`Zn|Zn^(2+)(C_1)||Zn^(2+)(C_2)|Zn`. For this cell `Delta G` is negative if

Options

A. `C_1 = C_2`
B. `C_1 gt C_2`
C. `C_2 gt C_1`
D. None of these

Answer

To determine when ΔG is negative for the given electrochemical cell represented as `Zn|Zn^(2+)(C_1)||Zn^(2+)(C_2)|Zn`, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Cell Representation**:
- The cell consists of two half-cells involving zinc. The left half-cell has a zinc electrode in a solution with concentration C1 of Zn²⁺ ions, while the right half-cell has a zinc electrode in a solution with concentration C2 of Zn²⁺ ions.

2. **Identifying the Reactions**:
- At the anode (left side), zinc is oxidized:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- At the cathode (right side), zinc ions are reduced:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]

3. **Using the Nernst Equation**:
- The Nernst equation relates the cell potential (E_cell) to the concentrations of the reactants and products:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
- Here, \( Q \) is the reaction quotient, which for this cell can be expressed as:
\[ Q = \frac{[\text{Zn}^{2+}]_{anode}}{[\text{Zn}^{2+}]_{cathode}} = \frac{C_1}{C_2} \]

4. **Determining the Sign of E_cell**:
- For ΔG to be negative, E_cell must be positive:
\[ \Delta G = -nFE_{cell} \]
- Thus, we need:
\[ E_{cell} > 0 \implies 0.0591 \log Q < E^\circ_{cell} \]
- Rearranging gives:
\[ \log Q < \frac{E^\circ_{cell}}{0.0591} \]

5. **Analyzing the Logarithm**:
- Since \( Q = \frac{C_1}{C_2} \), we want:
\[ \log \left(\frac{C_1}{C_2}\right) < 0 \]
- This implies:
\[ \frac{C_1}{C_2} < 1 \implies C_1 < C_2 \]
- Therefore, for ΔG to be negative, the concentration of Zn²⁺ at the cathode (C2) must be greater than the concentration at the anode (C1).

6. **Conclusion**:
- Thus, the condition for ΔG to be negative is:
\[ C_2 > C_1 \]
- The correct answer is option 3.