Page 124 - Basic physical chemistry for the atmospheric sciences

1 1 0 Basic physical chemistry
The reductant in Reaction (6.7) is H2S(aq , because it contains S(s),
)
which undergoes an increase in oxidation number [from - 2 on the
right side of Reaction (6.7) to zero on the left side] . Therefore, the
unbalanced oxidation half-reaction is
(6.7b)

Step 3. We must now balance the oxygen atoms in Reaction (6.7a).
Since this is an acidic solution, and there are two excess oxygen atoms
on the right side of Reaction (6.7a), we do this by first adding 2H 20(1)
to the right side of Reaction (6.7a)

Since there are now four excess hydrogen atoms on the right side of
the last reaction , we add 4H + (aq) to the left side to obtain the bal­
anced reduction half-reaction

NO 3 (aq) + 4H + (aq)� NO(g) + 2H20(1) (6.7c)
To balance the oxidation half-reaction (6.7b), since there are two
excess hydrogen atoms on the left side, we add 2H + (aq) to the right
side

H2S(aq)� S(s) + 2H + (aq) (6.7d)
Step 4. In Reaction (6.7c) the net charge on the left side is + 3, and on
the right side it is 0 . Therefore, we add three electrons to the left side
NO 3 (aq) + 4H + (aq) + 3e - � NO(g) + 2H20(l) (6.7e)

In Reaction (6.7d) the net charge on the left is 0 and on the right it is
+ 2. Therefore, two electrons must be added to the right side

(6.7f)
Step 5. To equalize the electrons lost by the reductant and gained by
the oxidant, we must multiply Reaction (6.7e) by 2 and Reaction (6.7f)
by 3 to give

8H + (aq) + 2N0 3 (aq) + 6 e - � 2NO(g) + 4H20(1) (6.7g)
and,

(6.7h)
[Note that in Reaction (6. 7h) the oxidation number of each of the three
sulfur atoms increases by two (from -2 in H 2S to zero in S ) , for a total