The correct option is A 10FeSO4+2KMnO4+8H2SO4→5Fe2(SO4)3+2MnSO4+8H2O+K2SO4
Step-1 : Assign the oxidation number to each element present in the reaction. +2 +6 −2FeSO4++1 +7 −2KMnO4++1 +6 −2H2SO4→+3 +6 −2Fe2(SO4)3++2 +6 −2MnSO4++1 −2H2O
Step 2. Remove all the spectator ions and convert the reaction into its ionic form.
Spectator ions - The ions whose oxidation numbers do not change during the reaction Fe2+++7MnO−4→Fe3++Mn2+
Step 3 : Split the reaction into two halves i.e. the oxidation half and the reduction half
Determine the oxidation states of the atoms and balance the reaction atom-wise and then electrons-wise. Fe2+oxidation−−−−−→Fe3+∣∣ MnO−4Reduction−−−−−−→Mn2+
Step 4: Balance the atom other than oxygen and hydrogen atom in both half reactions Fe2+→Fe3+| MnO−4→Mn2+
Fe and Mn atoms are balanced on both side.
Step 5: Now balance O and H atom by H2O and H+
respectively by the following way: For one excess oxygen atom, add one H2O
on the other side and two H+
on the same side. Fe2+→Fe3+
(no oxygen atom) ...(i) 8H++MnO−4→Mn2++4H2O
Step 6: Equation (i) and (ii) are balanced atom wise. Now balance both equations charge wise. To balance the charge, add electrons to the electrically positive side. Fe2+oxidation−−−−−→Fe3++e−
Step 7: The number of electrons gained and lost in each half reaction are equalised by multiplying both the half reactions with a suitable factor and finally the half reactions are added to give the overall balanced reaction.
Here, we multiply equation (1) by 5 and (2) by 1 and add them: Fe2+→Fe3++e−
(Here, at this stage, you will get balanced redox reaction in Ionic form)
Step IX: Now convert the Ionic reaction into molecular form by adding the elements or species, which are removed in step (2).
Now, by some manipulation, you will get: 5FeSO4+KMnO4+4H2SO4→52Fe2(SO4)3+MnSO4+4H2O+12K2SO4