JEE Advanced Chemistry question

Base on the E° values gives ½F2 + e⁻ → F⁻ E° = 2.87 V ½Cl2 + e⁻ → Cl⁻ E° = 1.40 V ½Br2 + e⁻ → Br⁻ E° = 1.09 V ½I2 + e⁻ → I⁻ E° = 0.62 V

(A)Oxidizing power of F2 is maximumCorrect
(B)Oxidizing power of I2 is maximum
(C)Oxidizing power is in order F2 > Cl2 > Br2 > I2Correct
(D)Oxidizing power is in order F2 < Cl2 < Br2 < I2

Answer: (A),(C)

Solution

A higher standard reduction potential corresponds to a stronger tendency to be reduced, i.e. a stronger oxidising agent. Since F2 has the highest listed E° (2.87 V) and I2 has the lowest (0.62 V), F2 is the strongest

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Question ID VK-159475