Please help! Anodes/Cathodes.

Question

A standard galvanic cell is constructed with Fe2+ | Fe and Cr3+ | Cr half cell compartments connected by a salt bridge. Which are correct?

 

 

(Choose all that apply.)
Fe2+ is reduced at the cathode.
In the external circuit, electrons flow from the Fe2+|Fe compartment to the Cr3+|Cr compartment.
The anode compartment is the Cr3+|Cr compartment.
As the cell runs, anions will migrate from the Fe2+|Fe compartment to the Cr3+|Cr compartment.
The cathode compartment is the Cr3+|Cr compartment.

 

I have have been able to work out one of these, but get this one and the second one wrong every time.

Answer 1

Hello Sam,

 

I noticed you've been posting a few Galvanic Cell/Redox Half Reactions questions. As I mentioned in another question for you, I made you and a student of mine named Chastity a video on our YouTube channel about Galvanic Cells: https://www.youtube.com/watch?v=W_G6Go6VVOA.

 

What James was saying (above) was very helpful! 

 

He means the following:

• See how both of the potentials are negative? This means you need to reverse the sign and order of the highest number in order to couple or pair it with the other reaction. This way, you end up with a positive cell voltage and the electrons can flow. 

 

Let's break this down a bit: 

 

3e− + Cr3+ → Cr E0 = −0.74 v (Let's reverse this one, originally it was gaining electrons and becoming reduced. Let's oxidize instead and flip the sign). 

 

Cr →  Cr3+ + 3e− = +0.74 v (Now that the largest potential is flipped and we can oxidize Chromium, we can couple it with the Iron reduction reaction below)

2e− + Fe2+ → Fe E0 = −0.44 v

 

You can assemble a Galvanic cell now and use it to answer your homework questions. See the video for more specifics or let me know if you'd like for me to tutor you in this topic. I could make you an expert in an hour! 

Feel free to ask any additional homework problems to be reviewed at http://www.drawntochemistry.com.

Hope this helps a bit!

Answer 2

Once you figure out which reaction is has the most positive potential, the rest is very straightforward.

 

3e⁻ + Cr³⁺ → Cr E⁰ = −0.74 v

 

2e⁻ + Fe²⁺ → Fe E⁰ = −0.44 v

 

Obviously, neither of these is spontaneous, but if you reverse one of them (which changes the sign of E⁰), then add the new potentials together, the one that gives a positive cell voltage is the way to go.

At that point, it's red cat, an ox.