## 6L.5

Ryan Chang 1C
Posts: 105
Joined: Sat Aug 24, 2019 12:17 am

### 6L.5

Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following skeletal equations:
(d) Au+(aq) -> Au(s) + Au3+(aq)

How do you know what the cathode and anode is?

Betania Hernandez 2E
Posts: 107
Joined: Fri Aug 02, 2019 12:15 am

### Re: 6L.5

Oxidation occurs in the anode and Reduction occurs in the cathode. In this problem, Au is both the oxidizing and reducing agent. Au+ is being oxidized when it goes to Au3+ since it goes from having an oxidation state of +1 to +3. It is being reduced when Au+ goes to Au(s) since it goes from having an oxidation state of +1 to 0.

DLee_1L
Posts: 103
Joined: Sat Aug 17, 2019 12:17 am

### Re: 6L.5

For this question, why is Au in both half reactions and why are Au and Au3+ on different sides of the anode half reaction if in the original equation has them on the same side

BeylemZ-1B
Posts: 95
Joined: Thu Jul 25, 2019 12:17 am

### Re: 6L.5

Betania Hernandez 2E wrote:Oxidation occurs in the anode and Reduction occurs in the cathode. In this problem, Au is both the oxidizing and reducing agent. Au+ is being oxidized when it goes to Au3+ since it goes from having an oxidation state of +1 to +3. It is being reduced when Au+ goes to Au(s) since it goes from having an oxidation state of +1 to 0.

I understand the basics of the anode and cathode, but what I don’t understand is how the oxidation half-reaction given in the textbook answers lists Au(s) as a reactNt when it is a product in the given equation. Someone help me understand!

Megan Cao 1I
Posts: 103
Joined: Sat Sep 07, 2019 12:18 am

### Re: 6L.5

The cathode is where the reduction is happening, and the anode is where the oxidation is happening. (A small way to remember this is that Cathode and Reduction start with constants (C&R), whereas Anode and Oxidation start with vowels (A&O).

for the reaction, since there's only one reactant, Au(aq) is both the oxidizing and reducing agent. It reduces by going from a charge of +1 to 0, and it oxidizes by going from a charge of +1 to +3.