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?
6L.5
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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.
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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!
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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.
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.
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