Writing simple electronic half equations (electrolysis)

Question

2Al2O3 -> 4Al + 3O2

At the anode :

At the cathode:

CuBr2 -> Cu + Br2

At the anode :

At the cathode:

2NaCl -> 2Na +Cl2

At the anode:

At the cathode:

ZnCl2 -> Zn + Cl2

At the anode:

At the cathode:

Answer 1

The Anode is where oxidation takes place, in other words the added is where electrons are lost.

The Cathode is where reduction takes place, in other words it is where electrons are gained.

To figure out which is which you need to look at the transfer of electrons across a reaction.

For example:

CuBr₂ --> Cu + Br₂

The left side of the equation has the ionic compound CuBr made up of

1. a Cu⁺² ion, and
2. 2 Br^- ions.

The right side of the equation has pure Cu and Br₂

1. In order for the Cu⁺² ion to become pure Cu, it had to gain 2 electrons.
2. In order for the 2 Br^- ions to become Br₂ they each had to lose an electron (for a total of two electrons)

We can write the two half-reactions as:

1. Cu⁺² + 2e^- --> Cu. and
2. 2 Br^- --> 2e^- + Br₂

This shows that:

The Cu⁺² ion is gaining electrons and so is being reduced which makes it the Cathode.

The Br^- ion lose electrons and so are oxidized which make it the Anode.

The other reactions in your question can all be done the same way.

Hope this helped.

Please leave a comment if you need more follow up or if this was helpful.

David

Answer 2

2Al₂O₃ -> 4Al + 3O₂

The Al goes from ⁺3 to 0 charge — reduced @cathode

The O goes from ^-2 to 0 charge — oxidized @anode

Anode: 6 O^-2 → 6 O₂ + 12e^-

Cathode: 4 Al⁺³ + 12e^- → 4 Al

CuBr₂ → Cu + Br₂

The Cu goes from ⁺2 to 0 charge — reduced @cathode

The Br goes from ^-1 to 0 charge — oxidized @anode

Anode: 2 Br^-1 → 2 Br₂ + 2e^-

Cathode: Cu⁺² + 2e^- → Cu

2 NaCl → 2 Na + Cl₂

The Na goes from ⁺1 to 0 charge — reduced @cathode

The Cl goes from ^-1 to 0 charge — oxidized @anode

Anode: 2 Cl^- → Cl₂ + 2e^-

Cathode: 2 Na⁺ + 2e^- → 2 Na

ZnCl₂ → Zn + Cl₂

The Zn goes from ⁺2 to 0 charge — reduced @cathode

The Cl goes from ^-1 to 0 charge — oxidized @anode

Anode: 2 Cl^- → Cl₂ + 2e^-

Cathode: Zn⁺² + 2e^- → Zn