# Question 15bad

Oct 25, 2015

Manganese is being reduced and chlorine is being oxidized.

#### Explanation:

The first thing to do here is assign oxidation numbers to the atoms that take part in the reaction.

I assume that you're familiar with the rules for assigning oxidation states, so I won't go into that here.

$\stackrel{\textcolor{b l u e}{+ 4}}{\text{Mn") stackrel(color(blue)(-2))("O"_2) + 4stackrel(color(blue)(+1))("H") stackrel(color(blue)(-1))("Cl") -> stackrel(color(blue)(+2))("Mn") stackrel(color(blue)(-1))("Cl"_2) + stackrel(color(blue)(0))("Cl"_2) + 2stackrel(color(blue)(+1))("H"_2) stackrel(color(blue)(-2))("O}}$

Your goal now is to identify elements that change their oxidation state in the reaction.

Notice that the oxidation state of manganese goes from $\textcolor{b l u e}{+ 4}$ on the reactants' side, to $\textcolor{b l u e}{+ 2}$ on the products' side, which means that manganese is being reduced, i.e. its oxidation state decreases.

On the other hand, chlorine goes from an oxidation state of $\textcolor{b l u e}{- 1}$ on the reactants' side, to an oxidation state of $\textcolor{b l u e}{0}$ on the products' side, which means that it is being oxidized, i.e. its oxidation number increases.

The half-reaction will look like this

• oxidation half-reaction

$\stackrel{\textcolor{b l u e}{- 1}}{{\text{Cl"^(-)) -> stackrel(color(blue)(0))("Cl}}_{2}}$

Balance the chlorine atoms by multiplying the left-hand side of the equation by $2$

$2 \stackrel{\textcolor{b l u e}{- 1}}{{\text{Cl"^(-)) -> stackrel(color(blue)(0))("Cl}}_{2}}$

Notice that two chlorine atoms are going from an oxidationstate of $\textcolor{b l u e}{- 1}$ to one of $\textcolor{b l u e}{0}$, which means that a total of two electrons are being lost..

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

• reduction half-reaction

stackrel(color(blue)(+4))("Mn")"O"_2 -> stackrel(color(blue)(+2))"Mn"""^(2+)

Notice that the atoms of oxygen that you have on the reactants' side are not balanced. Since you're in acidic solution, you can balance oxygen atoms by adding water and hydrogen atoms by adding protons, ${\text{H}}^{+}$.

In this case, you need $2$ oxygen atoms on the products' side, so add $2$ water molecules to get

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

Now you need $4$ hydrogen atoms on the reactants' side, so add $4$ protons

$4 \text{H"^(+) + stackrel(color(blue)(+4))("Mn")"O"_2 -> stackrel(color(blue)(+2))"Mn"""^(2+) + 2"H"_2"O}$

Notice that one manganese atom gains two electrons to go from an oxidation state of $\textcolor{b l u e}{+ 4}$ to one of $\textcolor{b l u e}{+ 2}$. This means that you have

$4 \text{H"^(+) + stackrel(color(blue)(+4))("Mn")"O"_2 + 2"e"^(-) -> stackrel(color(blue)(+2))"Mn"""^(2+) + 2"H"_2"O}$

The two half-reactions are now balanced.