NO2- +Al-->NH3 +Al(OH)4-
This may help.
Al(OH)4^-. So OH is -1 each for -4; therefore, Al is +3 so that +3 + (-4) = -1 charge on the complex ion.
Then the half rxn for Al is
Al + 4OH^- ==> Al(OH)4^- + 3e
The N in NO2^- is +3 (-2 for each O is -4 so +3 + (-4) = -1 charge on the ion. The N in NH3 is -3 so change in e is 6
NO2^- + 6e ==> NH3
Count up the charge; it is -7 on the left and zero on the right, add OH- to the right
NO2^- + 6e ==> NH3 + 7OH^-
Then add H2O to the other side.
NO2^- + 6e + 5H2O ==> NH3 + 7OH^-
Mulitiply the NO2^- half by 1 and the Al half by 3 and add them.
2Al + 8OH^- + 5H2O + NO2^- ==> 2Al(OH)4^- + 7OH^- and cancel 7 OH- obtain
2Al + OH^- + 5H2O + NO2^-==> 2Al(OH)4^- + NH3
I think that balances but check it. It's past my bed time.
all the methodology I see is to
1. Balance other compounds
2. Balance O, by adding h2o to side that is missing O
3. Balance H by adding H+ to other side to counter H20s
4. Balalnce coefficients in e-
Not sure how you came up wiht adding 5H20?
How are you able to simply add 7OH-'s in the beginning?
Your eq. is right, b/c I have hte answer but I need to understand this stuff. Final Exam Monday!
So if you could explain the add H2O to other side step a little more in reasoning, Id appreciate it. Is it a type you said multiply by 3 and not 2?
Here is how I do it. It's the long way bUT it teaches the use of oxidation numbers.
I'll use the example in this post.
1. Separate into half cells.
2. To either half, calculate oxidation state on both sides and add electrons on the appropriate side to balance the change in oxidation state..
Al ==> Al(OH)4^-
Al is zero on left; +3 on right.
Therefore, add 3e to right side.
Al ==> Al(OH)4^- + 3e
3. Count up the charge on the left and
a. add H^+ to balance charge if acid solution or
b. add OH^- to balance charge if basic solution.
Left side is zero; right side is 4-(1- for complex and 3- for 3e). This is a basic solution; therefore, add 4OH^- to left side to balance the charge.
Al + 4OH^- ==> Al(OH)4^- + 3e.
4. Add H2O to the appropriate side to balance the O and H. In this case the last line in step 3 is already balanced so this isn't necessary.
5. Check to see that everything balances in this half rxn.
a. change in oxidation state.
Next is NO2^- ==> NH3.
1. Oxidation state N on left is +3; on right is -3 so add 6e to left to balance change in oxidation state.
NO2^- + 6e ==> NH3
2. Count up the charge and add OH^^- to balance charge.
Left side is 7-. Right side is zero; therefore, add 7 OH^- to right to balance the 7- on the left
NO2^- + 6e ==> NH3 + 7OH^-
3. Now add H2O to the appropriate side to balance H (or O--it makes no difference).
NO2 + 5H2O + 6e ==> NH3 + 7OH^-
4. Check to see
a. oxidation state change balances.
b. charge balances.
c. atoms balance.
Then multiply each half rxn by a whole number to make the electrons equal and add the two half rxns.
Steve, there are several ways to balance these things; this method probably is the longest and most involved but I used this to teach my students because it MADE them use oxidation states. Some other methods don't do that.
SAW THIS!GREAT THANK YOU! my whole problem was not looking at OH right and not counting H+ when I added H+ then it threw off OH addition. Thanks!!! Really appreciate the long example and explanation. Generous of your time
Glad to help. Good luck on the exam.
And btw. honestly at this point not looking for short cut i really need to understand. That is how most tutorials teach is, yours is slightly different.
You add OH or H+ first and most say to add H2O first , then balance with H+ then come back after all doen for acidic and add OHs.... but i guess your way obv. works.
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