K2 Cr2 at KI + H₂ SO 4 K ₂ SO 4 +Cr2
(so4) 3 + I 2+ H₂O
Balance the equation by oxidation number method
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3k2 CR#HKLJCILK3-++PNBCCC
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ANSWER
(i) K
2
Cr
2
O
7
+KI+H
2
SO
4
→K
2
SO
4
+Cr
2
(SO
4
)
3
+I
2
+H
2
O
Ref image
Equate increase and decrease of oxidation number
K
2
Cr
2
O
7
+6KI+H
2
SO
4
→K
2
SO
4
+Cr
2
(SO
4
)
3
+3I
2
+H
2
O
Equate number of potassium and then SO
4
2−
and then hydropen
K
2
Cr
2
O
7
+6KI+7H
2
SO
4
→4K
2
SO
4
+Cr
2
(SO
4
)
3
+3I
2
+7H
2
O
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