a. Tl 2 O 3(s) +NH 2 OH (aq) ® TlOH (s) +N 2(g) (basicsolution) b. Cr 2 O 7 2 -

Question

a.   Tl2O3(s) +NH2OH(aq) ® TlOH(s) +N2(g)       (basicsolution)

b.   Cr2O72-(aq)+ I-(aq) ®Cr3+(aq) +IO3-(aq)         (acidic solution)

Explanation / Answer

balancing these reactions 8 H+ + (Cr2O7)2- + I- ---> 2 Cr3+ + IO3- + 4 H2O 2 H+ + 2 H2O + 2 NO3- + As2O3 ---> 2 H3AsO4 + N2O3 Not too sure about this one, and I'll explain why 2 Tl2O3 + 6 NH2OH ---> 3 N2 + 4 H2O + 2 TlOH 8 H+ + 2 NO2- + (Cr2O7)2- ---> 3 NO3- + 2 Cr3+ + 4 H2O When balancing half reactions in acidic solution, you separate theoverall reaction based on the oxidation and reduction halfreactions. Then you balance them individually by this process: 1) Balance everything except hydrogen and oxygen 2) Balance oxygen by using water 3) Balance hydrogen by using H+ 4) Balance change by using electrons If the electrons lost don't equal the electrons gained, you have tomultiply one of the reactions by some number in order for thenumber of electrons to be equal. Cancel out things that appear on opposite sides of the now balancedindividual half reactions, which will be the electrons for sure,and sometimes water and H+ You are done if it says acidic solution. If it says basic solution, you need to account for the fact thatthere will be no H+ in the solution, it will be attracted to theOH-, and it will form water as a result. In this case, you need toadd however many OH- to each side of the overall reaction for theH+ that you have. This is basically an extension of the acidicsolution, with one extra step. For example, if you had 3H+, youwould add 3 OH- to both sides, and you would have a net 3 H2O and 3OH- left on the opposite side. This is why I am not sure about the basic solution reaction thatyou gave. There were no H+ left in the reaction, there were 6 oneach side that would cancel out. Maybe someone else can help youout with that one, because I'm just not sure of it.

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