How do you balance this redox equation? Au O2 CN- --> Au(CN)2 ^1- H2O2

Question

in the first step when you split it apart into

Au 2CN- --> Au(CN)2 ^-1

how many electrons do you add? if you take CN as one species then the charge isn't changing so you only consider Au which is going from 0 to 1 so you add an e- to the right side. that doesn't seem to give me the right answer to balance out the OH and H2O at the end

Answer 1

This is a good reaction to know. The process is called gold leaching, its used to mine for gold.

First write out the equation.

Au(s) + O2(g) + CN-(aq) --> Au(CN)2-1(aq) + H2O2(aq)

Write the half reactions and balance them for all elements other than hydrogen and oxygen.

Au + 2CN- --> Au(CN)2-1

O2 --> H2O2

Next balance the reaction for oxygens.

Au + 2CN- --> Au(CN)2-1

O2 --> H2O2

Next balance the reaction for hydrogens.

Au + 2CN- --> Au(CN)2-1

2H+ + O2 --> H2O2

Next balance the reaction for electrons.

Au + 2CN- --> Au(CN)2-1

2- charge --> 1- charge so one electron is needed

Au + 2CN- --> Au(CN)2-1 + e-

2H+ + O2 --> H2O2

2+ charge --> 0 charge so 2 electrons are needed

2e- + 2H+ + O2 --> H2O2

Make sure the number of electrons transfered equals the number accepted. To do this, you multiply the top reaction by two to match the two electrons that are reduced.

2(Au + 2CN- --> Au(CN)2-1 + e-)

2e- + 2H+ + O2 --> H2O2

Now just add the half reactions without the electrons.



2Au + 4CN- + 2H+ + O2 --> 2Au(CN)2-1 + H2O2

The reaction above is for a reaction in an acidic solution, but we need a basic. Add hydroxide to both side to cancel out the H+.

2Au + 4CN- + 2H+ + 2OH- + O2 --> 2Au(CN)2-1 + H2O2 + 2OH-

Now simplify and thats your answer.

2Au(s) + 4CN-(aq) + 2H2O(l) + O2(g) --> 2Au(CN)2-1(aq) + H2O2(aq) + 2OH-(aq)

Next time I suggest that you use addition symbols between chemicals and don't forget the sign conventions like (s),(l),(g), and (aq).