# Electrochemical cell

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1. May 19, 2015

### AnkurGarg

In our textbook it is written that -voltmeter measures the electrode potential difference between 2 half cells (in an electrochemical cell) (electrode potential refers to the potential difference between electrode and its salt solution in which it is dipped)

it should come-voltmeter measures the potential difference between the two metal strips i.e; the cathode and the anode.i am really confused!!!

2. May 19, 2015

### Staff: Mentor

When you dip the metal strip into the solution it becomes a half cell (or, if the metal is inert, it will have potential of a half cell/redox system present in the solution).

3. May 19, 2015

### James Pelezo

One way to visualize an electrochemical voltaic cell consisting of two different electrodes is to relate the voltage difference to rolling a ball down an incline. The negative electrode (anode) is the top of the incline and the positive electrode (cathode) is the bottom of the incline. The greater the voltage differential, the steeper the incline. Charge flow (electron flow through the metallic conductors) is always from the anode (top of the incline) to the cathode (bottom of the incline). This, in essence, is why the Galvanic/Voltaic cell is a spontaneous reaction process. That is, if you were to release a ball on an incline, it would 'spontaneously' roll down to the bottom without outside assistance. Such is true when connecting the electrodes of a battery.

4. May 19, 2015

### AnkurGarg

James you didn't understood my question.I am asking for the proof -Potential difference between the two electrodes = Their electrode potential difference.

5. May 20, 2015

### James Pelezo

From a table of Standard Electrodes (see link: http://www.chemunlimited.com/Table of Reduction Potentials.pdf ), Select any two half reactions. Rule 1: electron flow is from the more negative electrode (anode) to the more positive electrode (cathode). This makes the anode the site of oxidation and the cathode the site of the reduction half-reaction. To calculate the potential difference between two electrodes, Eocell(net) = Eoreduction - Eooxidation. Choose any pair of half-reaction, substitute in the Eo formula. Eo(cell) is always > 0. If it comes out negative, switch the potential values. Then you'll have reduction - oxidation and the net voltage > 0.00

6. May 20, 2015

### Staff: Mentor

You need to try harder explaining what the problem is. Yes, potential difference is the difference in potentials, but it is so obvious I guess that's not what you are confused about.

7. May 20, 2015

### AnkurGarg

Hm Sir you are right..So here is my confusion-A metal strip has its own potential,and the solution in which it is dipped has its own..And the difference in these 2 potential is called the electrode potential of the half cell..Similarly there is another half cell with its own electrode potential...

Now, my confusion is-voltmeter should measure the difference between the potentials of the 2 metal strips and not the electrode potential difference. But our textbook says-Voltmeter measures the electrode potential difference between 2 half cells..

Our textbook is wrong??

8. May 20, 2015

### AnkurGarg

i Got it sir!!Ty very much!!!

9. May 20, 2015

### Staff: Mentor

No, that's not the way it works.

If the metal is inert, it doesn't have its own potential* - so it will have a potential of the half cell present in the solution. In a way it acts just as a connector between the half cell in the solution and the voltmeter.

If the metal is not inert and is a part of the redox half cell (say, copper dipped in the Cu2+ solution) it will have the potential of this redox system (it will work both as a connector and the half cell part).

*That's not exactly true, every object has some potential, but it doesn't matter here.

10. May 20, 2015

### AnkurGarg

What do you mean sir,when you said-it will have potential of this redox system!!???

11. May 20, 2015

### Staff: Mentor

Imagine having a solution that contains Fe(II)/Fe(III). Dip a platinum electrode into this solution. Pt is inert and it won't react, but the potential measured using this electrode will be that of Fe(II)/Fe(III) system present in the solution.

12. May 20, 2015

### James Pelezo

To refer to a substance as an electrode in a Galvanic process, it must be coupled with another substance so that a 'spontaneous' reaction (flow of electrons) occurs. Unless, the substance of interest is connected to a reactive counterpart it has no meaning relative to the Galvanic Process. It might be helpful to understand that Galvanic processes are, for instructional purposes, divided into two general types;i.e., Uncontrolled Galvanic Process and Controlled Galvanic Process. The Uncontrolled Galvanic Process is, for example, is a solid Zinc bar being submerged/dipped into a solution of Copper Sulfate. The copper ions (Cu+2)) coming in direct contact with the zinc metal's surface undergo immediate, spontaneous reduction into their basic standard metallic state and proceed to coat the zinc bar with a layer of Copper metal. At some point in time, the coating of reduced copper will be so thick that other copper ions will be unable to come into contact with the zinc metal, and the entire Galvanic process stops. In order to achieve a 'Controlled Galvanic Process', the oxidation process (Zn(s) => Zn+2(aq) + 2e-) must be separated from the reduction process (Cu+2 + 2e- => Cuo(s)) so that charge flow can be sustained. The two processes are connected with metallic conductors, a salt bridge or semi-permeable plate that will allow cations to flow from the oxidation cell into the reduction cell. So when you say, 'The metallic electrode has a potential and the solution has a potential', then this would be true ONLY if the ions of the solution react spontaneously with the metallic electrode. If an electrode (like Platinum) and salt ions (Ferric and Ferrous ions) do not react, as Borek inicated, then the concept of 'reduction potential' does not exist. There has to be an oxidation-reduction reaction occurring before any kind of potential difference issue can be defined.

13. May 21, 2015

### Staff: Mentor

So how do you measure the reduction potential of Fe(II)/Fe(III) system if it doesn't exist? Where does the number +0.77 V listed in the redox potential tables come from?

What do you use inert ORP electrodes for?

14. May 21, 2015

### James Pelezo

What I said was ... "Unless, the substance of interest is connected to a reactive counterpart (cell) it has no meaning relative to the Galvanic Process." I didn't say 'the assembly didn't exist. You said that if an inert electrode were inserted into an Fe(II)/Fe(III) solution it wouldn't react, and that's true, and it has a reduction potential of 0.77 volts as published. However, if this assembly is not connected to another electrode assembly, no redox processes will occur. Fe+3 + e- => Fe+2 does not spontaneously undergo a redox process without being connected to an electrode with with a different reduction potential. The published electrode potentials for individual half-reactions are the potentials generated relative to the Standard Hydrogen Electrode who's Eo = 0.00 volts. I can elaborate more on this issue if you wish. Now, if you are referring to common ORP electrodes that are Ag/AgCl electrodes in an electrolyte gel matrix, this system is the counter electrode for measuring potential differences between soloution EMF nd the ORP EMF, BUT an oxidation process and a reduction process is still present. So, a galvanic circuit will exist, but if the ORP electrode is removed, no redox rxns occur and the Platinum-in-electrolyte solution can not be defined. It has to be connected to something with a different electrode potential to be defined as having a redox electrode.

15. May 21, 2015

### Staff: Mentor

So basically what you are saying is that when the redox system is not connected to the reference electrode its potential doesn't exist? Something like "if nobody hears the falling tree it means it doesn't make a noise"?

16. May 21, 2015

### James Pelezo

Interesting analogy, but yes.

17. May 21, 2015

### Staff: Mentor

No - potential doesn't exist because we are measuring it. It exists on its own. Otherwise it would have to magically appear the moment you close the circuit. The only thing you modify the moment you close the circuit you allow the charge to flow - but the charge would not flow without the potential difference.

These are physical properties of circuits, often overlooked in chemistry, but the electrochemical cell follows exactly the same physics every other electrical system does.

This is just another aspect of the discussion we have had here not long ago.

18. May 21, 2015

### AnkurGarg

OMG i am confused whom to believe or not!!Well TY both for your admirable efforts!

19. May 21, 2015

### Staff: Mentor

When in doubt, a mentor or advisor is the one to believe, so that would be Borek.