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If I put two electrodes of two different metals in 1 solution containing ions for both of the metals and connect these electrodes with a wire, do I still have an electrochemical cell ? Or does it have to be made of two half cells in 2 different solutions joined by a salt bridge? If so,why ?

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  • $\begingroup$ If you mix everything together, the redox reaction happens right then and there. The whole point of the cell is to force the reaction to become circuitous such that the transferred electrons have to go the long away around in a circuit, doing useful work. $\endgroup$
    – Zhe
    Feb 13 '19 at 15:35
  • $\begingroup$ I'm sorry if this is a stupid question, I'm just starting out with electrochemical cells, but what do you mean the redox reaction would happen right then and there ? Suppose I'm working with Zn and Cu electrodes, Can the electrons that are a result of the oxidation of zinc atoms move in the solution and reach the Cu ions ? $\endgroup$
    – Jaja bae
    Feb 13 '19 at 15:49
  • $\begingroup$ Welcome to Chemistry.SE! If you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and question policies. $\endgroup$
    – A.K.
    Feb 13 '19 at 16:14
  • $\begingroup$ If I set up a functional cell, take it apart, and mix it all together, there would have to be a reaction. It's the reaction that creates electricity, but it's only useful to force the process of reacting to happen via moving electrons through a wire. $\endgroup$
    – Zhe
    Feb 13 '19 at 16:18
  • $\begingroup$ chemistry.stackexchange.com/questions/66123/… $\endgroup$
    – Mithoron
    Feb 13 '19 at 20:31
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Yes, if you dip two different metals in an aqueous solution containing ions, you will have an electrochemical cell. This is how all lemon batteries work, which consist of zinc or copper inserted into lemons (approximately an aqueous citric acid solution). You can see a potential difference between zinc and copper. The key point is that one cannot determine Nernstian potential in such cases. Half cell is just a mental construct as it cannot exist alone. If there is a reduction, something must be oxidized. Electrochemists usually separate the "half-cells" in order to conveniently calculate Nernst potential.

Imagine a hypothetical cell, which is partitioned into half cells with a plantinum sheet. Each compartment contains an electrode Cu and Zn, dipped in their respective salt solutions. What will happen in that case?

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    $\begingroup$ RE: "Electrochemical usually separate "half-cells" in order to conveniently calculate Nernst potential." -- This just isn't true. Separate "half-cells" are needed to prevent the redox reaction taking place on the electrode itself. If Cu and Zn electrodes were added to a $\ce{Cu^{+2}/Zn^{+2}}$ solution then the $\ce{Zn^{+2}}$ would react at the Cu electrode itself coating it with Zn and there would be very very little to no current flow between the electrodes. $\endgroup$
    – MaxW
    Feb 13 '19 at 17:55
  • $\begingroup$ Thanks for adding this point, however, my point was that one cannot calculate or even define Nernst potential for Zn ions when Cu and Zn ions are also present in the solution. It is important to separate half-cells for that reason as well. By the way, one does not draw any current during potential measurements. This is one of the basic requirements of potentiometry. $\endgroup$
    – M. Farooq
    Feb 13 '19 at 18:07
  • $\begingroup$ In order to measure the potential of a half cell you must draw a very tiny amount of current. The idea is that you draw so little that there are no double barriers or diffusion implications. $\endgroup$
    – MaxW
    Feb 13 '19 at 18:11
  • $\begingroup$ You miss the point. So long as the reaction continues you can calculate the Nernst potential for Zn ions when Cu and Zn ions are also present in the solution. What you can't do is to reliably measure the potential between the electrodes which requires current to flow between the electrodes. $\endgroup$
    – MaxW
    Feb 13 '19 at 18:20
  • $\begingroup$ I am not sure how would one calculate the half-cell potential of Zn when both Zn and Cu are present in the solution. One would see some value on a potentiometer, but it will have nothing to do with equilibrium potential. Do you remember old-school null point potentiometry? Under those conditions, the current flow was indeed zero (or for semantic reasons extremely small). $\endgroup$
    – M. Farooq
    Feb 13 '19 at 18:45

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