Question

Balance in neutral solution: MnO4- + S2O32- ---> SO42- + MnO2 I'm specifically having trouble with using H2O, H+, and OH- in balancing the half reactions.

Answer 1

step1: You split them into half reactions when ever you see that its a redox reaction

MnO4(-) ---> MnO2
SO3(2-) ---> SO4(2-)

As for what side H2O and H+ go on its useful to remember the pneumonic Major Hydroxide, or Major OH-. This tells us that after we have split it into half reactions we first balance the major species, for this example the major species are balanced (the Mn and the S).

step2: balance the O, to balance oxygen we add H2O to the opposite side of the equation (so that there are an equal number of O on either side ).

MnO4(-) --> MnO2 + 2H2O
SO3(2-) + H2O--> SO4(2-)

step3: balance the H+. To do this we add H+ to one side of the equation so that the number of hydrogen on either side of the arrow are equal.
MnO4(-) + 4H+ --> MnO2 + 2H2O
SO3(2-) + H2O--> SO4(2-) + 2H+

step4: the only thing left to do is balance electrons (e-), we do this by looking at the charge on either side of the equation,adding all (+) and( -) charges.
MnO4(-)+ 4H+ +3e- --> MnO2+ 2H2O (Before balance LHS +3, RHS 0, so add 3e- to LHS)
SO3(2-) + H2O--> SO4(2-) + 2H+ +2e- (Before balance LHS 2-, RHS 0 so add 2e- to RHS)

step5: we multiple each reaction by an integer so that when the two half reactions are added the e- cancel. Here,

(MnO4(-)+ 4H+ +3e- --> MnO2+ 2H2O) x2
(SO3(2-) + H2O--> SO4(2-) + 2H+ +2e) x3

2MnO4(-)+ 8H+ +6e- --> 2MnO2+ 4H2O
3SO3(2-) + 3H2O--> 3SO4(2-) + 6H+ +6e-
................................................................................
2 MnO4- + 3 SO32- +2H+ = 2 MnO2 + 3 SO42- + H2O