toxin - 29-10-2005 at 12:38
1)I am asked to balance this redox equation yet I am not given wether the reaction takes place in an acidic or aqueous solution, how would you go
about balancing this ?
MnO4^-1 + Cl^-1 + H^+1 ---> Mn^+2 + Cl(g) + H2O(l)
2)
how would you split this equation into half reactions ?
ClO2(aq) + OH-(aq) + ---> ClO3-(aq) + Cl-(aq) + H2O(l) (in a basic solution)
scince the oxidation numbers in OH- and H2O don't change should I omit from including them in the half reactions like this ?
ClO2 --> ClO3-
ClO2 --> Cl-
[Edited on 29-10-2005 by toxin]
The_Davster - 29-10-2005 at 12:50
I will not give you the answer, but I will tell you how to go about doing these questions.
For the first one, you are saying that it is not given to you in the question whether it occurs in acidic or basic medium, however, if you look at the
equation you are asked to balance there is a H+ entity, therefore the rxn is in acidic medium. Then go from there to get your answer.
For the second it is a disproportionation reaction so ClO2 will be in both half reactions. Like you said the (skeleton) half reactions for this
reaction will be
ClO2-->ClO3- and ClO2-->Cl-, and you create the half reactions from there assuming a basic medium as it says in the question. You are correct
in thinking the OH- is removed while creating the skeleton half reactions, but you will have to add it in later in order the get the full half
reactions.