Question

1. (a) Balance the two reduction half - reactions shown under basic condi tions: (i) N2-> NH2OH E

Answer 1

Hi,

This is indeed a redox reaction. The first step is to separate this equation into two half reactions. Do this based on the reactants and products that have similar elements other than O or H. In this case, separate based on Tl and N:

Tl2O3(s) -----> TlOH(s) (1)
NH2OH(aq) -----> N2(g) (2)

Lets take reaction (1)
First, balance for elements other than O or H. In this case, Tl:
Tl2O3(s) -----> 2TlOH(s)
Next, add an OH- to the right hand side to compensate for the O that is missing:
Tl2O3(s) ------> 2TlOH(s) + OH-
Next, balance for hydrogens. Right hand side has 3 hydrogens. Add 3H+ on the left hand side:
Tl2O3(s) + 3H+ -----> 2TlOH(s) + OH-
Now balance for charge by adding electrons (e-). Add enough to one side such that it matches the charge on the other. In this case, we have 3+ charge on one side and -1 charge on the other. The easiest way is to add 4e- to the products side, thus giving both sides of the reaction a net charge of -1.
Tl2O3(s) + 3H+ + 4e- -----> 2TlOH(s) + OH- (1*)
This half reaction is now balanced for both charge and matter.

Next, lets look at the the second half reaction:
NH2OH(aq) -----> N2(g)
Like last time, balance for N's first:
2NH2OH(aq) -----> N2(g)
Next, balance by adding OH- to the side clearly lacking oxygen and hydrogen (the product side):
2NH2OH(aq) -----> N2(g) + 2OH-
Add H+ to the product side to balance the hydrogens:
2NH2OH(aq) -----> N2(g) + 2OH- + 4H+
OH- and H+ will react in equal quantities to form water:
2NH2OH(aq) -----> N2(g) + 2H2O + 2H+
Again, add e-'s where necessary to balance charge:
2NH2OH(aq) -----> N2(g) + 2H2O + 2H+ + 2e-
Both sides have a net charge of 0.
To continue, we want basic conditions. To do so, we must eliminate the H+'s on the products side fully. Do this by adding 2OH- to both sides:
2NH2OH(aq) + 2OH- ------> N2(g) + 2H2O + 2H+ 2OH- + 2e-
2NH2OH(aq) + 2OH- ------> N2(g) + 2H2O + 2H2O + 2e-
2NH2OH(aq) + 2OH- ------> N2(g) + 4H2O + 2e- (2*)
Notice how the half reaction is still balanced for matter and charge?

We now have our two BALANCED half reactions:
Tl2O3(s) + 3H+ + 4e- -----> 2TlOH(s) + OH- (1*)
2NH2OH(aq) + 2OH- ------> N2(g) + 4H2O + 2e- (2*)
As you can see, 1* gains electrons. That half reaction is the REDUCTION half reaction.
2* loses electrons and is the OXIDATION half reaction.
Now, to get your overall reaction, combine the two reactions together. To do this, you must even out the electrons on both sides of the equation such that they balance out. (A proper balanced redox reaction must have no electrons present!)
To do this, we can multiply equation 2* by 2, thus giving:
4NH2OH(aq) + 4OH- -----> 2N2(g) + 8H2O + 4e- (2**)
Now, add 2** and 1* together:

4NH2OH(aq) + 4OH- -----> 2N2(g) + 8H2O + 4e- (2**)
Tl2O3(s) + 3H+ + 4e- -----> 2TlOH(s) + OH- (1*)
______________________________________...
4NH2OH(aq) + 4OH- + Tl2O3(s) + 3H+ + 4e- -----> 2N2(g) + 8H2O + 4e- + 2TlOH(s) + OH-
4NH2OH(aq) + 3H2O + OH- + Tl2O3(s) --------> 2N2(g) + 8H2O + 2TlOH(s) + OH-
4NH2OH(aq) + Tl2O3(s) -------> 2N2(g) + 5H2O + 2TlOH(s)

Thus, your overall equation is:
4NH2OH(aq) + Tl2O3(s) -------> 2N2(g) + 5H2O + 2TlOH(s)
(This equation was balanced in basic solution, but could also be done in acidic).
As you can see, this reaction is balanced both in charge and in matter. This is your final answer.

Hope this helps!