Balance the redox reaction by using Oxidation number method in acidic medium.
Cr2O7
2-
(aq) + SO3
2-
(aq) → Cr3+
(aq) + SO4
2-
(aq)
Answers
Answer:
Answer : The balanced reaction will be,
Cr_2O_7^{2-}+3SO_3^{2-}+8H^+\rightarrow 2Cr^{3+}+3SO_4^{2-}+4H_2OCr
2
O
7
2−
+3SO
3
2−
+8H
+
→2Cr
3+
+3SO
4
2−
+4H
2
O
Explanation :
The given unbalanced reaction is,
Cr_2O_7^{2-}+SO_3^{2-}\rightarrow Cr^{3+}+SO_4^{2-}Cr
2
O
7
2−
+SO
3
2−
→Cr
3+
+SO
4
2−
Step 1: Separate the skeleton equation into two half-reactions.
Oxidation : SO_3^{2-}\rightarrow SO_4^{2-}SO
3
2−
→SO
4
2−
Reduction : Cr_2O_7^{2-}\rightarrow Cr^{3+}Cr
2
O
7
2−
→Cr
3+
Step 2: Balance all atoms other than H and O.
Oxidation : SO_3^{2-}\rightarrow SO_4^{2-}SO
3
2−
→SO
4
2−
Reduction : Cr_2O_7^{2-}\rightarrow 2Cr^{3+}Cr
2
O
7
2−
→2Cr
3+
Step 3: Balance O.
Oxidation : SO_3^{2-}+H_2O\rightarrow SO_4^{2-}SO
3
2−
+H
2
O→SO
4
2−
Reduction : Cr_2O_7^{2-}\rightarrow 2Cr^{3+}+7H_2OCr
2
O
7
2−
→2Cr
3+
+7H
2
O
Step 4: Balance H.
Oxidation : SO_3^{2-}+H_2O\rightarrow SO_4^{2-}+2H^+SO
3
2−
+H
2
O→SO
4
2−
+2H
+
Reduction : Cr_2O_7^{2-}+14H^+\rightarrow 2Cr^{3+}+7H_2OCr
2
O
7
2−
+14H
+
→2Cr
3+
+7H
2
O
Step 5: Balance the charge.
Oxidation : SO_3^{2-}+H_2O\rightarrow SO_4^{2-}+2H^++2e^-SO
3
2−
+H
2
O→SO
4
2−
+2H
+
+2e
−
Reduction : Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2OCr
2
O
7
2−
+14H
+
+6e
−
→2Cr
3+
+7H
2
O
Step 6: Equalize electrons transferred.
Oxidation : SO_3^{2-}+H_2O\rightarrow SO_4^{2-}+2H^++2e^-SO
3
2−
+H
2
O→SO
4
2−
+2H
+
+2e
−
]× 3
Reduction : Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2OCr
2
O
7
2−
+14H
+
+6e
−
→2Cr
3+
+7H
2
O ]× 1
Step 7: Add the two half-reactions.
Cr_2O_7^{2-}+3SO_3^{2-}+8H^+\rightarrow 2Cr^{3+}+3SO_4^{2-}+4H_2OCr
2
O
7
2−
+3SO
3
2−
+8H
+
→2Cr
3+
+3SO
4
2−
+4H
2
0