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
Answers
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
