nedddddddd help in balancing equations

Yellow-Jellow said:

nedddddddd help in balancing equations

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hi everybody...i really have a lot of trouple balancing equations in acidic and basic conditions so i need your help in these ...thanx alot !! .. ...

balance the following in acidic conitions

1) Br2(l) + SO2(g) -----> Br-(aq) + SO4 2- (aq)



2) Cu(s) + NO3 - (aq)------> Cu2+ (aq) + NO(g)



3) PbO2(s) + Cl- (aq) -----> PbCl2(s) + Cl2 (g) ( balance Cl in both half-reactions)



4) Cl2(g) --------> ClO3 - (aq) + Cl- (aq) (same reactant in each half-reaction)



Balance the following in basic conditions



5) N2H4(aq) + Cu(OH)2 (S) ------> N2(g) + Cu(s)



6) MnO4 - (aq) + NO2 - (g) -------> MnO2(S) + NO3 - (aq)



7) Cr(OH)3 (s) + H2O2(aq) --------> CrO4 2- (aq) + H2O (aq)



8) Al(s) + NO3 - (aq) --------> Al(OH)4 - (aq) + NH3(aq)



also i like to know what is the half-reaction methode and how can i do it..thanx alot for the help!!

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Yellow-Jellow said:

nedddddddd help in balancing equations

--------------------------------------------------------------------------------

hi everybody...i really have a lot of trouple balancing equations in acidic and basic conditions so i need your help in these ...thanx alot !! .. ...

balance the following in acidic conitions

1) Br2(l) + SO2(g) -----> Br-(aq) + SO4 2- (aq)



2) Cu(s) + NO3 - (aq)------> Cu2+ (aq) + NO(g)
Here's how I'd do this:
Cu(s) -> Cu(2+) + 2e- x3
NO3 + 3 e- + 4 H+ -> NO + 2 H2O x2

3Cu + 2NO3- + 6e- + 8H+ -> 3 Cu(2+) + 6e- + 2NO + 4H20

3Cu + 2NO3- + 8H+ -> 3Cu(2+) + 2NO + 4H20
Total charge: 6+ = 6+
That should be correct.


General Chemistry come back to me!
In basic solution there's one extra step. Notice how there's 8 H+ ions there, you'd add 8OH- to both sides to make water and excess OH- on one side.

So you'd get it
3Cu + 2NO3- + 8H20 -> 3Cu2+ + 2NO + 4H20 + 8OH-

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