For the given concentration cell
` Cu(s)|Cu^(2+) (C_2 M) || Cu^(2+)(C_1 M) |Cu(s)`
Gibbs energy `triangle`G is negative if:

Options

A. `C_2 = C_1//sqrt2`
B. `C_2 = 2C_1`
C. `C_2 = sqrt2C_1`
D. `C_1 = C_2`

Answer

To determine when the Gibbs energy (ΔG) is negative for the given concentration cell, we can follow these steps:

### Step 1: Understand the Concentration Cell
The concentration cell is represented as:
\[ \text{Cu(s)} | \text{Cu}^{2+} (C_2 \, M) || \text{Cu}^{2+} (C_1 \, M) | \text{Cu(s)} \]
This indicates that we have copper solid at both electrodes, with different concentrations of copper ions (\(Cu^{2+}\)) in the two half-cells.

### Step 2: Identify the Reactions at the Electrodes
At the anode (oxidation):
\[ \text{Cu(s)} \rightarrow \text{Cu}^{2+} + 2e^- \]
At the cathode (reduction):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \]

### Step 3: Write the Expression for Gibbs Energy
The relationship between Gibbs energy and cell potential is given by:
\[ \Delta G = -nFE_{cell} \]
Where:
- \(n\) = number of moles of electrons transferred (which is 2 in this case),
- \(F\) = Faraday's constant,
- \(E_{cell}\) = cell potential.

For ΔG to be negative, \(E_{cell}\) must be positive.

### Step 4: Use the Nernst Equation
The Nernst equation relates the cell potential to the concentrations:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Since both electrodes are the same (copper), \(E^\circ_{cell} = 0\). Thus:
\[ E_{cell} = -\frac{0.0591}{2} \log Q \]

### Step 5: Determine the Reaction Quotient (Q)
The reaction quotient \(Q\) for this cell is given by:
\[ Q = \frac{[\text{Cu}^{2+}]_{anode}}{[\text{Cu}^{2+}]_{cathode}} = \frac{C_2}{C_1} \]

### Step 6: Set the Condition for \(E_{cell}\) to be Positive
For \(E_{cell}\) to be positive:
\[ -\frac{0.0591}{2} \log \left( \frac{C_2}{C_1} \right) > 0 \]
This implies:
\[ \log \left( \frac{C_2}{C_1} \right) < 0 \]
Which means:
\[ C_2 < C_1 \]

### Step 7: Analyze the Options
We need to find the condition among the options where \(C_1 > C_2\):
- Option A: \(C_2 = \frac{C_1}{\sqrt{2}} \) → \(C_1 = \sqrt{2}C_2\) (True, \(C_1 > C_2\))
- Option B: \(C_2 = 2C_1\) → \(C_1 < C_2\) (False)
- Option C: \(C_2 = \sqrt{2}C_1\) → \(C_1 < C_2\) (False)
- Option D: \(C_2 = C_1\) → \(C_1 = C_2\) (False)

### Conclusion
The only option that satisfies \(C_1 > C_2\) is:
**Option A: \(C_2 = \frac{C_1}{\sqrt{2}}\)**