For a copper/copper sulfate half reaction, do sulfate anions move to the anode and lose electrons?

Question

In my (high school) textbook, there's an example on finding the charge on 1 mole of electrons, which involves performing the electrolysis of an aqueous copper (II) sulfate using a copper anode and a copper cathode. I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.

However, in this example, the copper anode loses mass as the copper goes into solution as copper ions. What I don't understand is, since the sulfate ions in this case would go to the anode, wouldn't it combine with the copper ions to form copper sulfate? Or wouldn't it donate two electrons to the copper ion so it becomes copper metal again? Since the sulfate ion moves to the anode, shouldn't this happen?

Answer 1

I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.

This is one way of achieving charge balance, but there are many other ways.

Charge balance of half-reactions

Half-reactions have electrons either as reactants (reduction half-reaction at the cathode) or as products (oxidation half-reaction at the anode). For the cell in question, the oxidation half-reaction at the anode is:

$$\ce{Cu(s) -> Cu^2+(aq) + 2e-}$$

There is no physical state associated with the electrons, but they are removed through the wire. So if electrons appear in the half-reaction, you need some ions to balance it out. In a oxidation, we produce electrons, so either we need to produce cations at the same time or consume anions (or have ions on both sides of the equation, but of unequal charge). The ions (in solution) are also responsible for the charge transport between anode and cathode.

Oxidation state imbalance of half-reactions

At least one atom (as element, simple ion, or part of a compound or complex ion) has to undergo a change in oxidation state. The species that balance the charges, however, might or might not undergo a change in oxidation state. In the half-reaction in question, copper changes oxidation states, and the copper ions balance out the charge of the electrons so that both sides of the half-reaction have equal charge (zero, in this case).

To give a different example, here is a half-reaction involving lead:

$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + H+(aq) + 2e-}$$

Here, the lead gets oxidized (from element to +II), but remains at the electrode. The charge balance is achieved by hydrogen sulfate ions traveling to the electrode and remaining at the electrode as part of the solid lead(II)sulfate, and the released hydrogen ion.

Direction of ion flow

In the reaction in question, copper cations travel from anode to cathode in the solution to balance the charge transport due to electrons traveling from anode to cathode via the wire. We can write the equation showing this explicitly by combining the half-reactions and keeping anode and cathode species labeled.

$$\ce{Cu(ano) + Cu^2+(cat) + 2e-(cat) -> Cu^2+(ano) + 2e-(ano) + Cu(cat)}$$

Because we can't have electrons or copper ions accumulate in or at the electrodes, we need a transport mechanism for both species (through the wire and through the solution, respectively).

Here are two more examples with different ion transport.

Water electrolysis

$$\ce{4H+(aq) + 4e- -> 2H2(g)}$$ $$\ce{2H2O -> O2 + 4H+(aq) + 4e-}$$

At low pH, the cathode consumes hydrogen ions while the anode produces hydrogen ions. At neutral pH, you would say the cathode consumes hydrogen ions and the anode consumes hydroxide ions produced from auto-dissociation of water (i.e. anions travel to the anode and cations travel to the cathode - the classic case giving anions and cations their name).

Lead-acid battery

$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + 2e- + H+}$$ $$\ce{PbO2(s) + 3H+(aq) + HSO4^-(aq) + 2e- -> PbSO4(s) + 2H2O}$$

The element that changes oxidation state is lead in various solid compounds (elemental, lead(IV) in lead oxide and lead(II) in lead sulfate). The ion transport is due to hydrogen sulfate going from solution to both electrodes, forming lead sulfate, and hydrogen ions going to the cathode to combine with oxygen to form water (and hydrogen ions being released from hydrogen sulfate at the anode). So here, ion transport is accomplished with anions going to both electrodes, but an excess of cations (hydrogen ions) going to the cathode. To emphasize, hydrogen sulfate anions are traveling to the cathode. This is an counter example to "anions travel to the anode".

Cells with salt bridge

None of the cells in my three examples had salt bridges. If half-reactions have separate solutions connected by a salt bridge, ions don't travel the entire way between electrodes; instead, redox-inert ions such as potassium cations and chloride anions negotiate the charge transport between solutions of the half cells.

Answer 2

What is occurring is primarily the electrolysis of water in an electrolyte of CuSO4 with Copper electrodes. The detailed mechanics of the reaction is, however, more complex than is usually discussed.

To begin, I start with the seemingly simple electrolysis of water with a small amount of an electrolyte in acidic conditions as previously presented on StackExchange, to quote:

At the cathode the following mechanism is proposed in acidic media: \begin{align}\ce{ 2H+ + 2e- &-> H^{$*$} + H+ + e- \\ &-> H2 }\end{align} The asterisk superscript denotes, that the formed Hydrogen $\ce{H}^*$ is chemically adsorbed to the active site at the metal surface. They continue that hydrogen is only weakly bound to the metal surface and therefore the hydrogen evolution is much faster than the rate of oxygen production. Hence they focus on the reaction on the anode.

The referred to $\ce{H^{$*$}}$ has also been referred to as 'Hchemisorbed'. To quote from a reference:

The cathodic reaction produces Hchemisorbed by picking up an electron that released in the anodic reaction (H+ + e- = Hchemisorbed ) in Al corrosion in HCl. In such acidic solutions, the Hchemisorbed on the metal surface reacts by combining with other adsorbed Hchemisorbed to form H2 gas molecule, which bubbles from the metal surface. A very small amount of the uncombined Hchemisorbed will remain; however, this amount does not affect the whole process.

Apparently, the Hchemisorbed operational behaves as an (e-,H+) pair, to quote a source (bottom Page 818 here) operating on PbS as follows:

$\ce{ Pb(2+)S(2-) + 2 H^{$*$} -> Pb + H2S (g) }$

Now, with aqueous cupric sulfate, a corresponding postulated reaction:

$\ce{ Cu(2+)SO4(2-) + H^{$*$} -> Cu(+) + HSO4- }$

where the Cu+ concentration is not expected to be stable per this reference and the formation of HSO4- may lower the reaction cell's pH.

Further, in the current context with a cuprous oxide coating on the copper electrode:

$\ce{ Cu(+)O(2-) + 2 H^{$*$} -> Cu + H2O (g) }$

where the above theory corresponds to what is observed, per my experience, the copper anode has been cleaned and also the cathode possibly from increased acidity.

Note: If the Copper electrode is not pure Cu, an alloy, perhaps containing noble Silicon, you may be performing an ascribed path to H2SO4 preparation (see, for example, https://www.instructables.com/id/Make-Sulfuric-Acid-by-Copper-Sulfate-Electrolysis/ ), so they will be a definite rise in pH along with problematic explosive hydrogen generation.