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

  1. Jan 3, 2010 #1
    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.

    1. The problem statement, all variables and given/known data
    "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.


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



    3. 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!
     
  2. jcsd
  3. Jan 9, 2010 #2
    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 [tex]U=\epsilon_{cathode}-\epsilon_{anode}[/tex]

    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 lets 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.
     
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