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Homework Help: Electrochemistry - Galvanic Cell

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Two learners, Wade and Tang are working together on a project. They were asked by their teacher to assemble a galvanic cell that would produce a large enough voltage to light up a 1.5V bulb.

    They are allowed to use two electrodes, one of which must be a zinc electrode. After having done the necessary calculations they finally agress on the second electrode.

    When the second electrode is connected to positive terminal of the voltmeter , the voltmeter shows a reading of -2.11V.

    But to their disappointment, Wade and Tang discover that the bulb won't light up

    1) Which electrode is the cathode? The zine electrode or the 2nd electrode. Motivate answer

    2) Identify the 2nd electrode.

    2. Relevant equations

    3. The attempt at a solution

    1) Ok i think the second electrode is the cathode. I believe so because it is connected to the postive end of the voltmeter.

    2) E cell = Ecathode - Eanode
    -2.11= Ecathode -(-0.76)
    -Ecathode=2.11 + 0.76
    Ecathode = -2.87

    Therefore according to Table of standard reduction potential Ecathode is Cadmium.

    The problem is face is that...how can cadmium be the Cathode as the cathode is were reduction takes place... should it be Zinc that gets reduced as it have a higher E value than cadmium?

    So i just need to know how to approach this question and any other questions in the future..i am writing my prelim exam on monday and this question i still cant figure out..

    Thanks Guys
  2. jcsd
  3. Sep 11, 2009 #2
    Hi...has anyone seen this thread as yet?
  4. Sep 11, 2009 #3
    First thing, I assume you mean calcium and not cadmium. Everything else you said is right: Calcium will be oxidized, electrons will travel to the zinc solution, zinc will be reduced. The calcium cell is the anode, the zinc cell is the cathode.

    Heres the reason you are confused: When you plug the electrodes into the voltmeter, if you get a negative reading it means that you have reversed the anode and the cathode. So if the second electrode is plugged into the positive terminal and the reading is "-2.11V", this means if you plug the zinc electrode into the positive terminal the reading will be: "2.11V". By definition, the voltage will be positive when the cathode is plugged into the positive terminal and negative when the anode is plugged into the positive terminal.

    Cathode in positive terminal: +2.11V
    Anode in positive terminal: -2.11V

    Since you plugged the second electrode into the positive terminal and got a negative number, it means the second electrode is the anode. Remember, if you measure a negative voltage all it means is that you reversed the polarities. If you do your Ecell equation using Zinc as the cathode and the unkown as the anode, and Ecell=2.11V (positive), you will get the same result for the second electrode (-2.87V)
  5. Sep 12, 2009 #4
    thanks alot man! im still a bit confused by i will make my way to understand it :)

    will it be safe to say if an electrode is connected to the positive terminal of the voltmeter it will be the cathode?
  6. Sep 12, 2009 #5
    Only if the measured EMF is positive.
  7. Sep 12, 2009 #6
    so when the emf is negative and the electrode is connected to the positive side it is the anode

    thanks alot guys for your help.
  8. Sep 13, 2009 #7
    If i give you a battery with no markings and tell you to find the voltage, you wont know which side to plug into positive and which side to plug into negative. You just have to guess, and if the emf is negative it means you plugged the anode into the positive side .
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