posted by Tom on .
Balance the equation:
_MnO4– + _H+ + _Fe2+ => _Mn2+ + _H2O + _Fe3+
How do you do balance this redox reaction?
Mn goes from +7 oxidation state on the left to +2 on the right. Fe goes from +2 on the left to +3 on the right.
Therefore, Mn gains by 5e and Fe loses 1 e. You must keep electrons gained and lost the same; therefore, multiply the Fe half cell by 5 and the Mn half cell by 1 (which gives 5e gained and 5e lost), then balance the others by inspection. (meaning the coefficient for MnO4^- is 1 and Mn^+2 is 1 and the coefficient for Fe on each side is 5. Then go from there. I shall be happy to check your final result.
1MnO4– + 8H+ + 5Fe2+ => 1Mn2+ + 4H2O + 4Fe3+
This is what I got. Is it correct?
Correct. Do you understand what I did to balance the gain and loss of electrons? Do you understand how to determine the oxidation state of an element? If not, I can give you a web site that is fairly good for the oxidation state. And for balancing redox equations. Here they are anyway.
Use them if you need them.
(Broken Link Removed)
By the way, I'm sure you understand that it isn't necessary to place the number 1 in front of MnO4^- and Mn^2+. It is understood to be 1 if no other number is there. And you made a typo: I'm sure you intended to write 5Fe3+ on the right and not 4.
Yea it was a typing mistake. Thanks alot I got it correct, also the website is very helpful. Thanks again