MnO4- (aq)+ ClO2- (aq)+ H2O(l) -> MnO2 + ClO4 (aq) +OH- Balance the redox equati
ID: 686382 • Letter: M
Question
MnO4- (aq)+ ClO2- (aq)+ H2O(l) -> MnO2 + ClO4 (aq) +OH-Balance the redox equation using the half reaction method. Getoxidation numbers, than half reaction. Balance # electrons. Check Hand O.
What species is oxidized?
What species is reduced?
How many electron are transferred?
What species is the oxidizing agent?
I don't think I balance the equation right.
Reduce MnO4- + 3e- -> MnO2
Oxidize ClO2- -> ClO4+4e-
What species is oxidized? I said Cl is oxidized
What species is reduced? I said Mn is reduced.
How many electrons is transferred? I said 12 electrons..
What species is is the oxidizing agent? I said Mn is the oxidizingagent.
Explanation / Answer
Let us write the oxidation numbers on the top of the chemicalspecies +7 -2 +3 -2 +1 -2 +4 -2 +7-2 -2 +1 MnO4- (aq) + ClO2- (aq)+ H2O(l) -> MnO2 + ClO4- (aq) + OH-The o n of Mn decreased from +7 to+4 (reduction) The o n of Cl increased from +3 to +7 (oxidation) MnO4- oxidizing the ClO2- ion, so it is oxidizingagent. ClO2_ reducing the MnO4- ion, so it is reducing agent. now let us balance 1. divide the eq in two oxidation halfreaction reduction half reaction ClO2- ------------> ClO4- MnO4- ------------------> MnO2 2.balance the other atoms except O & H, no change ClO2- ------------> ClO4- MnO4- ------------------> MnO2 3. add H2O atoms at 'O'shortage & add sufficientH+ ions to counter balance ClO2- +2H2O ------------> ClO4- +4H+ MnO4- + 4H+ ------------------> MnO2+2H2O 4.Now balance the charge by adding electrons ClO2- +2H2O------------> ClO4- +4H++4e- MnO4- +4H++3e- ------------------> MnO2+2H2O to make equal the number of electrons transfrd ,do as follows 3[ClO2- +2H2O------------> ClO4- +4H++4e-] 4[MnO4- +4H++3e- ------------------> MnO2+2H2O] then add the both equations,simplify by canceling the common termson both sides ,on rewriting ,we get it