This question involves calculating the Gibbs free energy change (ΔG) for an electrochemical cell under non-standard conditions. The cell is a Daniell cell with zinc and copper electrodes. The key concept is the relationship between ΔG, the standard cell potential (E°), and the reaction quotient (Q) via the Nernst equation.

Step 1: Write the cell reaction and identify the standard cell potential.
The cell notation is: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
The cell reaction is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
The standard cell potential, E°(cell), is given as 1.1 V.

Step 2: Recall the formula relating ΔG and the cell potential.
The Gibbs free energy change is given by: $\Delta G=-nF{E}_{\text{cell}}$
where:
n = number of electrons transferred in the reaction
F = Faraday constant
E_cell = cell potential under the given conditions

Step 3: Apply the Nernst equation to find E_cell.
The Nernst equation for this cell at temperature T is:
$E_{\text{cell}}=E^{°}\text{(cell)}-\frac{RT}{nF}\ln Q$
For the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), the reaction quotient Q is:
$Q=\frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}$
It is given that [Zn²⁺] = 10 [Cu²⁺]. Substituting this into the expression for Q:
$Q=\frac{10\left[{\mathrm{Cu}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}=10$
The number of electrons transferred, n, is 2. Substituting E°(cell) = 1.1 V, Q=10, n=2 into the Nernst equation:
$E_{\text{cell}}=1.1-\frac{RT}{2F}\ln \left(10\right)$
Since ln(10) = 2.303 log₁₀(10) = 2.303, the equation becomes:
$E_{\text{cell}}=1.1-\frac{RT}{2F}\left(2.303\right)=1.1-\frac{2.303RT}{2F}$

Step 4: Substitute E_cell into the ΔG formula.
$\Delta G=-nF{E}_{\text{cell}}=-2F(1.1-\frac{2.303RT}{2F})$
Distribute the -2F term:
$\Delta G=-2.2F+2.303RT$
Therefore, the expression for ΔG is $2.303RT-2.2F$ J mol⁻¹.

Final Answer: The correct expression is $2.303RT-2.2F$.

Related Topics & Formulae

Nernst Equation: This equation is used to calculate the cell potential under non-standard conditions. The general form is: $E_{\text{cell}}=E^{°}\text{(cell)}-\frac{RT}{nF}\ln Q$. At 298 K, it is often written using log base 10: $E_{\text{cell}}=E^{°}\text{(cell)}-\frac{0.059}{n}\log Q$.

Gibbs Free Energy and Cell Potential: The relationship $\Delta G=-nF{E}_{\text{cell}}$ shows that a positive cell potential (spontaneous reaction) corresponds to a negative ΔG. Under standard conditions, this becomes $\Delta G^{°}=-nF{E}^{°}$.

Reaction Quotient (Q): For a general reaction aA + bB → cC + dD, the reaction quotient is $Q=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$. Pure solids and liquids are not included in the expression for Q.