Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: granite350 on April 26, 2006, 01:22:05 PM
-
Balancing Redox Reactions. Balance the following redox reactions by the half-reaction method. This topic is discussed on pages 803 - 807 [840 - 845] in the textbook.
a. As2O3(s) + NO3?(aq) ? H3AsO4(aq) + N2O3(aq) (in acidic solution)
b. H2O2(aq) + Cl2O7(aq) ? ClO2?(aq) + O2(g) (in basic solution)
-
Please show us your attempts.
-
i can't seem to figure out the charges for AsO3 which i put as 0 H3Aso4 I put -3
just having trouble with the charges for N2O3(aq)
-
just having trouble with the charges for N2O3(aq)
I think that you mean the oxidation number.
Oxygen has the Ox-num -2, If you know this it is pretty easy to calculate the oxidation number of nitrogen.
-
http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc
-
my problem is trying to find the correct oxidation states
this has always been my problem
-
I think that you mean the oxidation number.
Oxygen has the Ox-num -2, If you know this it is pretty easy to calculate the oxidation number of nitrogen.
...and there are many other more detailed infos in Borek's link...
-
Wow thanks guys borek helped me out alot great site or link given.
saved me so much time
anyways is H3AsO4 oxidation number 5 since their are 4 O which is -8 and 3 H at +3 which gives the cetral atom a charge of 5
hope i have the hang of it and off to a good start thanks
-
anyways is H3AsO4 oxidation number 5 since their are 4 O which is -8 and 3 H at +3 which gives the cetral atom a charge of 5
Yeah! Correct! Oxidation number of Arsenic is +5! Now just write the second half-reaction and that's it! :)
-
for N2O3(aq) is the ox idation state this
O has 3 attoms which is 3*-2 which is has the central atom of 6
-
for N2O3(aq) is the ox idation state this
O has 3 attoms which is 3*-2 which is has the central atom of 6
How many nitrogen atoms in N2O3?
-
their are 2 nitrogen atoms this is where i got confused the equation does not work out cannot assign an oxidation and reduction equation
-
do u divide by 2
-
Wow this is what i got so far but its not much
AsO3-->H3AsO4+5e- oxidation half\
NO3- --> N2O3 + 3H2O(l)
I need to still balance out the hydrogents is the oxidaiton half don't know where to go from here i have a long way to go
-
AsO3-->H3AsO4+5e- oxidation half\
Should be As2O3.
Assign ON for all four substances involved. And write them here.
-
its a solid so the ON would be zero
-
2 H3AsO4 + 4 e- + 4 H+ -> As2O3 + 5 H2O
-
its a solid so the ON would be zero
State of aggregation doesn't matter. Do what I asked you to do.
-
albert whe i worked out H3AsO4 i get 5 electrons because O is -2 times 4 which is 8 and H is -1 times 3 which is -3 so i get 5 e-
-
for the reduction par is this correct 2NO3- + 2H + 2 e- -->N2O3 + 2H2O
-
Hydrogen is +1!
-
8-3 = 5 just wondering because if it is 4 e it would make the simplification much easier
-
for the reduction par is this correct 2NO3- + 2H + 2 e- -->N2O3 + 2H2O
Why only 2 electrons? How many NO3-s can you see?
-
I mean for H3AsO4=5 electrons
-
NO3- i see +5 O=-6 central atom is 5
wonder what i'm doing wrong???
the final answer i has was As2O3 + 5H2O+4NOH3- + 2 H + 4e- --> 2 H3AsO4 +4 H +4e- +2N2O3+4H20
-
NO3- i see +5 O=-6-1=7
???
NO3- oxygen = (-2)*3 = -6
nitrogen = (-6) - (-1) = +5
-
but to balance out the reduction part i need +2 electrons
wow i am confused
-
but to balance out the reduction part i need +2 electrons
Who said that?! >:(
You need to balance the half reaction (reduction). The charges aren't balanced and remember you have 2 NO3- groups: two electrons are not enough... ;)
-
wow i am confused
You are confusing yourself IMHO. Try to follow simple steps, don't try to do everything at the same time.
-
but N2O3 is +3
this is why i keep thinking i have to add 2 electrons to the other side.
this is where i get confused.
-
2NO3- is like having nitrogen +10 ( +5*2). On the other hand, N2O3 means you have +6 ( +3*2).
-
don't u divide nitrogen by 2 in N2O3 -6/2=3
-
Please, write the reduction reaction as I've, almost, told you. I know you can. :)
10 - 6
[/size]
How much is it?
-
2 NH3+2H +4e---> N2O3 + 2 H2O
-
2NO3- + 2H+ +4e- --> N2O3 + 2 H2O
Charges aren't balanced. Concentrate on H+: among the reagents there're 6 negative against 2 positive charges.
-
+7 e-
I might go to the chem tutor hope he's not packed this is getting me discouraged
-
+ 7 e- ?! Where does this come from?!
You have just to change the number of protons and, consequently, the number of molecules of water. The rest of the chemical equation is correct.