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Ru(s) + Cl-(aq) + NO3-(aq) -> RuCl63-(aq) + NO2(g)

How many H+ ions appear in the correctly balanced equation, and on which side of the arrow? 

As2S3(s) + NO3-(aq) ->H3AsO4(aq) +NO(g) + S8(s)

How many H+ ions appear in the correctly balanced equation,

Mar 18th, 2015

1 problem 

Ru(s) + Cl-(aq) + NO3-(aq) -> RuCl63-(aq) + NO2(g)

Eliminate the Cl- for the moment:

Ru ---> Ru3+ + 3e- 
e- + 2H+ + NO3- ---> NO2 + H2O 

Multiply bottom half-reaction by 3 and add: 

6H+ + 3NO3- + Ru ---> Ru3+ + 3NO2 + 3H2O 

Add in the chloride: 

6HCl + 3NO3- + Ru ---> RuCl63- + 3NO2 + 3H2O 

Add in three more H+, if you wish: 

6HCl + 3HNO3 + Ru ---> H3RuCl6 + 3NO2 + 3H2O 

2 problem 

As2S3(s) + NO3-(aq) ->H3AsO4(aq) +NO(g) + S8(s)

1) Write the half-reactions:

As26+ ---> H3AsO4 
S36¯ ---> S8 
NO3¯ ---> NO

2) Balance in acidic solution (because of the H3AsO4):

64H2O + 8As26+ ---> 16H3AsO4 + 80H+ + 32e¯ 
8S36¯ ---> 3S8 + 48e¯ 
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Combine the first two half-reactions:

64H2O + 8As2S3 ---> 3S8 + 16H3AsO4 + 80H+ + 80e¯ 
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

4) Multiply through by factors selected to balance the electrons:

3 [64H2O + 8As2S3 ---> 3S8 + 16H3AsO4 + 80H+ + 80e¯] 
80 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]

4) Show the two half-reactions with the factors applied:

192H2O + 24As2S3 ---> 9S8 + 48H3AsO4 + 240H+ + 240e¯ 
240e¯ + 320H+ + 80NO3¯ ---> 80NO + 160H2O

5) Add everything, eliminating only electrons:

320H+ + 80NO3¯ + 192H2O + 24As2S3 ---> 9S8 + 48H3AsO4 + 240H+ + 80NO + 160H2O

6) Eliminate hydrogen ion and water:

80H+ + 80NO3¯ + 32H2O + 24As2S3 ---> 9S8 + 48H3AsO4 + 80NO

7) Recombine (and rearrange):

24As2S3 + 80HNO3 + 32H2O ---> 9S8 + 48H3AsO4 + 80NO

Mar 19th, 2015

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