`Ag|AgCl|Cl^(-)(C_(2))||Cl^(-)(C_(1))|AgCl|Ag` for this cell `DeltaG` is negative if `:`

Options

A. `C_1=C_2`
B. `C_1gtC_2`
C. `C_2gtC_1`
D. Both (a) and (c)\

Answer

To solve the problem regarding the cell representation `Ag|AgCl|Cl^(-)(C_(2))||Cl^(-)(C_(1))|AgCl|Ag`, we need to analyze the conditions under which the Gibbs free energy change (ΔG) is negative. Here’s a step-by-step solution:

### Step 1: Understand the Cell Representation
The cell representation indicates a concentration cell where the same electrodes (Ag and AgCl) are used, but the concentrations of the chloride ions (Cl^(-)) are different: C2 on the left side and C1 on the right side.

**Hint:** Concentration cells involve the same materials but differ in concentration.

### Step 2: Identify the Anode and Cathode
In this concentration cell:
- The left side (C2) is the anode where oxidation occurs.
- The right side (C1) is the cathode where reduction occurs.

**Hint:** Oxidation occurs at the anode and reduction at the cathode.

### Step 3: Write the Half-Reactions
At the anode (left side):
- Oxidation: \( \text{Ag} \rightarrow \text{Ag}^+ + e^- \)

At the cathode (right side):
- Reduction: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)

**Hint:** Identify oxidation and reduction reactions based on electron transfer.

### Step 4: Determine the Cell Potential (E_cell)
For concentration cells, the standard cell potential (E°_cell) is zero because the same species are involved. Therefore, the cell potential can be expressed as:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{Ag}^+]_{right}}{[\text{Ag}^+]_{left}} \right) \]
Since \( E^{\circ}_{cell} = 0 \):
\[ E_{cell} = -\frac{0.0591}{n} \log \left( \frac{C_1}{C_2} \right) \]

**Hint:** The Nernst equation helps to calculate the cell potential based on concentration.

### Step 5: Relate E_cell to ΔG
The relationship between Gibbs free energy change (ΔG) and cell potential (E_cell) is given by:
\[ \Delta G = -nFE_{cell} \]
For ΔG to be negative (indicating a spontaneous reaction), E_cell must be positive.

**Hint:** A negative ΔG indicates a spontaneous process.

### Step 6: Analyze Conditions for ΔG < 0
From the equation for E_cell:
- For \( E_{cell} > 0 \):
\[ -\frac{0.0591}{n} \log \left( \frac{C_1}{C_2} \right) > 0 \]
This implies:
\[ \log \left( \frac{C_1}{C_2} \right) < 0 \]
Thus, \( C_1 < C_2 \).

**Hint:** The logarithm of a fraction is negative when the numerator is less than the denominator.

### Conclusion
For ΔG to be negative, the concentration C1 must be less than C2:
\[ \Delta G < 0 \text{ if } C_1 < C_2 \]

### Final Answer
Thus, the answer to the question is: ΔG is negative if \( C_1 < C_2 \).

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