Cr2O72-+S2O32-->Cr3+ (SO4)2-
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Answer:
Explanation:
We gots a redox equation....so we write half equation...
Dichromate is reduced:
C
r
2
O
2
−
7
+
14
H
+
+
6
e
−
→
2
C
r
3
+
+
7
H
2
O
(
l
)
Sulfite is oxidized:
S
O
2
−
3
+
H
2
O
→
S
O
2
−
4
+
2
H
+
+
2
e
−
We adds THREE of the latter to one of the former to eliminate the electrons....
C
r
2
O
2
−
7
+
3
S
O
2
−
3
+
3
H
2
O
+
14
8
H
+
+
6
e
−
→
2
C
r
3
+
+
7
4
H
2
O
(
l
)
+
3
S
O
2
−
4
+
6
H
+
+
6
e
−
....to get finally...
C
r
2
O
2
−
7
orange red
+
3
S
O
2
−
3
+
8
H
+
→
2
C
r
3
+
green
+
3
S
O
2
−
4
+
4
H
2
O
(
l
)
..the which is balanced with respect to mass and charge as required....
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