Question

Question #91e2a

Answer 1

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

Explanation:

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"`