Question

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

Answer 1

See diagram in explanation

Explanation:

Oxidation-Reduction Rxn => `MnO_2` in acidic medium.

Answer 2

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(IV)MnO_2(s)+4H^+ + 2e^(-)rarrMn^(2+)+2H_2O` `(i)`

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

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

`Cl^(-) rarr 1/2stackrel(0)Cl_2+e^-` `(ii)`

Any species whose `"oxidation number"` `"INCREASES"` is said to be `"oxidized"`, and any species whose `"oxidation number"` `"decreases"` is said to be `"reduced"`,

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

`(i) + 2xx(ii)` gives...............................

`MnO_2(s)+4H^(+)+ 2Cl^(-)rarrMn^(2+)+Cl_2(g) + 2H_2O(l)`

Given the half equation method, for every electron loss, `"oxidation"`, there is thus a FORMAL electron gain, `"reduction".`

The `(s)`, `(g)`, `(l)` represents the physical state of the reactant/product, i.e. `"solid/gas/liquid/"`.........The ions, i.e. `Mn^(2+)` etc. are assumed to be present in solution, and thus do not need a descriptor of state. Happy?