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)
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!
2 answers
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
Like in this example? : C + O2 = CO2
2 answers