The standard electrode potential of ` Zn^(2+) // Zn` is -0.76V and that of `Ca^(2+) // Cu` is 0.34V . The emf(V) and the free energy change ` (kJ "mol"^(-1))` , respectively , for a Daniel cell will be

Options

A. `-0.42 and 81`
B. `1.1 and -213`
C. `-1.1 and 213`
D. `0.42 and -81`

Explanation

To solve the problem, we need to calculate the EMF (electromotive force) of the Daniel cell and the free energy change (ΔG) associated with the cell reaction.

### Step 1: Identify the half-reactions and their standard electrode potentials

We have the following standard electrode potentials:
- For the reduction of zinc:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E^\circ = -0.76 \, \text{V} \]
- For the reduction of copper:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = 0.34 \, \text{V} \]

### Step 2: Determine the anode and cathode

In a galvanic cell:
- The anode is where oxidation occurs.
- The cathode is where reduction occurs.

For the Daniel cell:
- **Anode (oxidation)**:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
The oxidation potential is the negative of the reduction potential:
\[ E^\circ_{\text{oxidation}} = +0.76 \, \text{V} \]

- **Cathode (reduction)**:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
The reduction potential is:
\[ E^\circ_{\text{reduction}} = 0.34 \, \text{V} \]

### Step 3: Calculate the EMF of the cell

The EMF (E°cell) of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V} \]

### Step 4: Calculate the free energy change (ΔG)

The free energy change can be calculated using the formula:
\[ \Delta G = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred (2 for both reactions)
- \( F \) = Faraday's constant = 96500 C/mol
- \( E^\circ_{\text{cell}} = 1.10 \, \text{V} \)

Substituting the values:
\[ \Delta G = -2 \times 96500 \, \text{C/mol} \times 1.10 \, \text{V} \]
\[ \Delta G = -2 \times 96500 \times 1.10 = -212130 \, \text{J/mol} \]

To convert to kJ/mol:
\[ \Delta G = -212.13 \, \text{kJ/mol} \]

### Final Answers
- EMF of the cell: **1.10 V**
- Free energy change: **-212.13 kJ/mol**