Consider the following concentration cell`:`
`Zn(s)|Zn^(2+)(0.024M)||Zn^(2+)(0.480M)|Zn(s)`
which of the following statements is `//` are correct?

Options

A. The `EMF` of the cell at `25^(@)C` is nearly `+0.039V.`
B. The `EMF` of the cell at `25^(@)C` is nearly `-0.039V`.
C. If water is added in `LHE`, so that the `[Zn^(2+)]` is reduced to `0.012 M`, the cell voltage increases.
D. If water is added in `LH` , so that the `[Zn^(2+)]` is reduced to `0.012M` , the cell voltage decreases.

Explanation

To solve the problem regarding the concentration cell given by the equation:

\[ \text{Zn(s)} | \text{Zn}^{2+}(0.024 \, M) || \text{Zn}^{2+}(0.480 \, M) | \text{Zn(s)} \]

we will analyze the cell and calculate the electromotive force (EMF) of the cell.

### Step-by-Step Solution:

1. **Identify the Components of the Cell:**
- The left side (anode) has a concentration of Zn²⁺ at 0.024 M.
- The right side (cathode) has a concentration of Zn²⁺ at 0.480 M.

2. **Determine the Standard Cell Potential (E°):**
- For a concentration cell involving the same species, the standard cell potential (E°) is always 0 V because the half-reactions involve the same species (Zn²⁺).

3. **Use the Nernst Equation:**
- The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log K_{equilibrium} \]
- Here, \( n \) is the number of electrons transferred in the reaction. For Zn, \( n = 2 \).

4. **Calculate the Reaction Quotient (K):**
- The equilibrium constant (K) for the concentration cell can be calculated using the concentrations:
\[ K = \frac{[\text{Zn}^{2+}]_{cathode}}{[\text{Zn}^{2+}]_{anode}} = \frac{0.480}{0.024} = 20 \]

5. **Substitute Values into the Nernst Equation:**
- Substitute \( E^\circ_{cell} = 0 \), \( n = 2 \), and \( K = 20 \):
\[ E_{cell} = 0 - \frac{0.059}{2} \log(20) \]
- Calculate \( \log(20) \):
\[ \log(20) \approx 1.301 \]
- Therefore:
\[ E_{cell} = -\frac{0.059}{2} \times 1.301 \approx -0.0384 \, V \]

6. **Effect of Adding Water:**
- If water is added to the left electrode, the concentration of Zn²⁺ decreases to 0.012 M.
- Recalculate K with the new concentration:
\[ K' = \frac{0.480}{0.012} = 40 \]
- Substitute this back into the Nernst equation:
\[ E'_{cell} = 0 - \frac{0.059}{2} \log(40) \]
- Calculate \( \log(40) \):
\[ \log(40) \approx 1.602 \]
- Therefore:
\[ E'_{cell} = -\frac{0.059}{2} \times 1.602 \approx -0.0473 \, V \]

7. **Conclusion:**
- The EMF of the cell increased when the concentration of Zn²⁺ in the left half-cell decreased due to the addition of water.