For the reaction `:`
`A+2B^(3+) rarr 2B^(2+)+A^(2+)`
`E^(c-)` of the given redox reaction is `:`
`A^(2+)+2e^(-) rarr A` `E^(c-)=Xv`
`B^(3+)+e^(-) rarr B^(2+)` `E^(c-)=yV`

Options

A. `x-2y`
B. `x+y//2`
C. `x-y`
D. `y-x`

Explanation

To solve the problem, we need to determine the standard cell potential (E°cell) for the given redox reaction:

**Given Reaction:**
\[ A + 2B^{3+} \rightarrow 2B^{2+} + A^{2+} \]

**Half-Reactions:**
1. Oxidation (Anode):
\[ A \rightarrow A^{2+} + 2e^{-} \]
The standard reduction potential for this half-reaction is given as:
\[ E^{\circ}_{\text{oxidation}} = -X \, \text{V} \]
Therefore, the standard reduction potential for the reverse reaction (reduction) is:
\[ E^{\circ}_{\text{reduction}} = X \, \text{V} \]

2. Reduction (Cathode):
\[ 2B^{3+} + 2e^{-} \rightarrow 2B^{2+} \]
The standard reduction potential for this half-reaction is given as:
\[ E^{\circ}_{\text{reduction}} = Y \, \text{V} \]

**Calculating E°cell:**
The standard cell potential (E°cell) can be calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]

Substituting the values:
- The cathode reaction (reduction of B) has a potential of \( Y \, \text{V} \).
- The anode reaction (oxidation of A) has a potential of \( X \, \text{V} \).

Thus, we have:
\[ E^{\circ}_{\text{cell}} = Y - X \]

**Final Answer:**
The total standard cell potential for the given redox reaction is:
\[ E^{\circ}_{\text{cell}} = (Y - X) \, \text{V} \]

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