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 standard emf `(E^(c-))` of a half cell `Y^(2)|Y^(o+)` is `0.15V`, the standard emf of the half cell `Y^(o+)|Y` will be

Options

A. `0.19V`
B. `0.53V`
C. `0.49V`
D. `0.64V`

Explanation

To find the standard emf of the half cell \( Y^{+1} | Y \), we can follow these steps:

### Step 1: Understand the given data
We have the following standard reduction potentials:
- For the half cell \( Y^{2+} + 2e^- \rightarrow Y \), the standard reduction potential \( E^{\circ}_{Y^{2+}/Y} = 0.34 \, V \).
- For the half cell \( X^{2+} + 2e^- \rightarrow X \), the standard reduction potential \( E^{\circ}_{X^{2+}/X} = 0.25 \, V \).
- The standard emf of the half cell \( Y^{2+} | Y^{+1} \) is given as \( 0.15 \, V \).

### Step 2: Define the unknown
Let the standard emf of the half cell \( Y^{+1} | Y \) be denoted as \( A \, V \).

### Step 3: Use the relationship of standard reduction potentials
According to the Nernst equation, the standard emf can be expressed as:
\[ E^{\circ}_{cell} = E^{\circ}_{reduction} - E^{\circ}_{oxidation} \]
For the half cell \( Y^{2+} + 2e^- \rightarrow Y \):
\[ E^{\circ}_{Y^{2+}/Y} = 0.34 \, V \]
For the half cell \( Y^{+1} + e^- \rightarrow Y \):
\[ E^{\circ}_{Y^{+1}/Y} = A \, V \]
The relationship can be set up as follows:
\[ 0.34 = 0.15 + A \]

### Step 4: Solve for \( A \)
Rearranging the equation:
\[ A = 0.34 - 0.15 \]
Calculating gives:
\[ A = 0.19 \, V \]

### Step 5: Finalize the answer
Thus, the standard emf of the half cell \( Y^{+1} | Y \) is:
\[ A = 0.19 \, V \]

### Conclusion
The standard emf of the half cell \( Y^{+1} | Y \) is \( 0.19 \, V \).

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