A volataic cell consists of an electode of solide silver immerse in a `0.10M AgNO_(3)` solution and an electrode of unknown metal `'X'` immersed in a `0.10M` solution `X(NO_(3))_(2)`. A porous barrier separates the two half of the cell. Also given `:`
`E^(c-)._((Ag^(o+)|Ag))=0.80V` and `E^(c-)._(cell)=1.05V` at `25^(@)C`
`X(s)|X^(2+)(1.0M)||Ag^(o+)(0.1M)|Ag(s)`
If `Ag^(o+)|Ag` half cell in the above voltaic cell is replaced by `Zn^(2+)|Zn` half cell `(E^(c-)._(Zn^(2+)|Zn)=-0.76V)`

Options

A. The direction of current flow will remain same.
B. Polarity of the electrodes will be reversed.
C. Cell will stop working.
D. `EMF` of the cell will increase.

Explanation

To solve the problem, we need to analyze the voltaic cell consisting of a solid silver electrode in a 0.10 M AgNO3 solution and an unknown metal 'X' in a 0.10 M X(NO3)2 solution. We will also consider the effect of replacing the silver half-cell with a zinc half-cell.

### Step-by-Step Solution:

1. **Identify the Half-Cells:**
- The first half-cell consists of silver: \( \text{Ag}^+ | \text{Ag} \) with \( E^\circ = 0.80 \, \text{V} \).
- The second half-cell consists of the unknown metal: \( \text{X} | \text{X}^{2+} \).

2. **Determine the Cell Representation:**
- The cell representation is given as \( \text{X}(s) | \text{X}^{2+}(1.0 \, \text{M}) || \text{Ag}^+(0.1 \, \text{M}) | \text{Ag}(s) \).

3. **Calculate the Cell Potential (E_cell):**
- The overall cell potential is given as \( E_{cell} = 1.05 \, \text{V} \).
- The relationship for cell potential is given by:
\[ E_{cell} = E_{cathode} - E_{anode} \]
- Here, the cathode is the silver half-cell and the anode is the unknown metal half-cell.

4. **Set Up the Equation:**
- Let \( E_{X} \) be the standard reduction potential for the unknown metal X.
- The equation becomes:
\[ 1.05 = 0.80 - E_{X} \]

5. **Solve for E_X:**
- Rearranging the equation gives:
\[ E_{X} = 0.80 - 1.05 = -0.25 \, \text{V} \]

6. **Replace the Silver Half-Cell with Zinc Half-Cell:**
- The zinc half-cell is represented as \( \text{Zn}^{2+} | \text{Zn} \) with \( E^\circ = -0.76 \, \text{V} \).
- The new cell representation becomes \( \text{X}(s) | \text{X}^{2+} || \text{Zn}^{2+} | \text{Zn}(s) \).

7. **Calculate the New Cell Potential:**
- The new cell potential is:
\[ E_{cell} = E_{cathode} - E_{anode} \]
- Here, the cathode is now zinc, so:
\[ E_{cell} = -0.76 - E_{X} \]
- Substituting \( E_{X} = -0.25 \):
\[ E_{cell} = -0.76 - (-0.25) = -0.76 + 0.25 = -0.51 \, \text{V} \]

8. **Conclusion:**
- The new cell potential after replacing the silver half-cell with the zinc half-cell is \( -0.51 \, \text{V} \).

### Final Answer:
The cell potential after replacing the silver half-cell with the zinc half-cell is \( -0.51 \, \text{V} \).