For the Daniel Cell involving the cell reaction `Zn_((s))+Cu_((aq))^(+2)hArrZn_((aq))^(+2)+Cu_((s))` the standard free energies of formation of `Zn_((s)),Cu_((s)),Cu_((aq))^(+2)` and `Zn_((aq))^(+2)` are 0, 0, `64.4` KJ/Mole and `-154.0` KJ/Mole, respectively. Calculate the standard EMF of the cell

Options

A. `2.13` Volts
B. `1.13` Volts
C. `2.26` Volts
D. `3.42` Volts

Explanation

To calculate the standard EMF of the Daniel Cell involving the reaction:

\[ \text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightleftharpoons \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)} \]

we will follow these steps:

### Step 1: Write the standard free energy change equation
The standard Gibbs free energy change (\( \Delta G^\circ \)) for the reaction can be calculated using the standard free energies of formation of the reactants and products.

\[ \Delta G^\circ = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)} \]

### Step 2: Substitute the values
From the problem, we have the following standard free energies of formation:
- \( \Delta G_f^\circ (\text{Zn}_{(s)}) = 0 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Cu}_{(s)}) = 0 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Cu}^{2+}_{(aq)}) = 64.4 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Zn}^{2+}_{(aq)}) = -154.0 \, \text{kJ/mol} \)

Now substituting these values into the equation:

\[ \Delta G^\circ = \left( \Delta G_f^\circ (\text{Zn}^{2+}_{(aq)}) + \Delta G_f^\circ (\text{Cu}_{(s)}) \right) - \left( \Delta G_f^\circ (\text{Zn}_{(s)}) + \Delta G_f^\circ (\text{Cu}^{2+}_{(aq)}) \right) \]

\[ \Delta G^\circ = \left( -154.0 + 0 \right) - \left( 0 + 64.4 \right) \]

\[ \Delta G^\circ = -154.0 - 64.4 = -218.4 \, \text{kJ/mol} \]

### Step 3: Convert \( \Delta G^\circ \) to Joules
Since we need to use Joules for the calculation of EMF, we convert \( \Delta G^\circ \) from kJ to J:

\[ \Delta G^\circ = -218.4 \times 10^3 \, \text{J/mol} = -218400 \, \text{J/mol} \]

### Step 4: Use the relationship between \( \Delta G^\circ \) and EMF
The relationship between the Gibbs free energy change and the EMF of the cell is given by:

\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

Where:
- \( n \) = number of moles of electrons transferred (in this case, \( n = 2 \) for the reduction of \( \text{Cu}^{2+} \) to \( \text{Cu} \))
- \( F \) = Faraday's constant \( \approx 96500 \, \text{C/mol} \)
- \( E^\circ_{\text{cell}} \) = standard EMF of the cell

### Step 5: Solve for \( E^\circ_{\text{cell}} \)
Rearranging the equation gives:

\[ E^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{nF} \]

Substituting the values:

\[ E^\circ_{\text{cell}} = -\frac{-218400}{2 \times 96500} \]

Calculating this:

\[ E^\circ_{\text{cell}} = \frac{218400}{193000} \approx 1.13 \, \text{V} \]

### Final Answer
The standard EMF of the Daniel Cell is approximately **1.13 V**.

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