Balancing Complex Redox Reactions

Question

How would you go about balancing this type of redox reaction?

Au(s) + NaCN(aq) + O2(g)+ H2O(l)---> Na[Au(CN)2](aq) + NaOH (aq)

Answer 1

The previous answer is indeed correct, but if you are looking to balance this via 1/2 reactions of oxidation and reduction, you could do this as follows:

Au(s) + NaCN(aq) + O₂(g)+ H₂O(l)---> Na[Au(CN)₂](aq) + NaOH (aq)

Au has oxidation number of zero on the left and 1+ on the right. It has been oxidized

O in O2 has oxidation number of zero on the left and 2- on the right. It has been reduced

Oxidation half reaction: Au ==> Au⁺ + e-

Reduction half reaction: O₂ + 2H₂O + 4e- ==> 4 OH^-

Multiply oxidation reaction by 4 to equalize electrons and add the 2 half reactions to get...

4Au + O₂ + 2H₂O ==> 4Au⁺ + 4OH^-

Adding back the spectator ions we have...

4Au + O₂ + 2H₂O + 8NaCN ==> 4Na[Au(CN)₂] + 4NaOH

Answer 2

There are three things to notice about this reaction that will make the balancing simpler:

1. CN appears together on both sides of the equation and so can be thought of as a single species.
2. Au is a pure element on the left side, thus the coefficient can easily be determined after everything else is balanced.
3. O₂ and H₂O come together to form OH in the form of NaOH on the right.

Often, when trying to balance a complicated reaction, the task can be made simpler if we can break it into smaller pretend reactions that can be put together to make up the whole reaction.

What I mean to say is this:

1. Ignore Au for now.
2. Consider the reaction: O₂ + H₂O --> OH^-
3. If we balance this we get: O₂ + 2H₂O +4e^- --> 4OH^-
4. This tells us we will have 4OH^- which means our main reaction will have 4NaOH
5. Now we can look at this reaction: NaCN --> Na[Au(CN)₂] + 4Na⁺
6. Recall, we are ignoring the Au, and this reaction is also much easier to balance than the original.
7. If we balance we get: 8NaCN --> 4Na[Au(CN)₂] + 4Na⁺

At this point we know that we will need:

1. 8NaCN
2. 1O₂
3. 2H₂O
4. 4NaOH, and
5. 4Na[Au(CN)₂]

to balance this reaction. Now we can stop ignoring the Au and notice that 4Na[Au(CN)₂] tells us we will need 4Au to balance the reaction.

The balanced reaction therefore is:

4Au + 8NaCN + O₂ + 2H₂O. -->. 4Na[Au(CN)₂] + 4NaOH

This is how I approach a lot of complicated balancing problems. Hopefully this was helpful for you.

Please leave a comment if you found this helpful or if you need more clarification.

Thank you,

David