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Given : E°(I2/2I−) = 1.36 V, E°(Cr3+/Cr) = −0.74 V E°(Cr2O7^2−/Cr3+) = 1.33 V, E°(MnO4−/Mn2+) = 1.51 V The correct order of reducing power of the species (Cr, Cr3+, Mn2+ and Cl−) will be:

1. Mn2+ < Cl− < Cr3+ < Cr
2. Mn2+ < Cr3+ < Cl− < Cr
3. Cr3+ < Cl− < Mn2+ < Cr
4. Cr3+ < Cl− < Cr < Mn2+

Correct answer: Mn2+ < Cl− < Cr3+ < Cr

Solution

The reducing power of a species is inversely related to its standard reduction potential; lower potential indicates stronger reducing ability. Since Mn2+ has the lowest potential, followed by Cl−, then Cr3+, and finally Cr, which has the highest potential, the correct order is Mn2+ < Cl− < Cr3+ < Cr.

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