`Zn|Zn^(2+)(c_1M)||Zn^(2+)(c_2M)|Zn`. For this cell, `DeltaG` would be negative , if :

Options

A. 1. `c_1=c_2`
B. 2. `c_1 gt c_2`
C. 3. `c_2 gt c_1`
D. 4. None of these

Explanation

To determine the conditions under which ΔG would be negative for the given electrochemical cell, we can follow these steps:

### Step 1: Understand the relationship between ΔG and E_cell
The Gibbs free energy change (ΔG) is related to the cell potential (E_cell) by the equation:
\[ \Delta G = -nFE_{cell} \]
where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately 96485 C/mol)
- \( E_{cell} \) = cell potential in volts

### Step 2: Identify the half-reactions
For the given cell:
- At the anode (oxidation):
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- At the cathode (reduction):
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \]

### Step 3: Write the expression for E_cell
The Nernst equation can be used to calculate the cell potential (E_cell):
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}} \]
In this case, \( n = 2 \) (since 2 electrons are involved), and we denote:
- \( [Zn^{2+}]_{anode} = c_1 \)
- \( [Zn^{2+}]_{cathode} = c_2 \)

Thus, the equation becomes:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{c_1}{c_2} \]

### Step 4: Analyze the case when E_cell is positive
For ΔG to be negative, E_cell must be positive:
\[ E_{cell} > 0 \implies E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{c_1}{c_2} > 0 \]
Assuming \( E^{\circ}_{cell} = 0 \) (since both half-reactions involve the same species, zinc), we simplify:
\[ -\frac{0.0591}{2} \log \frac{c_1}{c_2} > 0 \]
This implies:
\[ \log \frac{c_1}{c_2} < 0 \implies \frac{c_1}{c_2} < 1 \implies c_1 < c_2 \]

### Step 5: Conclusion
Thus, for ΔG to be negative, the concentration of Zn²⁺ at the cathode (c_2) must be greater than the concentration of Zn²⁺ at the anode (c_1):
\[ c_2 > c_1 \]

### Final Answer
ΔG would be negative if \( c_2 > c_1 \).

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