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.
Balance the equation:
_MnO4– + _H+ + _Fe2+ => _Mn2+ + _H2O + _Fe3+
How do you do balance this redox reaction?
4 answers
1MnO4– + 8H+ + 5Fe2+ => 1Mn2+ + 4H2O + 4Fe3+
This is what I got. Is it correct?
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.
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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.
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