3) Balance the following equations by oxidation number method
i) K,Cr,O, + KI + H2SO4 → K,SO4 + Cr2(SO4)3 +12+H2O
ii) KMnO4 + Na,SO3 → MnO2 + Na2SO4 + KOH
iii) Cu+ HNO3 → Cu(NO3)2 + NO2+ H2O
iv) KMnO4+H2C2O4 + H2SO4 → K2SO4 + MnSO4 + CO2 + H2O
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11th
Chemistry
Redox Reactions
Balancing Redox Reactions
Balance the following equat...
CHEMISTRY
Balance the following equation by oxidation number method.
K
2
Cr
2
O
7
+FeSO
4
+H
2
SO
4
→Cr
2
(SO
4
)
3
+Fe
2
(SO
4
)
3
+K
2
SO
4
+H
2
O.
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ANSWER
Writing oxidation numbers of all the atoms.
K
2
+2
Cr
2
+6
O
7
−2
+
Fe
+2
S
+6
O
4
−2
+
H
2
+1
S
+6
O
4
−2
=
Cr
2
+3
(
S
+6
O
4
−2
)
3
+
Fe
2
+3
(
S
+6
O
4
−2
)
3
+
K
2
+1
S
+6
O
4
−2
+
H
2
+1
O
−2
Change in Ox. no. has occurred in chromium and iron.
K
2
Cr
2
+6
O
7
→
Cr
2
+3
(SO
4
)
3
...........(i)
Fe
+2
SO
4
→
Fe
2
+3
(SO
4
)
3
Decrease in Ox. no. of Cr per molecule =(2×6−2×3)=6 units
Increase in Ox. no. of Fe per molecule =1 unit
Hence, eq. (ii) should be multiplied by 6,
K
2
Cr
2
O
7
+6FeSO
4
→Cr
2
(SO
4
)
3
+3Fe
2
(SO
4
)
3
To balance sulphate ions and potassium ions, 7 molecules of H
2
SO
4
are needed.
K
2
Cr
2
O
7
+6FeSO
4
+7H
2
SO
4
=Cr
2
(SO
4
)
3
+3Fe
2
(SO
4
)
3
+K
2
SO
4
To balance hydrogen and oxygen, 7H
2
O should be added on RHS. Hence, balanced equation is,
K
2
Cr
2
O
7
+6FeSO
4
+7H
2
SO
4
=Cr
2
(SO
4
)
3
+3Fe
2
(SO
4
)
3
+K
2
SO
4
+7H
2
O.