Question

Balance the following equations by oxidation number method
i) K2Cr207+ KI+H2SO4_K2SO4+ Cr2(SO4)3 +I2+H2O
ii) KMnO4 + Na2so3 → MnO2 + Na2SO4+ KOH
iii) Cu+ HNO3 → Cu(NO3)2 + NO2+ H2O
iv) KMnO4+H2C204 + H2SO4 - K2SO4 + MnSO4 + CO2 +H2O​

please answer this question

Answer 1

Explanation:

(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

(ii) KMnO

4

+Na

2

SO

3

→MnO

2

+Na

2

SO

4

+OH

Ref. image

Equate increase and decrease 3×2.12×3

2KMnO

4

+3Na

2

SO

3

→2MnO

2

+3Na

2

SO

4

+KOH

Equate number of K and then balance number of hydrogen

2KMnO

4

+3Na

2

SO

3

+H

2

O→2MnO

2

+3Na

2

SO

4

+2KOH

(iii) Cu+HNO

3

→Cu(NO

3

)

2

+NO

2

+H

2

O

Ref. Image

equate increase and decrease and also equate nitrate

Cu+4HNO

3

→Cu(NO

3

)

2

+2NO

2

+2H

2

O

Equate hydrogen

(iv) KMnO

4

+H

2

C

2

O

4

+H

2

SO

4

→K

2

SO

4

+MnSO

4

+CO

2

+H

2

O

Ref. image

Equate increase and decrease oxidation number

KMnO

4

+5H

2

C

2

O

4

+H

2

SO

4

→K

2

SO

4

+MnSO

4

+10CO

2

+H

2

O

Equate K and SO

4

2−

and then hydrogen

2KMnO

4

+5H

2

C

2

O

4

+3H

2

SO

4

→K

2

SO

4

+2MnSO

4

+10CO

2

+8H

2

O