Question #bfe80

1 Answer
Oct 16, 2016

Answer:

#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:

#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: #SO_2->SO_4^(2-)#

Reduction: #ClO_3^(-)->Cl^(-)#

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

They are balanced.

Step 3 : balance oxygen using #H_2O#:

Oxidation: #SO_2+2H_2O->SO_4^(2-)#

Reduction: #ClO_3^(-)->Cl^(-)+3H_2O#

Step 4 : balance hydrogen using #H^(+)#:

Oxidation: #SO_2+2H_2O->SO_4^(2-)+4H^(+)#

Reduction: #ClO_3^(-)+6H^(+)->Cl^(-)+3H_2O#

Step 5 : balance charges using #e^(-)#:

Oxidation: #SO_2+2H_2O->SO_4^(2-)+4H^(+)+2e^(-)#

Reduction: #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: #color(red)(3xx)(SO_2+2H_2O->SO_4^(2-)+4H^(+)+cancel(2e^(-)))#

Reduction: #color(red)(1xx)(ClO_3^(-)+6H^(+)+cancel(6e^(-))->Cl^(-)+3H_2O)#

RedOx: #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^(+)# ions by adding #OH^(-)# to each side:

RedOx: #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_2O# molecules on both sides:

RedOx: #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)#

Here is a video that explains this topic in details:
Balancing Redox Reactions | Basic Medium.