Question #bfe80

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Question #bfe80

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Answer 1 · 2016-10-16T17:00:58

$C l O_{3}^{-} (a q) + 3 S O_{2} (g) + 6 O H^{-} (a q) \to 3 S O_{4}^{2 -} (a q) + C l^{-} (a q) + 3 H_{2} O (l)$ $ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(blue)(6)OH^(-)(aq)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+color(red)(3)H_2O(l)$

Explanation:

The reaction to balance is the following:

$C l O_{3}^{-} (a q) + S O_{2} (g) \to S O_{4}^{2 -} (a q) + C l^{-} (a q)$ $ClO_3^(-)(aq)+SO_2(g)->SO_4^(2-)(aq)+Cl^(-)(aq)$

Step 1 : we will first split the reaction into two half equations:

Oxidation: $S O_{2} \to S O_{4}^{2 -}$ $SO_2->SO_4^(2-)$

Reduction: $C l O_{3}^{-} \to C l^{-}$ $ClO_3^(-)->Cl^(-)$

Step 2 : balance the other elements other than oxygen and nitrogen.

They are balanced.

Step 3 : balance oxygen using $H_{2} O$ $H_2O$ :

Oxidation: $S O_{2} + 2 H_{2} O \to S O_{4}^{2 -}$ $SO_2+2H_2O->SO_4^(2-)$

Reduction: $C l O_{3}^{-} \to C l^{-} + 3 H_{2} O$ $ClO_3^(-)->Cl^(-)+3H_2O$

Step 4 : balance hydrogen using $H^{+}$ $H^(+)$ :

Oxidation: $S O_{2} + 2 H_{2} O \to S O_{4}^{2 -} + 4 H^{+}$ $SO_2+2H_2O->SO_4^(2-)+4H^(+)$

Reduction: $C l O_{3}^{-} + 6 H^{+} \to C l^{-} + 3 H_{2} O$ $ClO_3^(-)+6H^(+)->Cl^(-)+3H_2O$

Step 5 : balance charges using $e^{-}$ $e^(-)$ :

Oxidation: $S O_{2} + 2 H_{2} O \to S O_{4}^{2 -} + 4 H^{+} + 2 e^{-}$ $SO_2+2H_2O->SO_4^(2-)+4H^(+)+2e^(-)$

Reduction: $C l O_{3}^{-} + 6 H^{+} + 6 e^{-} \to C l^{-} + 3 H_{2} O$ $ClO_3^(-)+6H^(+)+6e^(-)->Cl^(-)+3H_2O$

Step 7 : sum the two half equations while cancelling the number of electrons by multiplying each half equation by the corresponding integer:

Oxidation: $3 \times (S O_{2} + 2 H_{2} O \to S O_{4}^{2 -} + 4 H^{+} + 2 e^{-})$ $color(red)(3xx)(SO_2+2H_2O->SO_4^(2-)+4H^(+)+cancel(2e^(-)))$

Reduction: $1 \times (C l O_{3}^{-} + 6 H^{+} + 6 e^{-} \to C l^{-} + 3 H_{2} O)$ $color(red)(1xx)(ClO_3^(-)+6H^(+)+cancel(6e^(-))->Cl^(-)+3H_2O)$

RedOx: $C l O_{3}^{-} (a q) + 3 S O_{2} (g) + 3 H_{2} O (l) \to 3 S O_{4}^{2 -} (a q) + C l^{-} (a q) + 6 H^{+} (a q)$ $ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(red)(3)H_2O(l)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+color(red)(6)H^(+)(aq)$

Step 8 : cancel the $H^{+}$ $H^(+)$ ions by adding $O H^{-}$ $OH^(-)$ to each side:

RedOx: $C l O_{3}^{-} (a q) + 3 S O_{2} (g) + 3 H_{2} O (l) + 6 O H^{-} (a q) \to 3 S O_{4}^{2 -} (a q) + C l^{-} (a q) + 6 H^{+} (a q) + 6 O H^{-} (a q)      6 H_{2} O$ $ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(red)(3)H_2O(l)+color(blue)(6)OH^(-)(aq)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+underbrace(color(red)(6)H^(+)(aq)+color(blue)(6)OH^(-)(aq))_(6H_2O)$

Step 9 : FINALLY, cancel the $H_{2} O$ $H_2O$ molecules on both sides:

RedOx: $C l O_{3}^{-} (a q) + 3 S O_{2} (g) + 6 O H^{-} (a q) \to 3 S O_{4}^{2 -} (a q) + C l^{-} (a q) + 3 H_{2} O (l)$ $ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(blue)(6)OH^(-)(aq)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+color(red)(3)H_2O(l)$

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