Question

All the energy released from the reaction \[\ce{X -> Y}\], Δ_rG° = −193 KJ mol⁻¹ is used for oxidizing \[\ce{M+ -> M^{3+} + 2e−}\], E° = −0.25 V.

Under standard conditions, the number of moles of M⁺ oxidized when one mole of X is converted to Y is [F = 96500 C mol⁻¹].

Answer 1

Given: Δ_r​G° = −193 kJ mol⁻¹

= −193000 J mol⁻¹

Oxidation half-reaction: \[\ce{M+ −> M^3+ + 2e−}\]

Standard electrode potential, (E°) = −0.25 V

Faraday constant, (F) = 96500 C mol⁻¹

Energy available = electrical energy required to oxidize M⁺:

ΔG = nFE

Where:

n = Number of moles of electrons transferred

F = Faraday constant

E = Standard electrode potential

Let x be the number of moles of M⁺ oxidized when one mole of X is converted to Y.

Each mole of M⁺ loses 2 electrons, so:

ΔG = x × 2 × F × E

Substitute values:

193000 = x × 2 × 96500 × 0.25

193000 = x × 48250

x = `193000/48250`

x = 4

So, 4 moles of M⁺ are oxidized when 1 mole of X is converted to Y under standard conditions.