3. Use the oxidation number method to balance the following equation and then identify the oxidizing and reducing agents:
HNO3(aq) + C2H6O(l) + K2Cr2O7(aq)--KNO3(aq) + C2H4O(l) + H2O(l) + Cr(NO3)3(aq)
Answers
Answer:
8
HNO
3
+
3
C
2
H
6
O
+
K
2
Cr
2
O
7
→
2
KNO
3
+
3
C
2
H
4
O
+
7
H
2
O
+
2
Cr
(
NO
3
)
3
Explanation:
You can find the steps for balancing redox equations here.
Your unbalanced equation is
HNO
3
+
C
2
H
6
O
+
K
2
Cr
2
O
7
→
KNO
3
+
C
2
H
4
O
+
H
2
O
+
Cr
(
NO
3
)
3
Step 1. Calculate the oxidation numbers of every atom.
Left hand side:
H= +1; N= +5; O = -2; C = -2; K = +1; Cr = +6
Right hand side:
K = +1; N = +5; O = -2; C = -1; H = +1; Cr = +3
Step 2. Calculate the changes in oxidation number:
C: -2 → -1; Change = +1
Cr: +6 → +3; Change = -3
Step 3. Balance changes in oxidation number.
You need 3 atoms of
C
for every 1 atom of
Cr
.
You have 2 atoms of
Cr
, so you need 6 atoms of
C
.
This gives you total changes of +6 and -6.
Step 4. Insert coefficients to get these numbers.
HNO
3
+
3
C
2
H
6
O
+
1
K
2
Cr
2
O
7
→
KNO
3
+
3
C
2
H
4
O
+
H
2
O
+
2
Cr
(
NO
3
)
3
Step 5a. Balance
K
.
HNO
3
+
3
C
2
H
6
O
+
1
K
2
Cr
2
O
7
→
2
KNO
3
+
3
C
2
H
4
O
+
H
2
O
+
2
Cr
(
NO
3
)
3
Step 5b. Balance
N
.
8
HNO
3
+
3
C
2
H
6
O
+
1
K
2
Cr
2
O
7
→
2
KNO
3
+
3
C
2
H
4
O
+
H
2
O
+
2
Cr
(
NO
3
)
3
Step 6. Balance
O
.
8
HNO
3
+
3
C
2
H
6
O
+
1
K
2
Cr
2
O
7
→
2
KNO
3
+
3
C
2
H
4
O
+
7
H
2
O
+
2
Cr
(
NO
3
)
3
Step 7. Balance
H
.
Done.
Step 8. Check that all atoms balance.
Atom
m
Left hand side
m
Right hand side
m
H
m
m
m
m
m
26
m
m
m
m
m
m
m
26
m
N
m
m
m
m
m
l
l
8
m
m
m
m
m
m
m
l
8
m
O
m
m
m
m
m
34
m
m
m
m
m
m
m
34
m
C
m
m
m
m
m
l
l
6
m
m
m
m
m
m
m
l
6
m
Cr
m
m
m
m
m
l
2
m
m
m
m
m
m
m
l
2
The balanced equation is
8
HNO
3
+
3
C
2
H
6
O
+
K
2
Cr
2
O
7
→
2
KNO
3
+
3
C
2
H
4
O
+
7
H
2
O
+
2
Cr
(
NO
3
)
3
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