What is the Gibbs energy of the following reaction?
`Zn(s)+Cu^(2+) (aq) rarr Zn^(2+)(aq)+Cu^(2+) (s), E_("cell")^(@)=1.1 V`

Options

A. `106135.75 J mol^(-1)`
B. `21227 J mol^(-1)`
C. `-212.27 kJ mol^(-1)`
D. `21227 J mol^(-1)`

Explanation

To find the Gibbs energy (ΔG°) of the given reaction, we can use the relationship between Gibbs free energy and the standard electrode potential (E°) of the cell. The equation we will use is:

\[ \Delta G° = -nFE° \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E° \) = standard electrode potential of the cell (given as \( 1.1 \, V \))

### Step 1: Determine the number of electrons transferred (n)
In the reaction:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]
Zinc (Zn) is oxidized from 0 to +2, which means it loses 2 electrons. Therefore, \( n = 2 \).

### Step 2: Use the Faraday's constant (F)
The Faraday's constant is \( F = 96500 \, C/mol \).

### Step 3: Substitute the values into the Gibbs energy equation
Now we can substitute \( n \), \( F \), and \( E° \) into the equation:

\[ \Delta G° = -nFE° \]
\[ \Delta G° = - (2) (96500 \, C/mol) (1.1 \, V) \]

### Step 4: Calculate ΔG°
Calculating the above expression:

\[ \Delta G° = - (2) (96500) (1.1) \]
\[ \Delta G° = - (2) (106150) \quad \text{(since } 96500 \times 1.1 = 106150\text{)} \]
\[ \Delta G° = -212300 \, J/mol \]

### Step 5: Convert to kilojoules
To convert joules to kilojoules, we divide by 1000:

\[ \Delta G° = -212.3 \, kJ/mol \]

### Final Answer
Thus, the Gibbs energy of the reaction is:

\[ \Delta G° = -212.3 \, kJ/mol \]