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

$$\ce{Cu(s) -&gt; Cu^2+(aq) + 2e-}$$

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:

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

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.

$$\ce{Cu(ano) + Cu^2+(cat) + 2e-(cat) -&gt; Cu^2+(ano) + 2e-(ano) + Cu(cat)}$$

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

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

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

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

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.