What will be the electrical potential energy in a Galvanic cell, when the concentration of Zn and Cu ions is unity (1 mol L^-1)?

1.1 V

To calculate the electrical potential energy in a Galvanic cell, we can use the Nernst equation. The Nernst equation relates the cell potential (\(E_{\text{cell}}\)) to the standard cell potential (\(E_{\text{°cell}}\)) and the concentrations of the ions involved in the cell reaction.

The Nernst equation is given as:

\[ E_{\text{cell}} = E_{\text{°cell}} - \frac{RT}{nF} \ln(Q) \]

Where:
\(E_{\text{cell}}\) = Cell potential
\(E_{\text{°cell}}\) = Standard cell potential (under standard conditions, usually at 25°C and 1 atm pressure)
\(R\) = Gas constant (8.314 J/(mol K))
\(T\) = Temperature in Kelvin
\(n\) = Number of moles of electrons exchanged in the balanced cell reaction
\(F\) = Faraday's constant (approximately 96,485 C/mol)
\(\ln\) = Natural logarithm
\(Q\) = Reaction quotient (ratio of products to reactants concentrations)

For a Galvanic cell with the reaction:

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

The number of moles of electrons exchanged (\(n\)) is 2 because two electrons are transferred in the balanced cell reaction.

At 25°C (298 K) and 1 atm pressure, the standard cell potential for this cell is:

\[ E_{\text{°cell}} = E_{\text{°red}}(\text{Cu}^{2+}/\text{Cu}) - E_{\text{°red}}(\text{Zn}^{2+}/\text{Zn}) \]

\[ E_{\text{°cell}} = 0.34 \text{ V} - (-0.76 \text{ V}) \]
\[ E_{\text{°cell}} = 1.10 \text{ V} \]

Now, we can calculate the concentration term (\(Q\)) using the given information that the concentration of Zn and Cu ions is unity (1 mol/L):

\[ Q = \frac{[\text{Zn}^{2+}(\text{aq})]}{[\text{Cu}^{2+}(\text{aq})]} \]
\[ Q = \frac{1 \text{ mol/L}}{1 \text{ mol/L}} \]
\[ Q = 1 \]

Next, we need to convert the temperature to Kelvin:

\[ T = 25°C + 273.15 \text{ K/°C} \]
\[ T = 298.15 \text{ K} \]

Now, we can calculate the electrical potential energy (\(E_{\text{cell}}\)) using the Nernst equation:

\[ E_{\text{cell}} = 1.10 \text{ V} - \left(\frac{8.314 \text{ J/(mol K)} \times 298.15 \text{ K}}{2 \times 96,485 \text{ C/mol}} \times \ln(1)\right) \]
\[ E_{\text{cell}} \approx 1.10 \text{ V} - 0 \]
\[ E_{\text{cell}} \approx 1.10 \text{ V} \]

Therefore, the electrical potential energy in the Galvanic cell, when the concentration of Zn and Cu ions is unity (1 mol L\(^{-1}\)), is approximately 1.10 V. Thus, the correct answer is option 1: 1.1 V.