How do I know where each component of a reaction goes within a galvanic cell?

[TravisB]

Trying to teach myself electrochemistry out of a textbook and it's kind of confusing at some parts without any teacher to ask questions, but I think I'm doing a pretty good job. I've got a test tomorrow on this stuff and I'm kind of burned out from studying (about 20 hours in the past 4 days) so if you guys would help me out I'd really appreciate it.

Homework Statement

"Describe completely the galvanic cell based on the following half-reactions under standard conditions:"

They want me to list:
1. the cell potential and the balanced cell reaction
2. the direction of electron flow
3. which is the anode and the cathode
4. the nature of each electrode and the ions present in each compartment.

Homework Equations

Ag(1+) + 1 electron ---> Ag Cell potential = 0.80 volts
Fe(3+) + 1 electron ---> Fe(2+) Cell potential = 0.77 volts

The Attempt at a Solution

Both of the half-reactions listed are reduction reactions. This is a redox reaction, so one has to be flipped. The reaction has to have a positive voltage, so it's the one with the iron, switching the sign of the cell potential and giving a balanced equation of Ag(1+) + Fe(2+) ---> Ag + Fe(3+)

Add the cell potentials from the half reactions to get a cell potential of .03 volts for the reaction.

This is where I'm getting confused. At first I listed the side with the Ag+ and the Fe 2+ on the left side of the cell as the anode, and the side with the Ag and the Fe 3+ as the cathode on the right side, since that's the way that it's written in the reaction, but that's not right.

Then I figured out which ones are being reduced and put them on the side that I labeled as "anode", and put the ones being oxidized on the cathode side, but that wasn't right either.

The book says that the Fe(2+) and the Fe(3+) are on one side as the anode with a platinum electrode and the Ag+ is on the other side as the cathode with a silver electrode but I can't figure out how they're getting that.

Also, how do I know which electrodes to use on which? I get that the anode has a platinum electrode because both of the elements are aqueous, but how am I supposed to know that the cathode is a piece of solid silver?

I've been wracking my head over this for about an hour and I can't figure it out. Thanks again!

 

[csopi]

By definition the cathode is the electrode at which REDUCTION is happening, and the anode is where the OXIDATION, so in your case the silver electrode is the cathode. The voltage of the cell can be calculated as $$U=\epsilon_{cathode}-\epsilon_{anode}$$

The Fe(2+)/(Fe(3+) electrode is a so called "redox electrode", and usually they use an inert metal, mostly Pt.
The other electrode must be Ag/Ag+, so called "electrode of the first kind" (considering the given 0.8 V electrod potencial), a common type. If other relevant materials were present (beyond Ag and Ag(+)), they must have added other reactions, I mean the electrode reaction wouldn't be so simple as Ag(+) + e(-) --> Ag (for example if inside the electrode your Ag is connected to let's say a Pt-rod, you must add the following reaction and the corresponding potencial (contact potential): e- (in the Ag) --> e- (in the Pt))

As for the present ions: Ag(+) at the cathode, Fe(2 and 3+) at the anode, but if you short circuit the cell with a resistor (so some current is flowing), the metal ions (and the counter ions as well) start to migrate and mix.

I hope I answered your questions.

 

[Blueshift5]

it is important to understand the principles behind electrochemistry and how to apply them in a practical setting. In order to determine the components of a galvanic cell, one must first understand the basics of a redox reaction. In this case, we have two half-reactions, one involving silver ions (Ag+) and the other involving iron ions (Fe3+ and Fe2+). These half-reactions can be combined to form a complete redox reaction by flipping one of the half-reactions and balancing the electrons on both sides. This gives us the balanced equation of Ag+ + Fe2+ ---> Ag + Fe3+ with a cell potential of 0.03 volts.

Next, we need to determine the direction of electron flow. Electrons always flow from the anode to the cathode, so in this case, they would flow from the iron electrode to the silver electrode.

Now, we can determine which is the anode and which is the cathode. The anode is the site of oxidation, where electrons are lost, while the cathode is the site of reduction, where electrons are gained. In this case, the iron electrode is the site of oxidation, making it the anode, and the silver electrode is the site of reduction, making it the cathode.

The nature of each electrode can be determined by considering the ions present in each compartment. The anode, where oxidation occurs, is typically made of a reactive metal or a non-metal, such as platinum, that can facilitate the oxidation reaction. In this case, the iron electrode would be made of platinum to allow for the oxidation of Fe2+ to Fe3+. The cathode, where reduction occurs, is typically made of a metal that is less reactive than the ions present in the solution. In this case, the silver ions are being reduced to solid silver, so the cathode would be a piece of solid silver.

In summary, to determine the components of a galvanic cell, one must understand the principles of redox reactions, the direction of electron flow, and the nature of each electrode based on the ions present in each compartment. I hope this explanation helps you better understand electrochemistry and good luck on your test tomorrow!