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 analyze the cell potentials using the Nernst equation. Here’s a step-by-step solution:

### Step 1: Understand 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 (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( n \) = number of electrons transferred in the reaction
- \( F \) = Faraday's constant (96485 C/mol)
- \( Q \) = reaction quotient

### Step 2: Identify the Cell Reactions
For the given cell reactions:
1. Zinc (Zn) is oxidized to zinc ions (Zn²⁺).
2. Copper ions (Cu²⁺) are reduced to copper (Cu).

The overall cell reaction can be written as:
\[ \text{Zn (s)} + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu (s)} \]

### Step 3: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) for the cell reaction is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]
Only the aqueous species are included in the reaction quotient, as solids do not contribute.

### Step 4: Analyze Given Concentrations
Assuming we have different concentrations for \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \):
- For example, let’s consider three cases:
1. \( [\text{Zn}^{2+}] = 0.2 \, M \) and \( [\text{Cu}^{2+}] = 0.2 \, M \) → \( Q = \frac{0.2}{0.2} = 1 \)
2. \( [\text{Zn}^{2+}] = 0.02 \, M \) and \( [\text{Cu}^{2+}] = 0.2 \, M \) → \( Q = \frac{0.02}{0.2} = 0.1 \)
3. \( [\text{Zn}^{2+}] = 0.2 \, M \) and \( [\text{Cu}^{2+}] = 0.002 \, M \) → \( Q = \frac{0.2}{0.002} = 100 \)

### Step 5: Determine the Effect on Cell Potential
From the Nernst equation, we see that as \( Q \) increases, the term \( \ln Q \) increases, which decreases the cell potential \( E_{cell} \). Therefore, to maximize \( E_{cell} \), we need to minimize \( Q \).

### Step 6: Compare the Values
- For \( Q = 1 \): \( E_{cell} \) is at a certain value.
- For \( Q = 0.1 \): \( E_{cell} \) is higher than for \( Q = 1 \).
- For \( Q = 100 \): \( E_{cell} \) is the lowest.

### Conclusion
The maximum current can be drawn from the cell with the lowest \( Q \) value, which corresponds to the highest \( E_{cell} \). In this case, the cell with \( [\text{Zn}^{2+}] = 0.02 \, M \) and \( [\text{Cu}^{2+}] = 0.2 \, M \) would provide the maximum current.