MnO4^– + SO3^2– + H2O → MnO2 + SO4^2– + OH^–
If the reaction occurs in an acidic solution where water and H+ ions are added to the half-reactions to balance the overall reaction, how many electrons are transferred in the balanced reduction half-reaction?
The answer I got is 3*2 which equals 6.
Mn goes from +7 to +4 : gained three.
S goes from +4 to +6: lost two
O is unchanged
H is unchanged
so above total is six, once you balance electrons. (2Mn, 3S)
Howefer, I'm having trouble with this problem because in an acidic solution KMnO4 goes to Mn^2+ and not MnO2.
I noticed that as well, Dr. Bob. I'm not quite sure why my teacher has it set up this way.
Thank you for clearing this up. So it isn't a typo. I suspect your teacher just goofed with the problem. As it stands, however, it isn't possible to answer the question becasue the question is flawed..
To determine the number of electrons transferred in the balanced reduction half-reaction, we need to examine the changes in oxidation states of the relevant elements involved. In this case, the element undergoing reduction is Mn (from MnO4^- to MnO2).
Initially, Mn in MnO4^- has a +7 oxidation state. In MnO2, the oxidation state is +4.
To balance the reduction half-reaction, we need to calculate the difference in oxidation states, which is 7 - 4 = 3. This tells us that 3 electrons are gained by each Mn atom during reduction.
Therefore, the correct answer is 3 electrons transferred in the balanced reduction half-reaction.