Synthetic Inorganic Chemistry

THE HALOGENS 165

this is compensated by the change of two atoms of chlorine from
— 1 to 0, so that algebraically the total changes of valence add to 0.
The reactions of potassium dichromate and potassium per-
manganate are represented by the equations

K 2 Cr 207 + 14HC1 -»2KC1 + 2CrCl 3 + 3C1 2 + 7H 2 O

2KMnO 4 + 16HC1 -> 2KC1 + 2MnCl 2 + 5C1 2 + 8H 2 O

These are likewise reactions of oxidation and reduction, and the

valence changes add up as follows:

2Cr +6 to+3 2 X (-3) = -6

6C1 -Ho 0 6 X (+1) = +6

2Mn

10C1

+7

-1

Total

to +2

to 0

change

2

10

X

X

(-5) =

(+1) =

= 0

-10

+ 10

Total change = 0

We may summarize this experiment by the statement that

free chlorine is liberated from hydrochloric acid by strong oxidiz-

ing agents. Whether or not the oxidizing agent is strong enough

to do this may perhaps be foretold by considering the element

which has the higher than ordinary valence; if the chloride of this

element in which the higher valence would be satisfied is unstable,

then the oxidizing agent will set chlorine free.

11. Bromine and Iodine from Bromides and Iodides.
    Test the action of any one of the oxidizing agents used in
    Experiment 10, say manganese dioxide, on hydrobromic and
    hydriodic acids.
    Add a few drops of chlorine water to 5 cc. of a bromide
    solution, for example NaBr. Then, in order to find whether
    bromine has been set free, add 1 cc. of carbon disulphide,
    shake vigorously, and let the heavier liquid settle to the
    bottom. The free halogen is more soluble in carbon di-
    sulphide than in water, consequently it dissolves in and im-
    parts its characteristic color to it. Note that the globule
    has acquired an orange-red color.
    Likewise add a few drops of chlorine water and of bro-
    mine water to separate portions of an iodide solution, for
    example KI, and test in each case with carbon disulphide.
    Note that in each case the globule becomes violet colored.