Gold exists in two common positive oxidation states, +1
and +3. T...

Question

Gold exists in two common positive oxidation states, +1
and +3. The standard reduction potentials for these oxidation states are
Au+1aq2 + e- ¡ Au1s2 Ered ° = +1.69 V
Au3+1aq2 + 3 e- ¡ Au1s2 Ered ° = +1.50 V
(c) Miners obtain gold by soaking gold-containing ores 
in an aqueous solution of sodium cyanide. A very soluble 
complex ion of gold forms in the aqueous solution because 
of the redox reaction 
4 Au1s2 + 8 NaCN1aq2 + 2 H2O1l2 + O21g2 
¡ 4 Na3Au1CN2241aq2 + 4 NaOH1aq2 
What is being oxidized, and what is being reduced in this 
reaction?

Accepted Answer

Hey everyone today, we're being asked to identify the reducing and oxidizing agents in the reaction between carbon and sulfuric acid to form carbon dioxide, sulfur dioxide and water. So, to identify the oxidation or oxidizing and reducing agents, we need to find out which molecules are being oxidized and reduced before and after the reaction has taken place. So let's go ahead and take a look at the react inside. First we have carbon and two sulfuric acids, H two, S. 04. So we need to find the oxidation states for these molecules. And for the carbon, it's pretty easy. Carbon is most stable. It is in its natural state alone. As C And recalling our oxidation rules, we know that a stable state element will have an oxidation state of zero An oxidation state of zero for sulfuric acid. However, it's a little more difficult. Now the molecule we wanted or elements that we need to pay attention to is the sulfur because it can have different oxidation states depending on what it is bonded to. Now recall, this is again a neutral element, which means it must counteract the some of the oxidation states of the other elements bonded to it in order for it to be neutral, recalling our oxidation rules, We know that oxygen unless it's bonded in hydrogen peroxide or bonded to a flooring, will have a charge of negative two. And since we have four oxygen's, that makes it a total charge of -8. Similarly, when hydrogen is bonded to a non metal in a molecule, it will have a charge of plus one an oxidation state of plus one. And since we have to hide regions that makes us plus two, therefore the total charge or or some of oxidation states in the molecule is negative six, it is negative eight plus two. So to counteract this, sulfur must have the oxidation state of plus six. So sulfur in sulfuric acid Has an oxidation state of Plus six. Taking a look at the reactant. Now we have a carbon dioxide C. 02, two, sulfur dioxides as 02 And two water molecules. Let's take a look at the carbon again. We remember our oxidation rules that oxygen has an oxidation state of negative two. And since we have two of those that makes us negative four. So the carbon must have an oxidation state of plus four. So it means it goes from 0 to Plus four. And sulfur similarly is also bonded to two oxygen's. So this is negative four, which means it also goes to plus four, but now it goes down from plus six to plus four. So what does this mean? Well, carbon lost electrons. It lost electrons. Meanwhile sulfur gained electrons after the reaction. And if we remember our oxidation and reduction rules for oil rig, let me write this. Better oil rig oxidation is losing and reduction is gaining which means carbon was oxidized it lost electrons and sulfur was reduced. It gained electrons. Which means that carbon here got oxidized oxidized and sulfuric acid got reduced. And because of this we know that carbon is therefore the reducing agent because it aids the reduction of sulfuric acid reducing agent, and sulfuric acid is the oxidizing agent as it aids the oxidation of carbon. I hope this helps, and I look forward to seeing y'all in the next one.

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