CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

Identify the correctly balanced redox reaction using ion-electron method.
FeSO4+KMnO4+H2SO4Fe2(SO4)3+MnSO4+H2O+K2SO4

A
10FeSO4+2KMnO4+8H2SO45Fe2(SO4)3+2MnSO4+8H2O+K2SO4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2FeSO4+10KMnO4+4H2SO42Fe2(SO4)3+5MnSO4+8H2O+10K2SO4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
8FeSO4+10KMnO4+2H2SO45Fe2(SO4)3+4MnSO4+H2O+2K2SO4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10FeSO4+8KMnO4+2H2SO44Fe2(SO4)3+2MnSO4+5H2O+K2SO4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 10FeSO4+2KMnO4+8H2SO45Fe2(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+++7MnO4Fe3++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+ MnO4Reduction−−−−−Mn2+

Step 4: Balance the atom other than oxygen and hydrogen atom in both half reactions
Fe2+Fe3+| MnO4Mn2+

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++MnO4Mn2++4H2O ...(ii)

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 ...(1)
5e+8H++MnO4Reduction−−−−−Mn2++4H2O ...(2)

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 ...(1) × 5

5e+8H++MnO4Mn2++4H2O ...(2) × 1
______________________________________________
5Fe2++8H++MnO45Fe2++Mn2++4H2O
(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+4H2SO452Fe2(SO4)3+MnSO4+4H2O+12K2SO4
or 10FeSO4+2KMnO4+8H2SO45Fe2(SO4)3+2MnSO4+8H2O+K2SO4.

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image