[Cu(H2O)6]2+ (aq) + 2OH- --> Cu(OH)2 (s) + 6 H2O (l)

What would the full equation be for this, rather than the ionic one?

I understand that OH- becomes NaOH but where does the Na go on the right hand side? Thanks

It goes with whatever anion you added to the [Cu(H2O)6]^2+ (i.e., chloride, bromide, sulfate, etc.

To write the full equation for the given ionic equation, we need to consider the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the chemical reaction. In this case, the spectator ion is the sodium ion (Na+), which comes from NaOH. The sodium ion does not react with any other species, so it is not included in the balanced chemical equation.

The full equation would look like this:

[Cu(H2O)6]2+ (aq) + 2OH- (aq) → Cu(OH)2 (s) + 6 H2O (l)

As you correctly mentioned, the hydroxide ion (OH-) combines with the copper(II) ion ([Cu(H2O)6]2+) to form copper(II) hydroxide (Cu(OH)2), which appears as a solid precipitate. The water (H2O) molecules remain in the liquid state.

However, it is important to note that the hydroxide (OH-) does not become NaOH. The hydroxide ion (OH-) is often used as a reagent, but it does not automatically combine with sodium (Na+) to form sodium hydroxide (NaOH) in this particular equation.