[habbababba]

1) According to the first ionic half-equation:
 
MnO4- + 8H+ + 5e-  --> Mn2+ + 4H2O

it seems to me that, in terms of exchange of oxygen atoms, the H+ ions are getting oxidized (gaining oxygen) and the Mn in MnO4- is reduced (since the H+ ions are stripping oxygen atoms off of Mn). Knowing that the main source of H+ in the solution is the sulfuric acid, can it be said that sulfuric acid is a reducing agent in this half-equation, even though it has the reputation of being an oxidizing agent? If it isn't, then what is its role in this reaction?

2) The e- in the half-equation above are coming from the Fe2+ ions according to the half-equation Fe2+ --> Fe3+ + e-.
So in terms of electron exchange, Fe2+ are oxidized (losing e-).
Look what we have here: One species is reduced (Mn) but 2 species are oxidized (H+ and Fe2+)

[Hunter2]

No. H⁺ is already oxidised. H⁺ change the position from the Acid to water nothing else.

The Redoxpairs are MnO₄^-/Mn²⁺ and Fe²⁺/Fe³⁺

[habbababba]

No. H⁺ is already oxidised. H⁺ change the position from the Acid to water nothing else.

But to change position from acid to water, H⁺ requires  gaining oxygen, which is defined as oxidation, hence the confusion.

[mjc123]

But H⁺ in water exists as H₃O⁺ - it's already bonded to oxygen. Even if it wasn't, its oxidation number doesn't change. That's what you want to go by - "gain/loss of oxygen" or "loss/gain of hydrogen" don't always work as definitions.

[habbababba]

"gain/loss of oxygen" or "loss/gain of hydrogen" don't always work as definitions.

1) What about "gain/loss of e^-"? In the reduction half-equation above, which species exactly is gaining the electrons? Is it the H⁺?

2) So after all, what is the role of sulfuric acid? Will the reaction between iron sulfate and potassium permanganate take place without the presence of sulfuric acid?

[Hunter2]

Additionally in Sulfuric acid is the H⁺ also bonded to Oxygen of the Sulfate ion.
The H⁺ has oxidation number +1 and it will not change, bonded to Oxygen in Water molecule or Sulfuric acid molecule.

In the half reaction Manganese-VII in Permanganate gain 5 Electrons to form Manganese-II.

The iron-II loose one electron to form iron-III.

[habbababba]

In the half reaction Manganese-VII in Permanganate gain 5 Electrons to form Manganese-II.

Thanks for the clarification. So if I want to highlight the role of sulfuric acid in this reaction, it would be as follows:

the manganese-VII in Permanganate can gain 5 e^- if the oxygen atoms are stripped off. This is achieved by the H⁺ ions released by sulfuric acid.

Is this correct?

[Hunter2]

Yes you can say so.

General receipe, if oxygen is involved:

Oxidation in acidic condition: on educt side you add water on product side you gain H⁺

Reduction in acidic condition goes vise versa like the Permanganate in your example: on educt side you add  H⁺ and on product side you gain water.

Oxidation in alcaline condition: on educt side you add  OH^- and on product side you gain water.

Reduction in alcaline condition goes vise versa : on educt side you add  water and on product side you gain OH^-.