Balancing redox reactions using the ion-electron method

[force]

I had some trouble balancing these using the "ion electron method", ordinarily I'd balance the half reactions (excluding H and O), add H20 to compensate for O, H+ to compensate for H, and electrons to compensate for charge.

P4(s) + NO3-(aq) ---> H2PO4-(aq) + NO(g) (in an acidic solution)

2CuS(s) + HNO3 --> Cu(NO3)2 + H20 + NO (g) + S (in an acidic solution)

ClO2(aq) + OH-(aq) + ---> ClO3-(aq) + Cl-(aq) + H2O(l) (in a basic solution)

1) how should the half reactions of these chemical equations look ?

2) aren't all types of chemical reactions also redox-reactions(chemical reactions in which a transfer of electrons occur) ?

 

[Cesium]

1) a) P4 --> H2PO4
NO3- --> NO

b) CuS --> Cu+2 + S
HNO3 --> NO

c) ClO2 basically oxides itself:
ClO2 --> ClO3-
ClO2 --> Cl-

2) No e.g. In double displacement reactions oxidation numbers will always stay the same.

But lots of reactions are like the combustion of hydrogen and fluorine

H2 + F2 --> 2HF

H2 → 2H+ + 2e-
+ 2e- + F2 → 2F-
---------------------
H2 + F2 → 2H+ + 2F-

Even though you would never review this equation, when learning about redox.

 

[force]

for b) what happened to the NO3- and H2O on the right ?

for c) what happened to the OH- on the left and the H2O on the right ?

also what if you are given a half reaction deffieicent of an element other than H and O, would you add ions of that element like for ex if you where defficient in Na on one side do you add Na+ to the other ?

 

[Borek]

All three equations can be balanced without using redox approach.

In the first case you must add water on the left and H+ on the right.

Second and third are ready to be balanced without further modifications.