Balance this reaction using ion electron method in shortest way possible . but must be clear .. #Fe^(2+)+MnO_4^(-)+H^(+) -> Fe^(3+)+Mn^(2+)+H_2O# ?
3 Answers
Explanation:
In ion electron method one reaction is divided into two reaction (half reaction) and then they are balanced and then added together
Figure the reducing(increase of electrons) and oxidizing(decrease of electrons) agent
Iron has already lost 2 electrons and in this reaction it is losing another one
Thus iron is being oxidized
In the ion MnO4, Mn has a charge of 7+ and it changing into 2+ in this reaction
Thus MnO4 is being oxidized
Now balance the the oxygen atoms
You can see in the reaction that oxygen is used to make water and no oxygen is let which is
thus 4 oxygen atoms can produce 4 water molecules
balance the reaction
Now for a redox reaction the addition of charges on both sides should be equal.So try it in this reaction
So both sides are not equal. To make them equal you should add electrons as you cant add protons.Thus if you think of adding 5 protons to the other side to make it equal to 7 it is wrong.But you can add 5 electrons to the other side to make it equal to 2
Where
Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. And this electrons come from the oxidizing agent that is
Again try to make the charges equal on same side and you would get
Now we want 5electrons not one so to get 5electrons you must have more
Add both the reactions
cancel the same terms
#bb(5stackrel(color(blue)(+2))("Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe"^(3+))(aq))#
Here are the primary steps for balancing in ACIDIC solution:
- Pick out the elements being oxidized and reduced and write out their unbalanced half-reactions (this should require almost zero effort as this is the easy part).
- Balance the non-
#bb"O"# and non-#bb"H"# atoms. - Balance
#bb"O"# using#"H"_2"O"# molecules, since the solution is aqueous. - Balance
#bb"H"# using#"H"^(+)# ions, since the solution is acidic. - Balance the remaining charge using electrons on the side with more positive charge, since electrons must cancel out in the overall reaction.
- Scale one or both reactions to cancel out the electrons.
- Add the reactions together.
In BASIC solution, add the step of adding
Usually you don't have to do all of these steps, and the nice thing is that if you have to balance another reaction that involves the same oxidation or reduction half-reaction, you can recycle the same half-reaction.
OXIDATION HALF-REACTION
Iron gets oxidized from a
#stackrel(color(blue)(+2))("Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe"^(3+))(aq)# (step 1)
Everything is balanced except for the charge, so we can skip straight to step 5.
#stackrel(color(blue)(+2))("Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe"^(3+))(aq) + e^(-)# (step 5)
The total charge on each side is now:
#(+2) = (+3) + (-1)# #color(blue)(sqrt"")#
Done with the oxidation half-reaction.
REDUCTION HALF-REACTION
In some respects, this is the easier one to identify. Reduction can sometimes be thought of as the loss of oxygen atoms, and oxidation the gain of oxygen atoms.
Manganese can be readily seen to have lost oxygen atoms going from
#stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq)# (step 1)
The non-oxygen and non-hydrogen atoms (i.e.
#stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)# (step 3)
Add
#8stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)# (step 4)
Balance the charge using electrons on the more positive side. Each side has:
#(??) + (+8) + (-1) = (+2)#
Hence, add
#5e^(-) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)# (step 5)
#(-5) + (+8) + (-1) = (+2) color(blue)(sqrt"")#
Done with the reduction half-reaction.
OVERALL REACTION
Now make the electrons cancel out. (Step 6)
#5(stackrel(color(blue)(+2))("Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe"^(3+))(aq) + cancel(e^(-)))#
#cancel(5e^(-)) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)#
Add the reactions to get: (Step 7)
#bb(5stackrel(color(blue)(+2))("Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe"^(3+))(aq))#
Explanation:
And we take
Charge and mass are balanced so this is kosher. And what would see in the reaction. The deep purple colour of permanganate would dissipate to give almost colourless