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