[kenny122]

i was asked to balance this equation using oxidation reduction methods.

AU(s) H+(aq) + No3-(aq) + Cl-(aq) --> AuCl4-(aq) + NO2(g) + H20(l)

i balanced by breaking the equation into two half reactions, and these are the coefficiants in front of each.

1,2,1,4 --> 1,1,1

then it asks. the reduction potential for the AuCl4 half reaction is 1.00 V. suggest a substance that could be used to convert the AuCl4 back into Au. Justify your choice. would you have to use a substance that would produce a spontaneous reaction?

thanks very much in advance for the help.

[Alpha-Omega]

This RXn is the dissolution of gold in Aqua Regia. 

Aqua regia can dissolve gold because each of its two component acids carries out a different function. The nitric acid is a good oxidizing agent. Chloride ions from the hydrochloric acid from coordination complexes with the gold ions, removing them from solution. Reducing the concentration of the Au3+ ions shifts the equilibrium towards the oxidized form.

Reaction equation:

Au(s ) + 3NO₃^-(aq ) + 6H⁺⁽aq ) ------> Au³⁺⁽aq ) + 3NO₂(g ) + 3H₂O(l )
Au³⁺(aq ) + 4Cl^-(aq ) -----> AuCl4^-(aq )

Aqua regia (Latin for "royal water") is a highly corrosive, fuming yellow or red solution. The mixture is formed by freshly mixing concentrated nitric acid and concentrated hydrochloric acid, usually in a volumetric ratio of one to three respectively.

[Alpha-Omega]

Aqua Regia:

One part HNO3 and three parts of HCl when mixed together, the mixture so obtained is called Aqua Regia. Noble metals such as gold,platinum are not dissolved either of the acid alone but in Aqua Regia they easily dissolve due to liberation of atomic Cl which is highly reactive. 

HNO₃ + 3HCl  NOCl + 2H₂O+2Cl 

Dissolving gold
Aqua regia dissolves gold, even though neither constituent acid will do so alone, because, in combination, each acid performs a different task. Nitric acid is a powerful oxidizer, which will actually dissolve a virtually undetectable amount of gold, forming gold ions (Au3+). The hydrochloric acid provides a ready supply of chloride ions (Cl-), which react with the gold to produce chloraurate anions, also in solution. The reaction with hydrochloric acid is an equilibrium reaction which favors formation of chloraurate anions (AuCl4-). This results in a removal of gold ions from solution and allows further oxidation of gold to take place, and so the gold is dissolved. In addition, gold may be oxidized by the free chlorine present in aqua regia. Appropriate equations are:

Au (s) + 3 NO₃^- (aq) + 6 H⁺ (aq) → Au³⁺ (aq) + 3 NO₂ (g) + 3 H₂O (l)
Au³⁺ (aq) + 4 Cl^- (aq) → AuCl₄^- (aq)
The oxidation reaction can also be written with nitric oxide as the product rather than nitrogen dioxide.

Au (s) + NO₃^- (aq) + 4 H⁺ (aq) → Au³⁺ (aq) + NO (g) + 2 H₂O (l)


REDUCTION POTENTIALS: 

The reduction potential for the AuCl4 half reaction is 1.00 V.

1.  The more positive the E°, the higher the tendency the species will undergo reduction.
2.  The more negative the E°, the higher the tendency the species will undergo oxidation.

Then you are asked:

Suggest a substance that could be used to convert the AuCl4 back into Au. So you need something that will reduce the AuCl₄ back to the SOLID Au (Au(s)).

To find this you need to go the Tables with Reduction Potenials.

The standard redustion potential for  Au⁺³ + 3e- ----> Au°   E° =1.50 V

Link To Standard Reduction Potentials:
 
http://www.adriandingleschemistrypages.com/apsrpotentials.pdf

Justify your choice. would you have to use a substance that would produce a spontaneous reaction?

I hope this Helps   

[Alpha-Omega]

This is a nicer table:  http://library.thinkquest.org/3659/reference/reductionpotentials.html

[Alpha-Omega]

Balanced half-reactions                         Eo /


Au3+ + 2e-  Au+                                 +1.41
Au3+ + 3e-  Au(s)                                +1.50
AuCl4- + 2e-  AuCl2- + 2Cl-                  +0.926
AuBr4- + 2e-  AuBr2- + 2Br-                  +0.805
AuCl4- + 3e-  Au(s) + 4Cl-                    +1.002
AuBr4- + 3e-  Au(s) + 4Br-                     +0.858
AuCl2- + e-  Au(s) + 2Cl-                      +1.154A
uBr2- + e-  Au(s) + 2Br-                          +0.963

[Alpha-Omega]

I was trying to decide the best approach to the last part of the question.  This is like an AP  HS Chemistry question and has a number of aspects that are addressed in the problem solving process.

You are given a reduction potential of 1.00 V for the AuCl4 half reation...this has to do with Voltaic cells.

So you have:  AuCl4- + 3e-  Au(s) + 4Cl-                  Eo =  +1.00 V


OK your sponteneity issue: 

So now you have to suggest a substance that could convert AuCl4 back to Au(s).  The Oxidized AuCl4 is going to be reduced back to Au (s). 

To do that you have to look at the Activity Series.  The SPONTENEITY of the reduction (if the process as driven by this chosen species produces a spontaneous RXN) has to be determined from the Activity Series...

So you are looking for some species that will oxidize more easily than gold.  In this process the gold is reduced to its solid/elemental state where the oxidation state on Au⁰.

You want to view the AUCl4 as the oxidiziing agent. An oxidizing agent causes another substance to be oxidized.  So you look at the AuCl4 as your oxidizing agent. 

An oxidizing substance ia always reduced in the process of oxidizing another species.

A reducing agent causes another substance to be reduced.  You want a substance that will reduce the Au to its elemental state.  As the reducing agent reduces the AuCl4 to Au it is going to be oxidized.

Gold:  Au is at the very bottom of the scale in terms os EASE OF OXIDATION.

The RXN:  Au   ----> Au^+[sup]3 [/sup] + 3e^-  is at the bottom of the list...

Gold has the hihest reduction potential and the lowest oxidation potential.

Here te gold cation will gain 3 electrons.

This HINT should make your process very easy....

You are going to have to look at the reduction potentials and the activity series to answer the last part of the question.

And remember:  A spontaneous redox reaction involves the transfer of electrons.

[AWK]

Practically zinc (preferable dust) is used.