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Manganese dioxide reacts with bromide ions to form manganese ions and bromate ions in an acidic aqueous solution. Write the balanced equation for this reaction.
Select one:

a. Br- + MnO2 + 3H2O Mn2+ + BrO3- + 2H3O+

b. Br- + 3MnO2 + 6H3O+ 3Mn2+ + 9H2O + BrO3-

c. 6Br- + MnO2 + 48H2O Mn2+ + 32H3O+ + 6BrO3-

d. Br- + 3MnO2 + 3H3O+ 3Mn2+ + BrO3- + 3H2O

e. 4Br- + 3MnO2 + 18H2O 3Mn2+ + 12H3O+ + 4BrO3-

MnO2 + Br^- ==> BrO3^- + Mn^2+
Here is a site that shows you how to balance the two half reactions.

http://www.chemteam.info/Redox/Redox.html

Hint: Mn changes from +4 in MnO2 to +2 in Mn^2+. Br changes from -1 in Br^- to +5 in BrO3^-

For MnO2 it says the product is Mn2+ (manganese ions), so write

For the second half-cell reaction, I got Br- + 3H2O ---> BrO3- + 6H+ 6e-. Now I believe that I have to multiply the manganese half-cell reaction by three to cancel out the number of electrons. Then, my answer would be Br- + 3MnO2 + 6H+ ---> 3MnO2+ + BrO3- + 3H2O. Is this correct?