whats is the reduction in 2KMnO4 + 10FeSO4 + 8H2SO4 = K2SO4 + 2MnSO4 + 5Fe2(So4)3 + H20
To determine the reduction in the given reaction of 2KMnO4 + 10FeSO4 + 8H2SO4 = K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + H2O, we need to identify the changes occurring to the manganese ions (Mn) in the reaction.
In this reaction, potassium permanganate (KMnO4) is being reduced, meaning it is losing electrons. The manganese ion (Mn) in KMnO4 has a +7 oxidation state. After the reaction, it is in the +2 oxidation state in MnSO4.
The number of electrons involved in the reduction can be determined by comparing the change in oxidation states of manganese in KMnO4 and MnSO4.
In KMnO4:
Mn has an oxidation state of +7.
In MnSO4:
Mn has an oxidation state of +2.
To calculate the reduction, we determine the difference in oxidation states:
+7 - (+2) = 5
Therefore, 5 electrons are involved in the reduction of manganese ions in the given reaction.