balancing equation for gen chem

[eravin99]

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(ClO3)- + (AsO2)- ----> (AsO4)3- + Cl-



Answer is

(ClO3)- + 3(AsO2)- + 3H2O ----> 3(AsO4)3- + Cl- + 6H+

 

[eravin99]

can someone show me the steps for this one. i keep getting it wrong

 

[mhamilton23]

1) Split the equation into half-reactions:

ClO3(-) --> Cl(-)
AsO2(-) --> AsO4(3-)

2) Do the following to both half-reactions:
a) Balance non O and H atoms first (in this case, they are already balanced)
b) Balance O atoms by adding H2O to the side that needs more O atoms

ClO3(-) --> Cl(-) + 3H2O
AsO2(-) + 2H2O --> AsO4(3-)

c) Balance H atoms by adding H+ to the side needing more hydrogens

ClO3(-) + 6H+ --> Cl(-) + 3H2O
AsO2(-) + 2H2O --> AsO4(3-) + 4H+

d) Balance the charge so that the charge on one side of the half-reaction equals the charge on the other side (charge of reactants=charge of products for each half-reaction). Do this by adding electrons to the more positively charged side:

ClO3(-) + 6H+ + 6e(-) --> Cl(-) + 3H2O <-- electrons on the reactant side indicates this half-reaction is being reduced.
AsO2(-) + 2H2O --> AsO4(3-) + 4H+ + 2e(-) <-- electrons on the product side indicates this half-reaction is being oxidized.

We can use this method to determine which half-reaction is being reduced and which is being oxidized. Although, you can usually tell just by looking at the half-reactions.

3) Since the amount of electrons released from oxidation must equal the number of electrons accepted by the reduced species, we need to multiply the half-reactions so the electrons added in each half-reaction equal each other:

1(ClO3(-) + 6H+ + 6e(-) --> Cl(-) + 3H2O)
3(AsO2(-) + 2H2O --> AsO4(3-) + 4H+ + 2e(-))

4) Combine the half-reactions back into one equation and simplify the equation:

ClO3(-) + 6H+ + 6e(-) + 3AsO2 + 6H2O --> Cl- + 3H2O + 3AsO4(3-) + 12H+ + 6e(-)

After canceling, we have the answer:

ClO3(-) + 3AsO2 + 3H2O --> Cl(-) + 3AsO4(3-) + 6H+

*NOTE: If this solution is acidic, then we stop here and the equation above is our answer, but if done in basic solution, we need to do a few more steps. Add OH- to the side with H+ to render H20 molecules. Then re-simplify the equation and you're done. You add OH- because the solution is basic, and H+ won't usually exist in a basic solution.

Hope this helps. Sorry if it is long or confusing.

 

[Danny289]

1 (ClO3- + 6H+ +6e- ======> Cl- + 3H2O)

3(AsO2- + 2H20 ======>(AsO4)-3 +4H+ +2e-)

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ClO3- + 6H+ +6e- + 3(AsO2)- + 6H20 ----> Cl- + 3H2O + 3(AsO4)-3 +12H+ +6e-
e- and water and H+ from both side equation simplies :


(ClO3)- + 3(AsO2)- + 3H2O ----> 3(AsO4)3- + Cl- + 6H+

hopefully it helped, further question plz ask! 🙂

 

[thefifthbeatle]

(ClO3)- + (AsO2)- ----> (AsO4)3- + Cl-



Answer is

(ClO3)- + 3(AsO2)- + 3H2O ----> 3(AsO4)3- + Cl- + 6H+

Step 1...Separate the two into half reactions...add water and H+ and e- to balance each if necessary to help balance
(ClO3)- + 6H+ + 6e- ---> Cl- + 3H2O

(AsO2)- + 2H2O -----> (AsO4)3- + 4H+ + 2e-


Step 2...we dont want electrons.....so multiply the bottom one **equation** by 3, so we have 6 on each side and we can cancel them out

3(AsO2)- + 6H2O -----> 3(AsO4)- + 12H+ (remember the 6e have been canceled)

now add the two equations together..everything on the left, everything the on the right

(ClO3)- + 6H+ +3(AsO2)- + 6H2O ----> Cl- + 3H20 + 3(AsO4)- + 12H+


Step 3, like any mathematical equation, subtract what stuff on both sides


(ClO3)- + 3(AsO2)- + 3H2O ----> 3(AsO4)3- + Cl- + 6H+

 

[eravin99]

thanks so much. i kept getting the wrong answer because i balanced the electrons of second half reaction wrong!! :0)

 

[thefifthbeatle]

Do you think these would actually be asked on the test..time consuming and how do you test it???..i mean, all you have to do is be able to balance the 5 answer choices, not actually do the work, therefore learning how to put electrons, H+ isnt necessary...we will jsut be shown 5 equations, four of which dont balance..know what i mean?