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Chemistry balancing redox equations

I don't understand why the balanced version is this( image below) and not this
MnO4- +fe2+ +8H+ +4e- --> Mn 2+ + fe3+ +4H2O
Because what I did is balance the oxygen with H2O and at H+ to balance the H and e- to balance the charges
Help where did they get the 5 from anf is my answer still correct

Reply 1

20240907_151838.jpg This is the unbalanced one

Reply 2

20240907_151846.jpg This is the balanced one they got

Reply 3

Original post
by Animal lover :))
I don't understand why the balanced version is this( image below) and not this
MnO4- +fe2+ +8H+ +4e- --> Mn 2+ + fe3+ +4H2O
Because what I did is balance the oxygen with H2O and at H+ to balance the H and e- to balance the charges
Help where did they get the 5 from anf is my answer still correct

Found a solution online, which asks you to balance the changes in oxidation states between Mn (from oxidation state +7 (due to having 4 x -2 oxidation state of the oxygen ions come out to a -1 charge on the MnO4 ion) to +2 on its own, a change of -5) and Fe (from +2 to +3, a change of +1). To balance this you need 5Fe : 1Mn to balance the oxidation states (+5 : -5), since the oxidation states of H and O remain the same.
https://socratic.org/questions/how-do-you-balance-this-redox-reaction-using-the-oxidation-number-method-fe2-aq-

Reply 4

Original post
by Animal lover :))
I don't understand why the balanced version is this( image below) and not this
MnO4- +fe2+ +8H+ +4e- --> Mn 2+ + fe3+ +4H2O
Because what I did is balance the oxygen with H2O and at H+ to balance the H and e- to balance the charges
Help where did they get the 5 from anf is my answer still correct

Basically you want to have the equation written so there are no electrons just sitting there on one side. It is far more useful to understand where they come from as this then makes it possible for the equation to be useful in calculations etc.

Let us start from the relevant half-equations:

MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O

Fe^2+ —> Fe^3+ + e^-

If we want to cancel out the electrons, we need the numbers of electrons in each half-equation to be equal. This means scaling up one or both of these equations.

It makes sense to scale up the oxidation equation by 5 so that there are 5 electrons involved to cancel out the 5 in the reduction equation:

MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O

5Fe^2+ —> 5Fe^3+ + 5e^-

Now adding these equations together:

MnO4^- + 8H^+ + 5e^- + 5Fe^2+ —> Mn^2+ + 4H2O + 5Fe^3+ + 5e^-

Now see that since both equations have 5 electrons each side, we can cancel them out completely as required?

MnO4^- + 8H^+ + 5Fe^2+ —> Mn^2+ + 4H2O + 5Fe^3+

This is the expected answer.
(edited 1 year ago)

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