Page 126 - Basic physical chemistry for the atmospheric sciences

112 Basic physical chemistry
Since there are now two excess hydrogen atoms on the right side of
the last equation, we add two O H - (aq) to the right side and two
H20(l) to the left side

OCl ( aq) + 2H20(l)- Cl - (aq) + H20(l) + 20H - (aq)
-
or, on canceling one H20(l) from each side
OCl - ( aq) + H20(1) - cl - (aq) + 20H - ( aq) (6.9c)

Since there is one excess oxygen atom on the right side of Reaction
(6.9b), as a first step in achieving atomic balance, we add one H20(1)
to the left side
Cr(OHh(s) + H20(l)- C r0� ( aq)
-
Since there are now five excess hydrogen atoms on the left side of the
last reaction , to achieve hydrogen balance we add five O H - (aq) to the
left side and five H20(1) to the right side
Cr(OHh(s) + H20(l) + 50H - (aq)- c ro� - ( aq) + 5H20(1)

or, on canceling one H20(1) from each side
Cr(OH h (s) + 50H - (aq)- C ro� - ( aq) + 4H20(l) (6.9d)

Step 4. In Reaction (6.9c) the net charge on the left side is 0 and on
the right side it is - 2. Therefore, we add two electrons to the left side

Ocl - ( aq) + H20(l) + 2e - - c 1 - (aq) + 20H - (aq) (6.9e)
In Reaction (6. 9 d) the net charge on the left side is - 5 and on the right
side it is - 2. Therefore, we add three electrons to the right side
Cr(OHh(s) + 5oH - (aq)- cro� - (aq) + 4 H20(l) + 3e - (6.9f)

Step 5. To equalize the electrons lost by the reductant and gained by
the oxidant, we must multiply Reaction (6.9e) by 3 and Reaction (6.9f)
by 2, to give
30cl - (aq) + 3H20(l) + 6e - - 3c1 - (aq) + 60H - (aq) (6.9g)

and ,
2Cr(0Hh(s) + 0 OH - (aq)- 2Cro�- ( aq) + 8 H20(1) + 6 e - (6.9h)
1
[The reader should check Reactions (6.9g) and (6.9h) for consistency
between electron losses or gains and changes in oxidation numbers -
6
see step 5 in Exercise . 2 . ]