Question

I'm a little confused about situations where multiple chemical compounds can be reduced/oxidized and the likelihood of this happening to one compound over the others, depending on the electrode potential of that compound. Here are some exercises I'm trying to figure out on this topic and their answers as given to me:

First exercise: electrolysis with inert electrodes and aqueous solution CuI₂, the question is: why do we observe I₂ gas release rather than O₂.

The answer is: If H₂O is oxidized at the anode (to form O₂) rather than I⁻ (to form I₂), we have E∘cell = E∘Cu⁺⁺/Cu - E∘H₂O/O₂=0.337−1.23=−0.893V which is lower than E∘cell = E∘Cu⁺⁺/Cu - E∘I2/I−=0.337−0.5355=−0.1985V and thus, since the potential is higher (in absolute value) for water, no O₂ should be released as long as their is I⁻ ions available

Second exercise: electrolysis with inert electrodes and aqueous solution at pH=0
, T=298,15K, p=1bar and containing 10⁻²M of Ag+ ions and 10⁻²M of Cu⁺⁺ ions, the question is: discuss the possibility of having a copper deposit without impurities from the silver.

The answer is: [I'm skipping some calculations but it can be provided if required] We get the standard electrode potentials: E∘Cu⁺⁺ / Cu = 0.337V and E∘Ag⁺/Ag = 0.799V and from Nernst equation (so calculation of the actual -non standard- cell potential) we get for H₂O/Ag and H₂O/Cu, respectively: Ecell = −0.549V and Ecell = −0.952V then we calculate the concentration of Ag⁺ at −0.952V (using Nernst law again) and we get: [Ag⁺] = 1,7×10⁻⁹M, which is very low compared to [Cu²⁺] = 10⁻²M, so almost only Cu will deposit on the electrode.

So here is my question on all this: I have read everywhere that more negative electrode potential means more likely to be oxidised, which actually correspond correctly to my first exercise (E∘I2/I−=0.5355V is more negative than E∘H2O/O2=1.23V), is it correct to apply this reasoning here though? And also that less negative electrode potential means more likely to be reduced and there I have a problem, in the second exercise E∘Cu⁺⁺ / Cu = 0.337V is not less negative than E∘Ag⁺/Ag=0.799V but quite the contrary, still we calculated that almost no Ag⁺ was deposited (so reduced, to my understanding) compared to Cu⁺⁺.

Could you help me to understand why we get this ? Simple and general guidelines for this kind of reasonings would really help.

Answer 1

The electrode potential determines whether a compound is likely to be oxidized or reduced in an electrolysis process. A more negative electrode potential indicates a higher likelihood of being oxidized, while a less negative electrode potential indicates a higher likelihood of being reduced. The specific compound that undergoes oxidation or reduction depends on its electrode potential and concentration relative to other compounds in the solution.

Step-by-step explanation:

When analyzing the likelihood of a chemical compound being reduced or oxidized in an electrolysis process, the electrode potential plays a crucial role. A more negative electrode potential indicates a higher likelihood of being oxidized, while a less negative electrode potential indicates a higher likelihood of being reduced.

In the first exercise, the electrode potential of water (-1.23V) is more negative than that of iodine (-0.5355V), resulting in the release of I2 gas instead of O2. In the second exercise, although the electrode potential of silver (0.799V) is more positive than that of copper (0.337V), the low concentration of Ag⁺ ions compared to Cu²⁺ ions leads to the deposition of copper without impurities from silver.