Calculate the emf of the cell, `Cd|Cd^(2+)(0.001M)||Fe^(2+)(0.6M)|Fe` at `25^(@)C`.
The standard reduction potential of `Cd//Cd^(2+)` and `Fe//Fe^(2+)` electrodes are -0.403 and -0.441 volt respectively.

Explanation

To calculate the emf (electromotive force) of the cell represented as `Cd|Cd^(2+)(0.001M)||Fe^(2+)(0.6M)|Fe`, we will use the Nernst equation. The Nernst equation relates the emf of a cell to the standard electrode potentials and the concentrations of the reactants and products involved in the half-reactions.

### Step-by-Step Solution:

**Step 1: Identify the half-reactions and their standard potentials.**

The half-reactions for the cell are:
1. For Cadmium (Cd):
\[ \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \quad E^\circ = -0.403 \, \text{V} \]

2. For Iron (Fe):
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad E^\circ = -0.441 \, \text{V} \]

**Step 2: Determine which reaction is oxidation and which is reduction.**

In the cell notation, the left side (anode) is where oxidation occurs, and the right side (cathode) is where reduction occurs. Thus:
- Cadmium (Cd) will be oxidized:
\[ \text{Cd} \rightarrow \text{Cd}^{2+} + 2e^- \]
- Iron (Fe) will be reduced:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \]

**Step 3: Calculate the standard cell potential (E°cell).**

The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.441 \, \text{V}) - (-0.403 \, \text{V}) = -0.441 + 0.403 = -0.038 \, \text{V} \]

**Step 4: Apply the Nernst equation.**

The Nernst equation is given by:
\[ E = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \) (25°C)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96485 \, \text{C/mol} \)
- \( Q \) is the reaction quotient.

**Step 5: Calculate the reaction quotient (Q).**

The reaction quotient \( Q \) for the cell can be calculated using the concentrations of the ions:
\[ Q = \frac{[\text{Cd}^{2+}]}{[\text{Fe}^{2+}]} = \frac{0.001}{0.6} \]

**Step 6: Substitute values into the Nernst equation.**

First, calculate \( Q \):
\[ Q = \frac{0.001}{0.6} = 0.0016667 \]

Now substitute into the Nernst equation:
\[ E = -0.038 \, \text{V} - \frac{(8.314)(298)}{(2)(96485)} \ln(0.0016667) \]

Calculating the second term:
\[ \frac{(8.314)(298)}{(2)(96485)} \approx 0.00414 \, \text{V} \]
Now calculate \( \ln(0.0016667) \):
\[ \ln(0.0016667) \approx -6.396 \]

Now plug it into the equation:
\[ E = -0.038 \, \text{V} - (0.00414)(-6.396) \]
\[ E \approx -0.038 + 0.0265 \approx -0.0115 \, \text{V} \]

**Final Answer:**
The emf of the cell is approximately **-0.0115 V**.