For the following cell,
`Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)||Cu(s)`
When the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG`
(in J `"mol"^(-1)`)
[F is Faraday constant, R is gas constant] T is temperaure, `E^(@)("cell")=1.1V`

Options

A. `2.303 RT+1.1F`
B. `1.1F`
C. `2.303RT-2.2F`
D. `-2.2F`

Answer

To solve the problem, we need to derive the expression for ΔG for the given electrochemical cell. Let's break down the steps:

### Step 1: Write the Cell Reaction
The cell representation given is:
\[ \text{Zn(s)} | \text{ZnSO}_4(aq) || \text{CuSO}_4(aq) | \text{Cu(s)} \]

The overall cell reaction can be written as:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

### Step 2: Identify Oxidation and Reduction Half-Reactions
From the overall cell reaction:
- **Oxidation half-reaction**:
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
- **Reduction half-reaction**:
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)} \]

### Step 3: Determine the Number of Electrons Transferred (n)
From the half-reactions, we see that 2 electrons are transferred. Thus, \( n = 2 \).

### Step 4: Use the Nernst Equation
The Nernst equation relates the standard cell potential (\( E^\circ_{\text{cell}} \)), the actual cell potential (\( E_{\text{cell}} \)), and the concentrations of the reactants and products:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
where \( Q \) is the reaction quotient.

### Step 5: Calculate the Reaction Quotient (Q)
Given that the concentration of \( \text{Zn}^{2+} \) is 10 times that of \( \text{Cu}^{2+} \):
Let \( [\text{Cu}^{2+}] = x \) and \( [\text{Zn}^{2+}] = 10x \).
Thus, the reaction quotient \( Q \) is:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{10x}{x} = 10 \]

### Step 6: Substitute Values into the Nernst Equation
Now substituting \( E^\circ_{\text{cell}} = 1.1 \, \text{V} \), \( n = 2 \), and \( Q = 10 \):
\[ E_{\text{cell}} = 1.1 - \frac{RT}{2F} \ln(10) \]

### Step 7: Relate ΔG to Ecell
The relationship between Gibbs free energy change (\( \Delta G \)) and cell potential is given by:
\[ \Delta G = -nFE_{\text{cell}} \]

### Step 8: Substitute Ecell into the ΔG Equation
Substituting the expression for \( E_{\text{cell}} \):
\[ \Delta G = -2F \left( 1.1 - \frac{RT}{2F} \ln(10) \right) \]
\[ \Delta G = -2F \cdot 1.1 + RT \ln(10) \]

### Step 9: Simplify the Expression
\[ \Delta G = -2.2F + RT \ln(10) \]

### Final Expression for ΔG
Thus, the final expression for \( \Delta G \) is:
\[ \Delta G = 2.303RT - 2.2F \]

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