Balance thefollowing equation by the Oxidation number
method
(Acidic medium)
Fe^2++H+cr2O7^2-< cr^3++Fe^3++H2O
Answers
Explanation:
step 1
Fe2+ + Cr2O72- + H+ → Fe3+ + Cr3+ + H2O
step 2
Fe+22+ + Cr+62O-272- + H+1+ → Fe+33+ + Cr+33+ + H+12O-2
step 3
O:Fe+22+ → Fe+33+(Fe)
R:Cr+62O-272- → Cr+33+(Cr)
step 4
O:Fe+22+ → Fe+33+
step 5
O:Fe2+ → Fe3+
R:Cr2O72- → 2Cr3+ + 7H2O
R:Cr+62O-272- → Cr+33+
step 6
O:Fe2+ → Fe3+ + e-
R:Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
step 7
O:Fe2+ → Fe3+ + e-| *6
R:Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O| *1
O:6Fe2+ → 6Fe3+ + 6e-
R:Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
step 8
6Fe2+ + Cr2O72- + 14H+ + 6e- → 6Fe3+ + 2Cr3+ + 6e- + 7H2O
step 10
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
step 11
doesn't matter what the charge is as long as it is the same on both sides.
6*2 + 1*-2 + 14*1 = 6*3 + 2*3 + 7*0
24 = 24
Since the sum
Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal we can write a balanced equation.
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O