What would be the reduction half reaction for the following redox reaction:
Au(s) + NaCN(aq) + O2(g)+ H2O(l)---> Na[Au(CN)2](aq) + NaOH (aq)
Since oxygen normally gets reduced to H2O, could it be 1/2O2 + 2H = H2O?
Thanks!
Yes, the O2 is reduced but not to H2O because there is no H2O on the right side. The O2 is reduced to O^2- in NaOH.
4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(l) = 4 NaAu(CN)2(aq) + 4 NaOH(aq)
Thank you for your response! Yea I saw that but I thought you could leave OH as a spectator ion and just assume that O2 would ultimately reduce to water. I just figured you could do that since in some reactions where it may not be clear to see what oxygen gets reduced to without OH being in the original equation.
Like in this example? : C + O2 = CO2
To determine the reduction half-reaction for a redox reaction, you'll need to identify the changes in oxidation numbers for each element involved. Here's how you can determine the reduction half-reaction for the given redox reaction:
Step 1: Identify the oxidized and reduced elements.
In the given reaction, the element that gets oxidized is Au (from Au(s) to Na[Au(CN)2](aq)), and the element that gets reduced is O (from O2(g) to H2O(l)).
Step 2: Write the half-reactions for each element.
The half-reaction for the oxidation process of Au is:
Au(s) → Na[Au(CN)2](aq)
The half-reaction for the reduction process of O is:
O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
Note that in the reduction half-reaction, we balanced the atoms (using H2O) and added the necessary electrons (e-) to balance the charges.
So, the reduction half-reaction for the given redox reaction is:
O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
Your proposed half-reaction, 1/2O2 + 2H = H2O, is not balanced and doesn't consider the number of electrons involved in the reduction process.