`Zn | Zn^(2+)(C_(1) || Zn^(2+)(C_(2)| Zn`. For this cell `DeltaG` is negative if:

Options

A. `c_(1)=c_(2`
B. `c_(1)gtc_(2)`
C. `c_(2)gtc_(1)`
D. None

Explanation

To determine when ΔG is negative for the given electrochemical cell, we can follow these steps:

### Step 1: Identify the Cell Components
The cell is represented as:
\[ \text{Zn} | \text{Zn}^{2+}(C_1) || \text{Zn}^{2+}(C_2) | \text{Zn} \]
Here, we have two zinc electrodes with different concentrations of zinc ions (\(C_1\) and \(C_2\)).

### Step 2: Determine Anode and Cathode
In this setup:
- The left side (Zn at concentration \(C_1\)) is the **anode**, where oxidation occurs.
- The right side (Zn at concentration \(C_2\)) is the **cathode**, where reduction occurs.

### Step 3: Write the Half-Reactions
**At the anode (oxidation):**
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]

**At the cathode (reduction):**
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]

### Step 4: Write the Overall Reaction
Combining the half-reactions, we get:
\[ \text{Zn} + \text{Zn}^{2+} \rightarrow \text{Zn}^{2+} + \text{Zn} \]
This simplifies to:
\[ \text{Zn} + \text{Zn}^{2+} \rightarrow \text{Zn}^{2+} + \text{Zn} \]

### Step 5: Relate E_cell to Concentrations
The Nernst equation relates the cell potential (E_cell) to the concentrations:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
For our reaction:
- \(E^\circ_{\text{cell}} = 0\) (since both half-reactions involve the same species)
- \(n = 2\) (the number of electrons transferred)

Substituting into the Nernst equation gives:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log \left( \frac{C_1}{C_2} \right) \]
Thus,
\[ E_{\text{cell}} = -0.02955 \log \left( \frac{C_1}{C_2} \right) \]

### Step 6: Relate ΔG to E_cell
The relationship between Gibbs free energy (ΔG) and cell potential (E_cell) is given by:
\[ \Delta G = -nFE_{\text{cell}} \]
Where:
- \(F\) is Faraday's constant.

### Step 7: Determine Conditions for ΔG to be Negative
For ΔG to be negative:
\[ E_{\text{cell}} > 0 \]
This occurs when:
\[ -0.02955 \log \left( \frac{C_1}{C_2} \right) > 0 \]
This implies:
\[ \log \left( \frac{C_1}{C_2} \right) < 0 \]
Thus:
\[ \frac{C_1}{C_2} < 1 \]
This means:
\[ C_1 < C_2 \]

### Conclusion
ΔG is negative if \(C_2 > C_1\).

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