- #1
TravisB
- 1
- 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.
"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.
Ag(1+) + 1 electron ---> Ag Cell potential = 0.80 volts
Fe(3+) + 1 electron ---> Fe(2+) Cell potential = 0.77 volts
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!
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!