Under acidic conditions, MnO4− becomes Mn2+. Write the balanced half-reaction for this transformation. (Omit states-of-matter from your answer.)
4 answers
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.
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
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
Balance the following reaction in an ACIDIC solution.