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+ HCl → CdCl
d. CdS + I
+ HI + S
2
2
Because one of the atoms undergoing oxidation or reduction has a subscript (I
) we will account for
2
the number of atoms of each element when preparing our summary chart:
no.
No.
Total
initial
final
change
Coefficient
-
-
atoms
e
e
×
S
-2
0
2
2
1
=
2
I
0
-1
1
×
2 (in I
)
=
2
×
=
2
1
2
Place the balancing coefficients into the equation in front of the elements undergoing oxidation and
reduction. For iodine, the 1 will go in front of the diatomic I
because these were the atoms being
2
counted.
+ HCl → CdCl
1 CdS + 1 I
+ HI + 1 S
2
2
Then balance the rest of the equation. First balance for iodine atoms, then for Cd and H:
+ 2 HCl → 1 CdCl
1 CdS + 1 I
+ 2 HI + 1 S
Answer:
2
2
→ HIO
e. I
+ HNO
+ NO
+ H
O
2
3
3
2
2
no.
No.
Total
initial
final
change
Coefficient
-
-
atoms
e
e
×
×
I
0
+5
5
2 (in I
)
10
1
=
10
2
×
N
+5
+4
1
=
1
10
=
10
Because of the subscript with iodine (I
), we multiply the change in oxidation number for iodine by 2
2
before we determine our coefficient multipliers.
The “1” for iodine is placed in front of the diatomic iodine; the “10” goes in front of both nitrogens.
Then balance for iodine on both sides of the equation, then for all other atoms.
→ 2 HIO
1 I
+ 10 HNO
+ 10 NO
+ 4 H
O
Answer:
2
3
3
2
2
-
+
-
2+
→ Mn
f.
MnO
+ H
+ Cl
+ Cl
+ H
O
4
2
2
no.
No.
Total
initial
final
change
Coefficient
-
-
atoms
e
e
×
Mn
+7
+2
5
5
2
=
10
×
×
Cl
-1
0
1
2 (in Cl
)
=
2
5
=
10
2
Because of the diatomic chlorine (Cl
) we multiply the change in oxidation number for chlorine by 2.
2
We then determine what coefficients are needed to balance for electrons. The “5” for chlorine will be
placed in front of the diatomic chlorine. Then balance both sides of the equation for chlorine, then for
all other atoms.
-
+
-
2+
→ 2 Mn
2 MnO
+ 16 H
+ 10 Cl
+ 5 Cl
+ 8 H
O
Answer:
4
2
2
Unit 6: Redox & Electrochemistry
Practice Set 3
Page 2 of 4

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