2007-04-30 10:25:16 · 4 answers · asked by adrian 1 in Science & Mathematics ➔ Chemistry
MnO- + 8H+ + C2O4 2- ---> Mn2+ +4H2O + 2CO2
2007-04-30 10:43:02 · answer #1 · answered by xpuredudex 1 · 0⤊ 1⤋
that's an oxidation-help (redox) equation and must be balanced in accordance to the strategies for redox equations. First, verify the oxidation numbers of each and every component whereever apparently interior the equation on the left section we've Mn^+7 O^-2 H^+a million Cl-a million on the right section we've Mn^+2 Cl^0 H^+a million O^-2 Mn decreases from +7 to +2. Cl will enhance from -a million to 0. Mn^+7 + 5 e --> Mn^+2 2 Cl^-a million --> Cl2 + 2 e Multiply the first equation through 2 and the 2d equation through 5 to get a stability of the electrons. 2 Mn^+7 + 10 e --> 2 Mn^+2 10 Cl^-a million --> 5 Cl2 + 10 e Now end off the equation. 2 MnO4- + 16 H+ + 10 Cl- --> 2 Mn2+ + 5 Cl2 + 8 H2O
2016-11-23 18:07:09 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The numbers you need are:
2, 16, 5, 2, 8, 10
2007-04-30 10:49:22 · answer #3 · answered by Gervald F 7 · 0⤊ 0⤋
2MnO4- + 16H+ +C2O42- ---> Mn2+ +8H2O +2CO2
2007-04-30 15:07:23 · answer #4 · answered by Lesya 1 · 0⤊ 0⤋