In order to fill in the missing parts in the problem, first, we need to identify each of the following species in the system.

• Anode - is the electrode where oxidation happens
• Cathode - is the electrode where reduction happens

To avoid confusion, cathode reaction is always drawn as the right electrode while the anode is the left electrode. Having said that, 

• Anode - `Ni _((s))`
• Cathode - `Cu _((s))`

What happens in both electrodes?

• Anode half-reaction

`Ni(OH)_2 _(s) + 2e^(-) hArr Ni _((s)) + 2OH^(-)`

• Cathode half reaction

`Cu(OH)_2 _(s) ++ 2e^(-) hArr Cu _((s)) + 2OH^(-)`

For the line notation:

`Ni _((s))|Ni(OH)_2 _(s) |KOH _((aq))|Cu(OH)_2 _(s) |Cu _((s))`

Anode half reaction is written on the left part and the cathode half reaction on the right side.