why does h2o in redox equations disappear
6MnO4– + 10Cr3+ + 11H2O → 6Mn2+ + 5Cr2O72– + 22H+
this is the redox equation of: 2 Cr3+ + 7 H2O Cr2O72– + 14 H+ + 6 e–
and
MnO4– + 8H+ + 5e- Mn2+ + 4H2O
i multiplied one by 5 and the other by 6 so that there are both 30 electrons
where did the h2os go and where are there 11 h+, 22h+ ?
this is the redox equation of: 2 Cr3+ + 7 H2O Cr2O72– + 14 H+ + 6 e–
and
MnO4– + 8H+ + 5e- Mn2+ + 4H2O
i multiplied one by 5 and the other by 6 so that there are both 30 electrons
where did the h2os go and where are there 11 h+, 22h+ ?
Original post by esmeralda123
6MnO4– + 10Cr3+ + 11H2O → 6Mn2+ + 5Cr2O72– + 22H+
this is the redox equation of: 2 Cr3+ + 7 H2O Cr2O72– + 14 H+ + 6 e–
and
MnO4– + 8H+ + 5e- Mn2+ + 4H2O
i multiplied one by 5 and the other by 6 so that there are both 30 electrons
where did the h2os go and where are there 11 h+, 22h+ ?
this is the redox equation of: 2 Cr3+ + 7 H2O Cr2O72– + 14 H+ + 6 e–
and
MnO4– + 8H+ + 5e- Mn2+ + 4H2O
i multiplied one by 5 and the other by 6 so that there are both 30 electrons
where did the h2os go and where are there 11 h+, 22h+ ?
The water molecules do not disappear.
6MnO4– + 10Cr3+ + 11H2O → 6Mn2+ + 5Cr2O72– + 22H+
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