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
To balance the redox equation, follow these steps:
Step 1: Write out the unbalanced equation:
_MnO4– + _H+ + _Fe2+ => _Mn2+ + _H2O + _Fe3+
Step 2: Determine the number of atoms on both sides of the equation:
Manganese (Mn):
On the left side: 1 (from MnO4–)
On the right side: 1 (from Mn2+)
Since the number of Mn atoms is already balanced, we move on to the next element.
Hydrogen (H):
On the left side: 1 (from H+)
On the right side: 2 (from H2O)
To balance hydrogen, add a coefficient of 2 in front of H2O on the right side of the equation.
Iron (Fe):
On the left side: 1 (from Fe2+)
On the right side: 1 (from Fe3+)
Since the number of Fe atoms is already balanced, we move on to the next element.
Oxygen (O):
On the left side: 4 (from MnO4–, each oxygen atom contributes 4 atoms)
On the right side: 4 (from H2O, each oxygen atom contributes 2 atoms)
The total number of oxygen atoms is already balanced.
Step 3: Balance the charges:
On the left side, the overall charge is -1 (from MnO4-).
On the right side, the overall charge is +3 (from Fe3+).
To balance the charges, we need to add electrons (e-) to the side with the higher positive charge. In this case, add 4 electrons to the left side of the equation.
The balanced equation is now:
_MnO4– + _H+ + _Fe2+ => _Mn2+ + _H2O + _Fe3+
(Note: The numbers in front of the elements/compounds are still blank, as we haven't balanced all the elements yet.)
Step 4: Balance the electrons:
On the left side, we have 4 electrons.
On the right side, we don't have any electrons yet.
To balance the electrons, we need to add 4 electrons to the right side of the equation.
The equation now becomes:
_MnO4– + _H+ + _Fe2+ => _Mn2+ + _H2O + _Fe3+ + 4e-
At this point, we have balanced all the elements and charges.
Step 5: Finalize the coefficients:
Calculate the total number of each element on both sides and adjust the coefficients accordingly:
Mn: 1 = 1
H: 1 = 2
Fe: 1 = 1
O: 4 = 4
e-: 4 = 4
The final balanced equation is:
MnO4– + 8H+ + Fe2+ => Mn2+ + 4H2O + Fe3+