Which Electrode is the Cathode in Wade and Tang's Galvanic Cell?

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Discussion Overview

The discussion revolves around identifying the cathode in a galvanic cell constructed by two learners, Wade and Tang, and understanding the implications of voltmeter readings in determining electrode roles. The scope includes theoretical concepts related to electrochemistry and practical application in a homework context.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Wade and Tang calculate the voltage of their galvanic cell and receive a negative reading when the second electrode is connected to the positive terminal of the voltmeter.
  • One participant suggests that the second electrode is the cathode because it is connected to the positive end of the voltmeter.
  • Another participant argues that a negative reading indicates the anode is connected to the positive terminal, implying that the second electrode is the anode.
  • There is a suggestion that the second electrode could be calcium instead of cadmium, with a discussion on the oxidation and reduction processes occurring at each electrode.
  • Participants clarify that the definition of cathode and anode is dependent on the polarity of the voltmeter reading, with a negative reading indicating a reversal of roles.
  • One participant expresses confusion about the relationship between the voltmeter reading and the identification of electrodes, seeking further clarification.

Areas of Agreement / Disagreement

Participants express differing views on the identification of the cathode and anode based on the voltmeter readings. There is no consensus on the correct identification of the second electrode or the implications of the negative voltage reading.

Contextual Notes

Participants discuss the relationship between electrode identification and voltmeter readings without resolving the underlying assumptions about the electrodes' materials and their standard reduction potentials.

Who May Find This Useful

Students studying electrochemistry, particularly those working on galvanic cells and seeking to understand the relationship between electrode roles and voltmeter measurements.

GrandMaster87
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Homework Statement


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.

Homework Equations





The Attempt at a Solution



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

2) E cell = Ecathode - Eanode
-2.11= Ecathode -(-0.76)
-Ecathode=2.11 + 0.76
-Ecathode=2.87
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 can't figure out..

Thanks Guys
 
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Hi...has anyone seen this thread as yet?
 
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)
 
kingkool said:
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)

thanks a lot man! I am 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?
 
Only if the measured EMF is positive.
 
so when the emf is negative and the electrode is connected to the positive side it is the anode

thanks a lot guys for your help.
 
GrandMaster87 said:
so when the emf is negative and the electrode is connected to the positive side it is the anode

thanks a lot guys for your help.

If i give you a battery with no markings and tell you to find the voltage, you won't 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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