`Co`|`Co^(2+)(C_(2))`||`Co^(2+)(C_(1))`|`Co`, for this cell, `DeltaG` is negative if :

Options

A. `C_(2)gtC_(1)`
B. `C_(1)gtC_(2)`
C. `C_(1)=C_(2)`
D. unpredictable

Answer

To determine when ΔG is negative for the given electrochemical cell, we need to analyze the cell reaction and apply the Nernst equation. Here’s a step-by-step solution:

### Step 1: Identify the Cell Components
The cell is represented as:
`Co | Co^(2+)(C2) || Co^(2+)(C1) | Co`

- Left side (anode): Cobalt (Co) is oxidized to Co²⁺ with concentration C2.
- Right side (cathode): Co²⁺ with concentration C1 is reduced to Co.

### Step 2: Write the Half-Reactions
The half-reactions can be written as:
- Anode (oxidation): Co → Co²⁺ + 2e⁻
- Cathode (reduction): Co²⁺ + 2e⁻ → Co

### Step 3: Determine the Overall Cell Reaction
Combining the half-reactions gives:
Co (s) + Co²⁺ (C1) ⇌ Co²⁺ (C2) + Co (s)

### Step 4: Relate ΔG to Cell Potential (E)
The Gibbs free energy change (ΔG) is related to the cell potential (E) by the equation:
ΔG = -nFE
Where:
- n = number of moles of electrons transferred (n = 2 for this reaction)
- F = Faraday's constant (approximately 96485 C/mol)
- E = cell potential

### Step 5: Determine When ΔG is Negative
ΔG is negative when E is positive. Therefore, we need to find conditions under which E > 0.

### Step 6: Apply the Nernst Equation
The Nernst equation is given by:
E = E° - (RT/nF) ln(Q)
Where Q is the reaction quotient:
Q = [Co²⁺ (C2)] / [Co²⁺ (C1)]

For our case:
E = E° - (0.0591/n) log([Co²⁺ (C2)] / [Co²⁺ (C1)])

### Step 7: Analyze the Reaction Quotient
Substituting n = 2, we get:
E = E° - (0.0591/2) log(C2/C1)

### Step 8: Determine Conditions for E > 0
For E to be positive:
E° - (0.0591/2) log(C2/C1) > 0

Rearranging gives:
E° > (0.0591/2) log(C2/C1)

This implies that if C1 > C2, the logarithm term becomes negative, making E positive.

### Conclusion
Thus, ΔG will be negative if the concentration of Co²⁺ at the cathode (C1) is greater than the concentration of Co²⁺ at the anode (C2):
**C1 > C2**