# How do we balance the oxidation of chloride anion to give chlorine gas by potassium permanganate?

May 28, 2017

See diagram in explanation

#### Explanation:

Oxidation-Reduction Rxn => $M n {O}_{2}$ in acidic medium. May 28, 2017

We use the redox method of balancing equations, where electrons are introduced as virtual particles...........

#### Explanation:

Using oxidation numbers we write the reduction equation:

$\stackrel{I V}{M} n {O}_{2} \left(s\right) + 4 {H}^{+} + 2 {e}^{-} \rightarrow M {n}^{2 +} + 2 {H}_{2} O$ $\left(i\right)$

Both mass and charge are balanced as required.........

And then we need a reduction half equation.........

$C {l}^{-} \rightarrow \frac{1}{2} \stackrel{0}{C} {l}_{2} + {e}^{-}$ $\left(i i\right)$

Any species whose $\text{oxidation number}$ $\text{INCREASES}$ is said to be $\text{oxidized}$, and any species whose $\text{oxidation number}$ $\text{decreases}$ is said to be $\text{reduced}$,

We add $\left(i\right)$ and $\left(i i\right)$ in such a way as to eliminate the electrons, and we get an equation that represents the observed stoichiometric change:

$\left(i\right) + 2 \times \left(i i\right)$ gives...............................

$M n {O}_{2} \left(s\right) + 4 {H}^{+} + 2 C {l}^{-} \rightarrow M {n}^{2 +} + C {l}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

Given the half equation method, for every electron loss, $\text{oxidation}$, there is thus a FORMAL electron gain, $\text{reduction} .$

The $\left(s\right)$, $\left(g\right)$, $\left(l\right)$ represents the physical state of the reactant/product, i.e. $\text{solid/gas/liquid/}$.........The ions, i.e. $M {n}^{2 +}$ etc. are assumed to be present in solution, and thus do not need a descriptor of state. Happy?