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.
A useful mnemonic is AN OX | RED CAT.
Anode is oxidation.
Reduction occurs at the cathode.
One way you can tell which half reaction is being reduced and which half reaction is being oxidized is based on the charges of the ions.
In oxidation, the element goes from `- -> +` . This can also mean that the charge goes from a small number to a larger number. Remember, you lose electrons in oxidation!``
In reduction, the element goes from `+ -> -` . This can also mean that the charge goes from a big number to a smaller number. Remember, you gain electrons in reduction!