Posted by Aubrey on .
Using the change in oxidation number method, balance the redox reaction shown below.
(NO2^2)+ (MnO4^-)---> (NO3^-)+ Mn^2+
Divide into half reactions. I will show you how to do MnO4^- and you do the other one.
MnO4^- ==> Mn^+2
Step 2. Determine the oxidation state of the element changing. In this case, Mn changes from +7 on the left to +2 on the right.
Step 3. Add electrons to the appropriate side to balance the change in oxidation state.
MnO4^- + 5e ==> Mn^+2
Step 4. Count the charge on each side and add H^+ (in an acid solution) or OH^- (in a basic solution) to balance the charge.
I see -6 on the left and +2 on the right. This is an acid solution, therefore, add 8H^+ to the left.
MnO4^- + 8H^+ + 5e ==> Mn^+2
Step 5. Add H2O (usually to the other side) to balance H and O.
MnO4^- + 8H^+ + 5e ==> Mn^+2 + 4H2O
Step 6. Check everything.
Post specific questions if you have them.