Page 125 - Basic physical chemistry for the atmospheric sciences

Oxidation-reduction reactions 1 1 1

increase of six, which is consistent with the six electrons on the right
side of reaction (6. 7h). Similarly, in Reaction (6. 7g) the oxidation
number of each of the two nitrogen atoms decreases by three (from
+ 5 in NO ) to + 2 in NO) , for a total decrease of six, which is
consistent with the six electrons on the left side of Reaction (6.7g).
This is a useful check on the half-reactions.]
S t ep 6. Adding Reactions (6.7g) and (6.7h) yields
8H + (aq) + 2NO ) (aq) + 3H2S(aq) + 6e - -
+
2NO(g) + 4H20(1) + 3 S (s) + 6H (aq) + 6e -
Canceling 6H + (aq) and 6e - from both sides of this reaction gives

2H + (aq) + 2NO 3 (aq) + 3H 2 S(aq)- 2NO(g) + 4H20(1) + 3S(s) ( . 8 )
6
Step 7 . Inspection shows that Reaction (6.8) conserves the various
atoms and electric charge.
Exercise 6.3. Balance the equation for the following redox reaction
in an aqueous solution
Cr(O H h (s) + OcJ - (aq) + OH - (aq)­
+
-
-
CrO� ( aq) C l ( aq) + H20(l) (6.9)
Solution.
Step I . The unbalanced equation is given by Reaction (6.9).
Step 2. The oxidant is OCI ( aq), because it contains Cl, which under­
-
goes a decrease in oxidation number [from + I on the right side of
Reaction (6.9) to - I on the left side] . Therefore, the unbalanced
reduction half-reaction is
oc1 - (aq - ) c 1 - ( aq) (6.9a)

The reductant in Reaction (6.9) is Cr(OHh(s), because it contains Cr
which undergoes an increase in oxidation number [from + 3 on the
right side of Reaction (6. 9 ) to + 6 on the left side] . Therefore, the
unbalanced oxidation half-reaction is

(6.9b)
Step 3 . We must now balance the oxygen atoms in Reaction (6.9a).
i
Since this s a basic solution, and there is one excess oxygen atom on
the left side of Reaction (6.9a , we first add one H 20(1) to the right side
)
of Reaction (6 .9a)
OCI - ( aq)- Cl - (aq) + H20(l)