Cell Diagrams

[Brian_Ho_2B]

Cell diagrams are giving me a bit of a hard time. I was doing this problem that asked me to write the diagram for a redox reaction with Au+ (aq) --> Au (s) + Au3+ (aq); note that this is unbalanced. I figured that since Au+ is reduced to Au(s) and Au+ is oxidized to Au3+, that the cell diagram should look like: Pt(s)|Au+(aq), Au3+(aq)||Au+(aq)|Au(s). However, the answer is actually Au(s)|Au+(aq)||Au3+(aq)|Au(s). Could someone explain the reasoning behind this cell diagram?

[GFolk_1D]

I am confused on this as well, when do we need to introduce Pt into the system?

[Brian_Ho_2B]

I am confused on this as well, when do we need to introduce Pt into the system?

We introduce Platinum as an electrode to the side(s) that have their reducing or oxidizing agents in aqueous phases. The platinum acts as a conductor of electrons for the aqueous agents. However, this question doesn't exactly follow that rule so I'm a bit confused.

[charleejohnson1L]

Since you were given a solid thing in the equation, I'm not sure the Pt is needed. You just need to figure out what's being oxidized vs reduced and put them on the correct side of the "salt bridge". I hope this helps :) I'm not entirely sure if I'm correct, but I remember a problem like this from AP chem in high school.

[Megan Vu 1J]

Since there is already a solid on both sides of the cell diagram for an anode and a cathode, you do not need to introduce platinum into the equation. Platinum does not interfere with the equation and is inert, but in this example, you do not need it since the Au (s) acts to balance the equation for the cell diagram.

[Angela Patel 2J]

Since you were given a solid thing in the equation, I'm not sure the Pt is needed. You just need to figure out what's being oxidized vs reduced and put them on the correct side of the "salt bridge". I hope this helps :) I'm not entirely sure if I'm correct, but I remember a problem like this from AP chem in high school.

But don't you need a solid on both sides to avoid Pt? Or does that not apply when it is the same element being oxidized and reduced?

[Jessica Esparza 2H]

I'm also confused on why the cell diagram was written this way!

[charleejohnson1L]

But don't you need a solid on both sides to avoid Pt? Or does that not apply when it is the same element being oxidized and reduced?

Since there's only one element, there will only be one beaker in the cell diagram if you were to draw it, if that makes sense. Everything is happening in one place, and this is why only one solid thing is needed. If there were multiple species being oxidized/reduced, like the examples in class, then you would need a solid on both sides to avoid using Pt. I hope this helps. :)