(a) CuSO
4
+4NH
3
→Cu(NH
3
)
4
]SO
4
(b) Na
2
SO
4
+BaCl
2
→BaSO
4
+2NaCl
(c) SO
2
+H
2
O→H
2
SO
3
(d) 2CuSO
4
+4KI→Cu
2
I
2
+2K
2
SO
4
+I
2
.
Solution
verified
Verified by Toppr
The reaction in which change in oxidation numbers of some of the atoms takes place is termed as a redox reaction.
(a)
CuSO
4
+2+6−2
+
4NH
3
−3+1
→[
Cu
+2
(
N
−3
H
3
+1
)
4
]
S
+6
O
4
−2
No change in oxidation number of any of the atoms.
(b)
Na
2
+1
S
+6
O
4
−2
+
Ba
+2
Cl
2
−1
→
Ba
+2
S
+6
O
4
−2
+
2Na
+1
Cl
−1
No change in oxidation number of any one of the atoms.
(c)
S
+4
O
2
−2
+
H
2
+1
O
−2
→
H
2
+1
S
+4
O
3
−2
No change in oxidation number of any one of the atoms.
(d)
2Cu
+2
S
+6
O
4
−2
+
4K
+1
I
−1
→
Cu
2
+1
I
2
−1
+
2K
2
+1
S
+6
O
4
−2
+
I
2
0
.
Oxidation number of Cu decreases from +2 to +1 and oxidation number of iodine increases from −1 to 0.
Thus, out of the above four reactions, the reaction (d) is a redox reaction.