The order of reducing potential is H⁺< Cu⁺² < I₂< Fe⁺³ < Au⁺³ < F₂ .
Given:
Fe⁺³ (aq) + e⁻ → Fe⁺² (aq) E° = +0.77 v
Cu⁺² (aq) + e⁻ → Cu⁺ (aq) E°= +0.15 V
F₂ (g) + 2e⁻ → 2F⁻ (aq) E° = +2.87 V
2H⁺ (aq) + 2e⁻ → H₂ (g) E° = 0.00 V
Au⁺³ (aq) + 3e⁻ → Au(s) E° = 1.50 V
I₂ (s) + 2e⁻ → 2I⁻ (aq) E° = 0.53 V
Smallest reducing potential means strongest oxidation potential.
So, the order of reducing potential is H⁺< Cu⁺² < I₂< Fe⁺³ < Au⁺³ < F₂ .
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