Under acidic conditions, MnO4− becomes Mn2+. Write the balanced half-reaction for this transformation. (Omit states-of-matter from your answer.)

Mn is +7 in MnO4^-. It is +2 in Mn^2+.

To balance the change in electrons we have
MnO4^- + 5e ==> Mn^2+. I do it that way and make this the first step of the process of balancing.

Some say there is a shorter way and you do this the LAST step and not the first. To do it that way you everything BUT the e and and you have this just before the last step.
MnO4^- + 8H^+ ==> Mn^2+ + 4H2O
Then you count the charge on the left (7+) and the charge on the right(2+). To balance the charge you add 5e to the left side so that 7+ +(-5) = 2+ and the final is MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O

MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O

Just for my understanding if you don't mind could you explain how you were able to get 5e? I was able to derive everything else except that portion.

Balance the following reaction in an ACIDIC solution.

Well, MnO4- must be feeling pretty acidic, because it's about to transform into Mn2+. In terms of balancing the half-reaction, let's get down to business:

MnO4- + H+ + 2e- -> Mn2+ + 2H2O

There you have it! Just remember, balancing equations is all about finding that perfect balance, just like in life. Keep on laughing!

To write the balanced half-reaction for the transformation of MnO4− to Mn2+ under acidic conditions, you need to follow certain steps:

1. Write the skeleton equation for the reaction:
MnO4− → Mn2+

2. Balance the atoms other than O and H:
MnO4− → Mn2+

3. Balance oxygens by adding H2O molecules to the opposite side of the Mn:
MnO4− → Mn2+ + H2O

4. Balance the hydrogens by adding H+ ions to the opposite side of the O:
MnO4− + 8H+ → Mn2+ + 4H2O

Here, we have balanced the equation under acidic conditions by adding H+ ions to neutralize the charge. This balanced half-reaction shows the transformation of MnO4− to Mn2+ in acidic conditions.