Question #91e2a

Equations of this sort can be easily balanced by using oxidation numbers for every atom involved in the reaction.

#stackrel(color(blue)(+1))("K")stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O")_4 + stackrel(color(blue)(+1))("H")stackrel(color(blue)(-1))("Cl") -> stackrel(color(blue)(+1))("K")stackrel(color(blue)(-1))("Cl") + stackrel(color(blue)(+2))("Mn")stackrel(color(blue)(-1))("Cl")_2 + stackrel(color(blue)(+1))("H")_2stackrel(color(blue)(-1))("O") + stackrel(color(blue)(0))"Cl"_2#

The next step is to look for atoms that have undergone a change in their oxidation numbers (ON), which correlates to either a reduction, or an oxidation of that particular element.

You can see that manganese, #"Mn"#, went from an ON of #color(blue)(+7)# on the reactants' side to an ON of #color(blue)(+2)# on the products' side, which means that it has been reduced, i.e. it gained electrons.

The reduction half-reaction looks like this

#stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+))#

Since you're in acidic solution, you can balance the oxygen and hydrogen atoms by adding water molecules to the side that needs oxygen, and protons, #"H"^(+)#, to the side that needs hydrogen.

In this case, you need four oxygen atoms on the products' side, so add four water molecules on that side

#stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+)) + 4"H"_2"O"#

To balance the hydrogen atoms, add eight protons on the reactants' side

#8"H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+)) + 4"H"_2"O"#

Chlorine, #"Cl"#, on the other hand, went from an ON of #color(blue)(-1)# on the reactants' side to an ON of #color(blue)(0)# on the products' side, which means it has been oxidized, i.e. it lost electrons.

The oxidation half-reaction looks like this

#2stackrel(color(blue)(-1))("Cl"^(-)) -> stackrel(color(blue)(0))"Cl"_2 + 2"e"^(-)#

Since one chlorine atom will lose one electron, it follows that two chlorine atoms will lose a total of two electrons.

During an oxidation-reduction reaction, the total number of electrons lost in the oxidation half-reaction must be equal to the total number of electrons gained in the reduction half-reaction.

This means that you must multiply the oxidation half-reaction by #5# and the reduction half-reaction by #2#, for a total of #10# electrons transferred.

#{ ([2stackrel(color(blue)(-1))("Cl"^(-)) -> stackrel(color(blue)(0))"Cl"_2 + 2"e"^(-)] xx 5), ([8"H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn"^(2+)) + 4"H"_2"O"] xx 2) :}#

Add the resulting half-equations to get

#{ (10stackrel(color(blue)(-1))("Cl"^(-)) -> 5stackrel(color(blue)(0))"Cl"_2 + 10"e"^(-)), (16"H"^(+) + 2stackrel(color(blue)(+7))("Mn")"O"_4^(-) + 10"e"^(-) -> 2stackrel(color(blue)(+2))("Mn"^(2+)) + 8"H"_2"O") :}#
#color(white)(aaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#
#10"Cl"^(-) + 16"H"^(+) + 2"MnO"_4^(-) + color(red)(cancel(color(black)(10"e"^(-)))) -> 5"Cl"_2 + color(red)(cancel(color(black)(10"e"^(-)))) + 2"Mn"^(2+) + 8"H"_2"O"#

Now, since hydrochloric acid, #"HCl"#, is a strong acid, it dissociates completely in aqueous solution. This means that the chloride anions, #"Cl"^(-)#, and the protons, #"H"^(+)#, present on the reactants' side are both coming from the hydrochloric acid.

Since you need #16# protons present in solution, you can say that you will also need #16# molecules of hydrochloric acid. This means that you have

#2"MnO"_4^(-) + 16"HCl" ->2"Mn"^(2+) + 5"Cl"_2 + 8"H"_2"O"#

Add the potassium cations and the rest of the chloride anions back to the equation to get

#2"KMnO"_4 + 16"HCl" -> "KCl" + 2"MnCl"_2 + 5"Cl"_2 + 8"H"_2"O"#

Now all you have to do is multiply the potassium chloride, #"KCl"#, by #2# to get the balanced chemical equation

#2"KMnO"_4 + 16"HCl" -> 2"KCl" + 2"MnCl"_2 + 5"Cl"_2 + 8"H"_2"O"#