The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:`
`Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V`
`X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V`
When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`.
If a half call `X|X^(2)(0.1M)` is connected to another half cell `Y|Y^(2+)(1.0M)` by means of a salt bridge and an external circuit at `25^(@)C`, the cell voltage would be

Options

A. `0.06V`
B. `0.12V`
C. `0.62V`
D. `0.72V`

Explanation

To solve the problem step by step, we will use the information provided about the standard reduction potentials of metals X and Y, and apply the Nernst equation to calculate the cell voltage.

### Step 1: Identify the Standard Reduction Potentials
We have the following standard reduction potentials:
- For Y: \( Y^{2+} + 2e^- \rightarrow Y \) with \( E^\circ = +0.34 \, \text{V} \)
- For X: \( X^{2+} + 2e^- \rightarrow X \) with \( E^\circ = +0.25 \, \text{V} \)

### Step 2: Determine the Anode and Cathode
Since electrons flow from X to Y, X is the anode (where oxidation occurs) and Y is the cathode (where reduction occurs).

### Step 3: Write the Half-Reactions
- Oxidation at the anode (X):
\[ X \rightarrow X^{2+} + 2e^- \]
- Reduction at the cathode (Y):
\[ Y^{2+} + 2e^- \rightarrow Y \]

### Step 4: Calculate the Standard Cell Potential (\( E^\circ_{\text{cell}} \))
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - 0.25 \, \text{V} = 0.09 \, \text{V} \]

### Step 5: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[X^{2+}]}{[Y^{2+}]} \right) \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( [X^{2+}] = 0.1 \, \text{M} \)
- \( [Y^{2+}] = 1.0 \, \text{M} \)

### Step 6: Substitute Values into the Nernst Equation
Substituting the known values:
\[ E_{\text{cell}} = 0.09 - \frac{0.059}{2} \log \left( \frac{0.1}{1.0} \right) \]

### Step 7: Calculate the Logarithm
Calculating the logarithm:
\[ \log \left( \frac{0.1}{1.0} \right) = \log(0.1) = -1 \]

### Step 8: Substitute Back into the Nernst Equation
Now substituting back:
\[ E_{\text{cell}} = 0.09 - \frac{0.059}{2} \times (-1) \]
\[ E_{\text{cell}} = 0.09 + 0.0295 \]
\[ E_{\text{cell}} = 0.1195 \, \text{V} \approx 0.12 \, \text{V} \]

### Final Answer
The cell voltage would be approximately \( 0.12 \, \text{V} \).