If `DeltaG^(@)` for the half cell `MnO_(4)^(-) | MnO_(2)` in an acid solution is xF then find the value of x.
(Given: `E_(MnO_(4)^(-) | Mn^(2+))^(@) = 0.34 V, E_(Sn^(4+)|Sn^(2+))^(@) = 0.15 V`)

Explanation

To solve the problem, we need to find the value of \( x \) in the equation \( \Delta G^\circ = xF \) for the half-cell reaction \( \text{MnO}_4^{-} | \text{MnO}_2 \) in an acidic solution. We will use the relationship between Gibbs free energy change, the number of electrons transferred, and the standard electrode potential.

### Step-by-Step Solution:

1. **Identify the Half-Reaction:**
The half-reaction for the conversion of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \) in acidic medium can be written as:
\[ \text{MnO}_4^{-} + 4 \text{H}^+ + 3 \text{e}^- \rightarrow \text{MnO}_2 + 2 \text{H}_2O \]
Here, \( n = 3 \) (the number of electrons transferred).

2. **Calculate the Standard Electrode Potential (\( E^\circ \)):**
We need to find the standard electrode potential for the half-reaction. The standard electrode potential for the half-cell \( \text{MnO}_4^{-} | \text{Mn}^{2+} \) is given as \( E^\circ = 0.34 \, \text{V} \).

However, we need the potential for the reaction involving \( \text{MnO}_2 \). We can use the following relationship:
\[ E^\circ_{\text{MnO}_4^{-} | \text{MnO}_2} = E^\circ_{\text{MnO}_4^{-} | \text{Mn}^{2+}} - E^\circ_{\text{MnO}_2 | \text{Mn}^{2+}} \]
Since \( E^\circ_{\text{MnO}_2 | \text{Mn}^{2+}} \) is not provided, we will assume that we are using the potential directly for the half-reaction involving \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \).

3. **Use the Gibbs Free Energy Equation:**
The relationship between Gibbs free energy change and standard electrode potential is given by:
\[ \Delta G^\circ = -nFE^\circ \]
Substituting \( n = 3 \) and \( E^\circ = 0.34 \, \text{V} \):
\[ \Delta G^\circ = -3F(0.34) \]

4. **Relate \( \Delta G^\circ \) to \( xF \):**
According to the problem, we have:
\[ \Delta G^\circ = xF \]
Therefore, we can set the two expressions for \( \Delta G^\circ \) equal to each other:
\[ xF = -3F(0.34) \]

5. **Solve for \( x \):**
Dividing both sides by \( F \) (assuming \( F \neq 0 \)):
\[ x = -3(0.34) = -1.02 \]

### Final Answer:
The value of \( x \) is \( -1.02 \).