Chemistry questions... :)

Question

69)

Balance each of the following redox reactions occuring in acidic aqueous solution. Use the half-reaction method.

a) ClO4- (aq) Cl- (aq) ---> ClO3- (aq) Cl2 (g)

b) MnO4- (aq) Al(s) ---> Mn^2 (aq) Al^3 (aq)

c) Br2 (aq) Sn (s) ---> Sn^2 (aq) Br- (aq)

I don't know how to use the half-reaction method... :(

Please explain so I can understand in the future! Thanks.

:)

Answer 1

With the half reaction method you are balancing the equation first by identifying the what is being oxidized and what is being reduced.

a) First write the reaction:

ClO4- (aq) + Cl- (aq) ---> ClO3- (aq) + Cl2 (g)

Next separate the equations by matching which compounds have the same set of elements. In this case ClO4- is match with ClO3- and Cl- with Cl2. (A side note: try and match the uncommon elements together first for example matching heavy metals) So:

ClO4- --> ClO3-

Cl- --> Cl2

Next balance the equations for all the atoms other than oxygen and hydrogen.

ClO4- --> ClO3-

2Cl- --> Cl2

Next balance the oxygens next by adding water to the side defficent oxygens.

ClO4- --> ClO3- + H2O

2Cl- --> Cl2

Next balance hydrogen atoms by adding H+ to the side defficent hydrogens.

2H+ + ClO4- --> ClO3- + H2O

2Cl- --> Cl2

Next balance the charge on both side by adding electrons to the more positive side. This is kinda tricky even for me sometimes.

2H+ +ClO4- --> ClO3- + H2O

1+ net charge --> -1 net charge

See how I counted both ions to find the net charge. To balance both sides you need 2 electrons on the more positive side, because you don't want a net charge of 0. You want the same net charge on both sides of the equation. Do the same for the other half reaction.

2Cl- --> Cl2 + 2e-

Now for both half reactions you half to make sure there is the same number of electrons given as there are accepted. To do this you look at the electrons. There are 2 for both half reactions so you are lucky. But you won't get so lucky on problem b) but I digress. The next thing you do is add both half reactions (forget about the electrons because they can't pressent on both sides of the equation).

The final answer should be:

2H+(aq) + 2Cl-(aq) + ClO4-(aq) --> ClO3-(aq) + Cl2(aq) + H2O(l)

b) I'm going to go a lot quicker this time.

MnO4-(aq) + Al(s) --> Mn2+(aq) + Al3+(aq)

Write the half reactions.

MnO4- --> Mn2+

Al --> Al3+

The elements other than oxygen and hydrogen are balance. Next add H2O to the side defficient oxygens.

MnO4- --> Mn2+ 4H2O

Al --> Al3+

Next add H+ to the side defficient hydrogen atoms.

8H+ + MnO4- --> Mn2+ + 4H2O

Al --> Al3+

Next add electron to the more positive side to balance the charge.

5e- + 8H+ + MnO4- --> Mn2+ + 4H2O

Al --> Al3+ + 3e-

This is something you haven't done before. This is a case when the  transfered electrons on the half reactions are not the same. No problem. Multiply each half reaction by the lowest number so the number of electrons transfered is the same. Kinda like finding the lowest common denominator in fractions.

3(5e- + 8H+ + MnO4- --> Mn2+ + 4H2O)

5(Al --> Al3+ + 3e-)

The lowest common denominator is 15 so you multiply one half reaction by 3 and the other by 5 and you get 15 electrons transfered.

Now add the reactions, not forgeting that you multiplyed them.

24H+(aq) + 3MnO4-(aq) + 5Al(s) --> 3Mn2+(aq) + 5Al3+(aq) + 12H2O(l)

c) The next one should be easy.

Br2(aq) + Sn(s) --> Sn2+(aq) + Br-(aq)

Write half reactions and balance the elements other than oxygen and hydrogen. Look how lucky you have no oxygens or hydrogens. So I guess you can go right to balancing electrons

Br2 --> 2Br-

Sn --> Sn2+ + 2e-

The electrons transfered are the same, too easy right. Now just add them.

Br2(aq) + Sn(s) --> 2Br-(aq) + Sn2+(aq)