All the energy released from the reaction `XtoY,Delta _(r)G^(@) = -193KJmol^(-1)` is used for oxidizing `M^(+)` as `M^(+)toM^(3+)+ 2e^(-)`, `e^(@) =-0.25V`. Under standard conditions, the number of moles of oxidized when one mole of X is converted to Y is `[F = 96500Cmol^(-1)]`

Explanation

To solve the problem, we need to determine how many moles of \( M^+ \) are oxidized when one mole of \( X \) is converted to \( Y \) using the energy released from the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Data
- The reaction \( X \to Y \) has a Gibbs free energy change \( \Delta_r G^\circ = -193 \, \text{kJ/mol} \).
- The oxidation reaction is \( M^+ \to M^{3+} + 2e^- \) with a standard reduction potential \( E^\circ = -0.25 \, \text{V} \).
- The Faraday constant \( F = 96500 \, \text{C/mol} \).

### Step 2: Calculate the Energy Available for Oxidation
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