-1
$\begingroup$

In the redox reaction chapter, we learn that oxidation and reduction go hand in hand. The oxidising agent oxidises by accepting the electron released by the reducing agent and gets reduced in the process.

However, in the half cell electrode (say, $\ce{Zn}$/$\ce{Zn^{2+}}$) only oxidation is taking place and no simultaneous reduction is happening in the scene. Also, in a $\ce{Cu}$/$\ce{Cu^{2+}}$ half cell electrode, only reduction is taking place. Until the half cells are joined by a saltbridge, oxidation and reduction go on violating the basic principle of redox reactions.

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ Half reactions cannot happen independently and (as you have already stated in the first line) a reduction is simultaneous with an oxidation. Half cells are just representations. Nothing violated. Reaction occurs only when 2 half reactions are present, and the sum of their potentials is positive. $\endgroup$
    – William R. Ebenezer
    Commented Apr 21, 2019 at 10:56
  • $\begingroup$ Is that true? In liquid ammonia solutions you can get electrons pulled out at the anode but they just pour into the liquid through the cathode. $\endgroup$
    – Oscar Lanzi
    Commented Nov 19, 2019 at 20:28
  • $\begingroup$ I think the liquid ammonia situation is a plain reaction, not a combination of two half cells. $\endgroup$
    – James Gaidis
    Commented Nov 20, 2022 at 14:53

3 Answers 3

Reset to default
0
$\begingroup$

Redox reactions consist of oxidation and reduction processes occurring simultaneously. What does this actually mean?

Redox reaction involves the transfer of electrons between species (Easy way to remember: increase in oxidation number = oxidation, hence decrease in oxidation number = reduction). These processes occur together at the same time at different interfaces (boundaries).

Considering your cell (Galvanic Cell): The main reacting species (excluding the spectator ions) are Copper and Zinc. The overall reaction that is taking place is:

$$ \ce {Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$$

Applying the previously mentioned trick, you can see that zinc is being oxidised from 0 oxidation number to 2+ (increase in oxidation number = oxidised), whereas copper ion is reduced from 2+ to 0 oxidation number (decrease in oxidation number = reduced). You can think of a half-cell equation as small components that make the overall reaction in redox reaction of Galvanic cell. They provide vital information on what is happening to the species in terms of electrons involved and the standard potential can be used to identify if the species is capable of operating as reducing agent or oxidising agent.

The half-cell equations is usually tabulated against standard hydrogen electrode as being the reference.(Note: half-cell equations are usually written as reductions). In order to identify which species is oxidising agent, you can think of it this way: the more positive the potential $\Rightarrow$ stronger oxidising agent $\Rightarrow$ oxidising agents by definition are reduced (reduction occurs at the cathode).

The reactions are only taking place when the circuit is closed (cells are connected). This leads to the flow of the electrons around the circuit and redox reaction can take place. To qualitatively prove that your reaction occurs, you have to consider the cell potential:

$$E_{cell} = E_{cathode} -E_{anode} $$

applying the concept of electrical work = Gibbs energy, thus

$$G=-nFE_{cell}$$

(Important: provided the concentration 1 M, if they are not you have to apply Nernst Equation to identify new cell potential) .

Improve this answer
$\endgroup$
2
  • $\begingroup$ Could the reaction $\ce {Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$ happen in a solution with the wire and the connecting electrodes. I mean if we just mixed a salt of $\ce{Cu^2+}$ with solid $\ce{Zn}$ will the reaction still happen? $\endgroup$
    – Anton
    Commented Mar 11, 2021 at 11:35
  • $\begingroup$ Yes The purpose of the cell is either to keep the materials separate or to use the energy of the reaction to do work. such as running your cell phone or your Tesla $\endgroup$
    – jimchmst
    Commented Nov 14, 2022 at 23:47
0
$\begingroup$

In a (standard) Zn/Zn++ half cell, no net oxidation (and no net reduction) takes place. A piece of zinc sits in a 1M solution of Zn++ ions (with their accompanying anions). On a micro scale, an equilibrium is occurring (could be occurring!) where Zn + Zn++ --> Zn++ + Zn. No net change occurs; this is a reference point. If it were not stable, some process would degrade the half cell and it would be react itself away. In reality, half cells are only useful to the extent that they don't react themselves away until they are connected to the other half of the overall cell.

Same for Cu/Cu++. However, when the half cells are connected (thru a salt bridge so ions can flow, and a wire so electrons can flow), two processes occur: electrons leave the zinc (the zinc oxidizes) and electrons flow to the copper ions (reducing them). No net "process" or reaction occurs in the half cells until they are connected.

The reality of net oxidation and reduction require both to occur simultaneously - WHEN THEY OCCUR - but we can speak of each process independently of the other. When electrons leave an atom (oxidation), they have to go somewhere (reduction); if electrons are not flowing thru a wire (no actual reaction) there is no net oxidation or reduction. The half cell potentials then define places from which electrons will flow, or locations to which they will flow.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nothing is simultaneous; oxidation and reduction can be and are separated in time and distance There probably are electrons from the big bang that have yet to find a proton $\endgroup$
    – jimchmst
    Commented Nov 14, 2022 at 23:41
  • $\begingroup$ If both oxidation and reduction are not essentially complete (~99%), we don't have equilibrium, we have a reaction in progress. Simultaneity here means that when we look at the system, it has stopped changing, so the total oxidation and the total reduction have both been accomplished to give an equilibrium by that time (however long it is). Many electrons cannot be lost (without finding a home) because the potentials would become incredibly high. $\endgroup$
    – James Gaidis
    Commented Nov 15, 2022 at 15:01
  • $\begingroup$ One of the requirements for equilibrium is propinquity. At equilibrium forward and reverse reactions are equal not 99%. To be reasonably confident it must be perturbed from both directions. If separated in distance it becomes hard to tell. $\endgroup$
    – jimchmst
    Commented Nov 20, 2022 at 3:18
  • $\begingroup$ I edited the answer to emphasize net reaction. A metal dipped into a standard electrolyte reaches its standard potential by donating electrons to necessarily propinquitous (nearby) acceptors until the metal acquires its standard negative potential and then the net reaction stops. No net measurable oxidation or reduction occurs in an unconnected half cell. If further net oxidation/reduction occurs, the metal is corroding; an equilibrium potential cannot then be measured. An example would be: Na in an aqueous solution of NaNO3. $\endgroup$
    – James Gaidis
    Commented Nov 20, 2022 at 15:55
0
$\begingroup$

Let's go back to the original text, that I copy here in italics :

Also, in a Cu/Cu$^{2+}$ half cell electrode, only reduction is taking place.

No ! Sorry to say no ! The $\ce{Cu/Cu^{2+}}$ half cell can work in the oxidation mode, if it is coupled to a more positive half cell, like the $\ce{Ag/Ag+}$ half cell. In the whole cell, both half reactions are : $$\ce{Anode : Cu -> Cu^{2+}+ 2 e- ; redox potential +0.34 V}$$ $$\ce{Cathode : Ag+ + e- -> Ag ; redox potential + 0.80 V}$$ so that the whole cell delivers : $0.80$ V - $0.34$ V = $0.46$ V

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.