QJEEM13Chd1e1: JEE Main 2013 Chemistry - Class 12 JEEM13Chd1e1 Solution: Given E C r 2 + / C r o = -0.74 V; E M n O 4 − / M n 2 + o =…

In electrochemistry, the strongest oxidizing agent will be the one with the highest standard electrode potential (E°), because a higher E° value means a greater tendency to gain electrons, i.e., get reduced. An oxidizing agent gains electrons and in doing so, oxidizes another species.

From the provided data, the species with the highest standard electrode potential (E°) is MnO₄⁻, with an E° of 1.51 V. This means that MnO₄⁻ has the greatest tendency to gain electrons and thus is the strongest oxidizing agent.

Therefore, the correct answer is :

Option C : $$Mn{O}_4^{-}$$ .