The maximum current can be drawn from which of the following cells?

Options

A. `Zn_((s))|Zn_((0.2M))^(2+)|| Cu_((0.2M))^(2+)| Cu_((s))`
B. `Zn_((s)) |Zn_((0.02M))^(2+)|| Cu_((0.2M))^(2+)| Cu_((s))`
C. `Zn_((s)) |Zn_((0.2M))^(2+) || Cu_((0.002M))^(2+)| Cu_((s))`
D. All of these

Answer

To determine which cell can draw the maximum current, we need to calculate the cell potential (E_cell) for each of the given cells using the Nernst equation. The higher the cell potential, the greater the current that can be drawn from the cell.

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]
where:
- \(E_{cell}\) = cell potential
- \(E^0_{cell}\) = standard cell potential
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin
- \(n\) = number of electrons transferred in the reaction
- \(F\) = Faraday's constant
- \(Q\) = reaction quotient

2. **Identify the Half-Reactions**:
In the given cells, zinc (Zn) is oxidized and copper (Cu) is reduced:
- Oxidation: \( \text{Zn (s)} \rightarrow \text{Zn}^{2+} + 2e^{-} \)
- Reduction: \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu (s)} \)

3. **Write the Overall Cell Reaction**:
The overall cell reaction can be written as:
\[ \text{Zn (s)} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu (s)} \]

4. **Calculate the Reaction Quotient (Q)**:
The reaction quotient \(Q\) is defined as:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]
For each cell, we will calculate \(Q\) based on the given concentrations.

5. **Calculate Q for Each Cell**:
- **Cell 1**:
- \([\text{Zn}^{2+}] = 0.2\), \([\text{Cu}^{2+}] = 0.2\)
- \(Q = \frac{0.2}{0.2} = 1\)
- **Cell 2**:
- \([\text{Zn}^{2+}] = 0.02\), \([\text{Cu}^{2+}] = 0.2\)
- \(Q = \frac{0.02}{0.2} = 0.1\)
- **Cell 3**:
- \([\text{Zn}^{2+}] = 0.2\), \([\text{Cu}^{2+}] = 0.002\)
- \(Q = \frac{0.2}{0.002} = 100\)

6. **Analyze the Effect of Q on E_cell**:
From the Nernst equation, we see that as \(Q\) increases, \(E_{cell}\) decreases (due to the negative sign in front of the logarithm). Therefore:
- Cell 1: \(Q = 1\) → \(E_{cell}\) is moderate.
- Cell 2: \(Q = 0.1\) → \(E_{cell}\) is higher (more favorable).
- Cell 3: \(Q = 100\) → \(E_{cell}\) is the lowest.

7. **Determine the Cell with Maximum Current**:
The cell with the highest \(E_{cell}\) will be able to provide the maximum current. Since Cell 2 has the lowest \(Q\) (0.1), it will have the highest \(E_{cell}\).

### Conclusion:
The cell from which the maximum current can be drawn is **Cell 2**.