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

Click For Summary
SUMMARY

The discussion centers on understanding the components of a galvanic cell, specifically the placement of electrodes and the direction of electron flow based on given half-reactions. The half-reactions provided are Ag(1+) + 1 electron → Ag with a cell potential of 0.80 volts and Fe(3+) + 1 electron → Fe(2+) with a cell potential of 0.77 volts. The correct balanced cell reaction is Ag(1+) + Fe(2+) → Ag + Fe(3+), with the Ag electrode functioning as the cathode and the Fe electrode as the anode. The cell potential is calculated to be 0.03 volts, and the use of platinum as the anode electrode is justified due to the aqueous nature of the reactants.

PREREQUISITES
  • Understanding of redox reactions and their components.
  • Familiarity with galvanic cells and electrode potentials.
  • Knowledge of half-reaction notation and balancing chemical equations.
  • Basic principles of electrochemistry, including the roles of anodes and cathodes.
NEXT STEPS
  • Study the Nernst equation for calculating cell potentials under non-standard conditions.
  • Learn about the construction and function of different types of electrodes in electrochemical cells.
  • Explore the concept of electrode kinetics and how it affects reaction rates in galvanic cells.
  • Investigate the role of concentration gradients in electrochemical reactions and their impact on cell performance.
USEFUL FOR

This discussion is beneficial for students studying electrochemistry, educators teaching the subject, and anyone involved in laboratory work with galvanic cells and electrochemical reactions.

TravisB
Messages
1
Reaction score
0
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!
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
12K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K