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I am preparing for my exam, and for some reason really struggle with half-reactions in the context of Galvanic cells. I do manage to solve the more straight forward questions, but really struggle with some.

Question asks to "write the half-reactions, balanced formula and Galvanic cell structure". Given reaction is

$$ \ce{Au+ (aq) -> Au (s) + Au^3+ (aq) } $$

I managed to successfully identify that the cathode is: $\ce{Au+ (aq) + e- -> Au (s) }$

But I got the anode wrong, I thought it should be $ \ce{Au+ (aa) -> Au^3+ (aq) + 2e- }$, while book says it should be $ \ce{Au (s) -> Au^3+ (aq) + 3e- } $

What am I missing?

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  • $\begingroup$ The book is wrong. Those half reactions can't be combined to give the total reaction. $\endgroup$
    – Zhe
    Mar 28 at 18:47

3 Answers 3

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Both half-reactions involving Au(III) are correct. However, one is easier to find in a table of standard potentials:

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Source: http://ch301.cm.utexas.edu/data/standard-potentials.pdf

If you combine your set or their set of half-reactions into a balanced chemical equation, you get the same answer. This is because you can combine two of the half reactions to get the third (they are linearly dependent):

$$\ce{Au(s) -> Au+ (aq) + e-}\tag{1}$$ $$\ce{Au+(s) -> Au^3+ (aq) + 2e-}\tag{2}$$ $$\ce{Au(s) -> Au^3+ (aq) + 3e-}\tag{1+2}$$

[OP] What am I missing?

You are missing the practical part. It is more straightforward to build a cell where the electrode is one of the reacting species. For the Au(I) to Au(III) reaction, you would need an inert electrode. Also, if you wanted to calculate the cell potential, you'd have extra work figuring out the missing electrode potential of half-reaction (2).

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The reaction is a disproportionation Au+ gives Au and Au+3

Oxidation: Au+ = Au+3 + 2e-; Reduction: Au+ + e- = Au. Multiply reduction by 2 to balance charge and add equations.

3Au+ = 2Au + Au+3 These are the correct equations and I do not think the book is correct about the anode reaction.

Lets try it their way!

Oxidation Au = Au+3 + 3e- ; Reduction Au+ + e- = Au Multiply reduction by 3 and add

Au + 3Au+ = Au+3 + 3Au Subtract a gold from each and... There is an extra oxidation of Au metal. The initial oxidation to Au+ has the highest potential about so should be inhibiting.

The problem is that to develop a cell the oxidant and reductant must be separated in space and that is not possible here since they are the same. The book's solution is to supply a different reductant, Au metal, for the anode. The only way I see to do this is a concentration cell where the cathode reaction is reduction of aurous ion 2Au+ +2e- = 2Au, Rate = Kr[Au+]^2; the anode reaction is oxidation of aurous ion Au+ = Au+++ + 2e-, rate = Ko[Au+]. When the concentrations are equal there is no cell potential and the disproportionation proceeds in both cathode and anode compartments. Increasing the concentration of Au+ in one of the compartments Increases the rate of the reduction reaction in that compartment removing electrons from the other compartment increasing the rate of the oxidation reaction in the other compartment. The compartment with the higher concentration loses Au+ ions thru reduction twice as fast as the other compartment loses Au+ from oxidation. The cell voltage decreases until the concentrations of Au+ are the same in each compartment and the cell voltage becomes zero. the disproportionation is continuing behind the scenes.

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The anodic reaction cannot be : $\ce{Au -> Au^{3+} + 3 e-}$, because there is no neutral gold atom $\ce{Au}$ in the original mixture to start the oxidation reaction of $\ce{Au}$. So the book is wrong, and the anodic reaction is $\ce{Au^+ -> Au^{3+} + 2 e-}$.

But the reaction $\ce{Au -> Au^{3+} + 3 e-}$ is necesary to know, just to get the potential of the anodic reaction $\ce{Au^+ -> Au^{3+} + 2 e^-}$. As Karsten states, the potential of $\ce{Au^+ -> Au^{3+} + 2e^-}$ is not given in the tables. But it can be deduced by combining the two potentials of the reaction ($1$) and ($1+2$) from Karsten's answer. As the potential of ($1$) is $+1.69$ V, and the potential of ($1$) + ($2$) = $1.42$V, the potential of ($2$) (which is also the potential of $\ce{Au^+ -> Au^{3+} + 2 e-}$), is $\frac{3·1.42 - 1,68}{2} = 1.285 $ V.

The necessity of this last calculation may have been at the origin of the wrong statement, contained in the textbook.

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  • $\begingroup$ A more complete table is needed Lange's Handbook gives "ca. 1.29v" suspiciously close to your calculated value. Please read my answer above; it was difficult to figure out how to separate something from itself and reunite it. $\endgroup$
    – jimchmst
    Mar 11 at 21:49

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