needed for the balanced half-equation?
2007-10-27 13:56:30 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
You need 8, each MnO4- provides 4 oxygens for water. So there are 4 H2O.
2007-10-27 14:04:00 · answer #1 · answered by cattbarf 7 · 0⤊ 1⤋
The balanced half-equation should be:
Mn(7+)O4(2-) + 6H(1+) + 11e(1-) ==> Mn(2+)O(2-) + H2(1+)O(2-)
You need 6 H+ to create water and 11 e- to reduce the hydrogen and the manganese.
2007-10-27 14:06:48 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Mn(+7) + 5e- ---> Mn(+2)
MnO4- + 5e- -> Mn(+2) [balance charge with H+, -6 on left, +2 on right]]
MnO4- + 8H(+) + 5e- -> Mn(+2) [balance H and O with H2O]
MnO4- + 8H(+) + 5e- -> Mn(+2) + 4H2O
2007-10-27 14:33:32 · answer #3 · answered by ChemistryMom 5 · 1⤊ 0⤋
its supposed to be 4H20 to balance out the oxygens on the left, so there should be 8H+ to balance out the Hydrogens on the left, with the Hydrogens on the right.
2007-10-27 14:01:52 · answer #4 · answered by Slava T 1 · 0⤊ 0⤋