Balance the following redox reactions by ion electron method. (in acidic medium)
${Cr}_{2} {O_{7}^{2 -}}_{(aq)} + {SO}_{2 (g)} \to {Cr}_{(aq)}^{3 +} + {SO}_{4}^{2 -}_{(aq)}$

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Detailed Solution

S.No.	Oxidation half reaction ${SO}_{2} \to {SO}_{4}^{2 -}$	S.No.	Reduction half reaction ${Cr}_{2} O_{7}^{2 -} \to {Cr}^{3 +}$
1)	Balancing other than oxygen and hydrogen atom ${SO}_{2} \to {SO}_{4}^{2 -}$	1)	${Cr}_{2} O_{7}^{2 -} \to 2 {Cr}^{3 +}$
2)	Balancing oxygen atoms ${SO}_{2} + 2 H_{2} O \to {SO}_{4}^{2 -}$	2)	${Cr}_{2} O_{7}^{2 -} \to 2 {Cr}^{3 +} + 7 H_{2} O$
3)	Balancing hydrogen atoms ${SO}_{2} + 2 H_{2} O \to {SO}_{4}^{2 -} + 4 H^{+}$	3)	${Cr}_{2} O_{7}^{2 -} + 14 H^{+} \to 2 {Cr}^{3 +} + 7 H_{2} O$
4)	Balancing charges ${SO}_{2} + 2 H_{2} O \to {SO}_{4}^{2 -} + 4 H^{+} + 2 e^{-}$	4)	${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$
5)	$3 ({SO}_{2} + 2 H_{2} O \to {SO}_{4}^{2 -} + 4 H^{+} + 2 e^{-})$	5)	${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$
6)	$3 {SO}_{2} + 6 H_{2} O \to 3 {SO}_{4}^{2 -} + 12 H^{+} + 6 e^{-}$	6)	${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$