# 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 ?

Apr 17, 2017

MnO4^-) + 8H^+ + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+)

#### 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
$\textcolor{w h i t e}{F {e}^{}} 2 \textcolor{w h i t e}{M \ln} {7}^{+} - 2 \textcolor{w h i t e}{l l l l l l l \times l l l} + 3 \textcolor{w h i t e}{l l x l l l l l} 2 +$
Fe^(2+) + MnO_4^-) + H^+ → Fe^(3+)+Mn^(2+) + H2O

Iron has already lost 2 electrons and in this reaction it is losing another one

$F {e}^{2 +} = F {e}^{3 +}$

Thus iron is being oxidized

MnO_4^-) = Mn^(2+)

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

MnO4^-) = Mn^(2+) + 4O

You can see in the reaction that oxygen is used to make water and no oxygen is let which is ${O}_{2}$

thus 4 oxygen atoms can produce 4 water molecules

MnO4^-) + H^+ = Mn^(2+) + 4H_2O

balance the reaction

MnO4^-) + 8H^+= Mn^(2+) + 4H_2O

Now for a redox reaction the addition of charges on both sides should be equal.So try it in this reaction

$\left(- 1\right) + 8 = + 7$
$\left(+ 2\right) + 0 = + 2$

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

MnO4^-) + 8H^+ + 5e^-)= Mn^(2+) + 4H_2O

Where $5 {e}^{-}$ is 5 electrons

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 $F {e}^{2 +}$

$F {e}^{2 +} = F {e}^{3 +}$

Again try to make the charges equal on same side and you would get

$F {e}^{2 +} = F {e}^{3 +} + {e}^{-}$

Now we want 5electrons not one so to get 5electrons you must have more $F {e}^{2 +}$

$5 F {e}^{2 +} = 5 F {e}^{3 +} + 5 {e}^{-}$

Add both the reactions

MnO4^-) + 8H^+ + 5e^-) + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+) + 5e^-

cancel the same terms

MnO4^-) + 8H^+ +cancel( 5e^-) + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+) + cancel(5e^-)

MnO4^-) + 8H^+ + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+)

Apr 17, 2017

$\boldsymbol{5 \stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)}$

Here are the primary steps for balancing in ACIDIC solution:

1. 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).
2. Balance the non-$\boldsymbol{\text{O}}$ and non-$\boldsymbol{\text{H}}$ atoms.
3. Balance $\boldsymbol{\text{O}}$ using $\text{H"_2"O}$ molecules, since the solution is aqueous.
4. Balance $\boldsymbol{\text{H}}$ using ${\text{H}}^{+}$ ions, since the solution is acidic.
5. Balance the remaining charge using electrons on the side with more positive charge, since electrons must cancel out in the overall reaction.
6. Scale one or both reactions to cancel out the electrons.
7. Add the reactions together.

In BASIC solution, add the step of adding ${\text{OH}}^{-}$ to both sides of the final reaction, combining ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ to make water, then canceling out common waters on each side.

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 $+ 2$ to $+ 3$ oxidation state. For pure ions, the charge is the oxidation state.

$\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)$ (step 1)

Everything is balanced except for the charge, so we can skip straight to step 5.

$\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right) + {e}^{-}$ (step 5)

The total charge on each side is now:

$\left(+ 2\right) = \left(+ 3\right) + \left(- 1\right)$ 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 ${\text{MnO}}_{4}^{-}$ to ${\text{Mn}}^{2 +}$, which is why I suggested this easier way to spot reduction if applicable.

$\stackrel{\textcolor{b l u e}{+ 7}}{{\text{Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn}}^{2 +}} \left(a q\right)$ (step 1)

The non-oxygen and non-hydrogen atoms (i.e. $\text{Mn}$) are balanced, so we skip to step 3. Add water to balance the oxygens.

$\stackrel{\textcolor{b l u e}{+ 7}}{\text{Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 3)

Add ${\text{H}}^{+}$ to balance the hydrogens.

$8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 4)

Balance the charge using electrons on the more positive side. Each side has:

(??) + (+8) + (-1) = (+2)

Hence, add $5 {e}^{-}$ to the left side.

$5 {e}^{-} + 8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (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)

$5 \left(\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right) + \cancel{{e}^{-}}\right)$
$\cancel{5 {e}^{-}} + 8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$

Add the reactions to get: (Step 7)

$\boldsymbol{5 \stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)}$

Sep 21, 2017

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 F {e}^{2 +} \rightarrow M {n}^{2 +} + 5 F {e}^{3 +} + 4 {H}_{2} O$

#### Explanation:

$F {e}^{2 +} \rightarrow F {e}^{3 +} + {e}^{-}$ $\left(i\right)$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i i\right)$

And we take $5 \times \left(i\right) + \left(i i\right)$ to eliminate the electrons.....

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 F {e}^{2 +} \rightarrow M {n}^{2 +} + 5 F {e}^{3 +} + 4 {H}_{2} O$

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 $M {n}^{2 +}$.