MnO4(aq) + 8H + (aq) + 5Fe2+ → Mn2+(aq) + 4H2O(I). The oxidation number of manganese in the above reaction changed from?

In the given chemical reaction, MnO₄^- ion is reduced to Mn²⁺ ion, indicating a decrease in the oxidation number of manganese. To determine the change in oxidation number, we need to know the oxidation state of manganese in the reactant and product side. In MnO₄^-, oxygen has an oxidation state of -2, so the oxidation state of Mn can be calculated as follows: (+4) + 4(-2) = -8 x + (-8) = -1 (the net charge of MnO₄^-) Solving for x, the oxidation state of Mn in MnO₄^- is +7. On the product side, there are two H₂O molecules, so the oxidation state of Mn can be calculated as follows: x + 4(+2) + 2(-2) = 0 (the net charge of Mn2+) Solving for x, the oxidation state of Mn in Mn²⁺ is +2. Therefore, the oxidation state of Mn changes from +7 to +2 in the given reaction, which means that the correct answer is "+7 to +2".