Chemistry, asked by shamraj21, 6 months ago

K2 Cr2 at KI + H₂ SO 4 K ₂ SO 4 +Cr2
(so4) 3 + I 2+ H₂O
Balance the equation by oxidation number method​

Answers

Answered by prativasahoo198021
0

3k2 CR#HKLJCILK3-++PNBCCC

Answered by srijabandi123
0

Answer:

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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