wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Balance by Ion electon method, Cr2O27+Fe+2Cr+3+H2O (in acidic medium).

A
Cr2O27+2Fe+2+14H+Cr+3+2Fe+3+7H2O
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Cr2O27+Fe+2+2H+2Cr+3+Fe+3+H2O
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Cr2O27+3Fe+2+14H+2Cr+3+3Fe+3+H2O
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Cr2O27+6Fe+2+14H+2Cr+3+6Fe+3+7H2O
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D Cr2O27+6Fe+2+14H+2Cr+3+6Fe+3+7H2O
Cr2O27+Fe+2Cr+3+Fe3++H2O
Reduction half :
+6Cr2O27+3Cr3+
Oxidation half :
Fe+2Fe+3
By taking reduction half reaction balance the Cr first.

Cr2O272Cr+3
Balance O by adding 7H2O to R.H.S. and balance H by adding 14 H+ to L.H. S.
Cr2O27+14H+2Cr+3+7H2O

Above equation is balanced atom wise. Now balance both equations charge wise. To balance the charge, add electrons to the electrically positive side.
Cr2O27+14H++6e2Cr+3+7H2O ----(1)

In oxidation half cell mass is balanced. So balance the charge :
Fe+2Fe+3+e----(2)

In equation (2) multiply by 6 and then add both equaiton (1) and (2).
Cr2O27+6Fe+2+14H+2Cr+3+6Fe+3+7H2O

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon