All the energy realesed from the reation `X rarr Y, Delta_(r) G^(@) = -193 kJ mol^(-1)`, is used for oxidizing `M^(+)` as `M^(+) rarr M^(3+) + 2e^(-), E^(@) = -0.25 V`. Under standard consistions, the number of moles of `M^(+)` oxidized when on e mol of `X` is converted to `Y` is `[F = 96,500 C "mol"^(-1)]`

Answer

4

Explanation

Given
`X to Y, Delta_(r)G^(@) = -193 kJ mol^(-1)`
`M^(+) to M^(3+) + 2e^(-), E^(@) = -0.25 V`
Let 193 kJ energy is used for oxidising .n. moles of `M^(+)` for 1 mol of `M^(+)`
`DeltaG^(@) = -n FE^(@)`
` = -2 xx 96500 xx (-0.25)`
48250 J mol or = 48.250 kJ
`therefore` Number of moles of `M^(+)` oxidised when X is converted to `Y = (193)/(48.250) = 4`