### Final Answer
\[ \Delta G^{\circ} = -213.072 \, kJ/mol \]