Final answer:
Iodine (I) is the element that is oxidized in the reaction I- + MnO4- + H+ → I2 + MnO2 + H2O, as its oxidation state increases from -1 to 0.
Step-by-step explanation:
The element that is oxidized in the reaction I- + MnO4- + H+ → I2 + MnO2 + H2O is iodine (I). In this reaction, the iodide ion (I-) is oxidized to molecular iodine (I2). Oxidation involves an increase in oxidation state, which in this case, is from -1 in I- to 0 in I2. Since the iodide ion loses electrons during the process, it is undergoing oxidation. On the other hand, manganese (Mn) is reduced from MnO4- (with a +7 oxidation state for Mn) to MnO2 (with a +4 oxidation state for Mn).
The concept of oxidation and reduction can be remembered by the mnemonic “OIL RIG”, which stands for “Oxidation Is Loss, Reduction Is Gain” of electrons. In the given redox reaction, multiple steps of electron transfer are implied, and by comparing the oxidation states of the elements involved in reactants and products, we can identify what is being oxidized and reduced.