47
$\begingroup$

In a galvanic (voltaic) cell, the anode is considered negative and the cathode is considered positive. This seems reasonable as the anode is the source of electrons and cathode is where the electrons flow.

However, in an electrolytic cell, the anode is taken to be positive while the cathode is now negative. However, the reaction is still similar, whereby electrons from the anode flow to the positive terminal of the battery, and electrons from the battery flow to the cathode.

So why does the sign of the cathode and anode switch when considering an electrolytic cell?

$\endgroup$
37
$\begingroup$

The anode is the electrode where the oxidation reaction

\begin{align} \ce{Red -> Ox + e-} \end{align}

takes place while the cathode is the electrode where the reduction reaction

\begin{align} \ce{Ox + e- -> Red} \end{align}

takes place. That's how cathode and anode are defined.

Galvanic cell

Now, in a galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative.

At the cathode, on the other hand, you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive.

Electrolytic cell

In an electrolytic cell, you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are "pushed out" of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak. So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode.

At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode.

A tale of electrons and waterfalls

Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a waterfall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole. The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into.

Ok, what happens at the electrolysis reaction? At the cathode, you have the high altitude situation. So the electrons flow to the "edge of their waterfall". They want to "fall down" because behind them the river is pushing towards the edge exerting some kind of "pressure". But where can they fall down to? The other electrode is separated from them by the solution and usually a diaphragm. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the "pushed out" electron immediately. And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket.

Now the anode: At the anode, you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little "water" that is added to it doesn't change the river much. It stays the same, unaltered so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance.

$\endgroup$
  • 1
    $\begingroup$ I'm still a little confused. You say that At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode . However, because the anode's electrons are sucked into the positive terminal of the battery, shouldn't the anode be considered negative? $\endgroup$ – 1110101001 Sep 27 '14 at 4:47
  • $\begingroup$ Also, since by definition the anode is where a loss of electrons occurs, won't there always be a buildup of negative charge and thus shouldn't the anode always be considered negative? $\endgroup$ – 1110101001 Sep 27 '14 at 4:50
  • 2
    $\begingroup$ @user2612743 In an electrolytic cell you are the person that determines which electrode is positive and which is negative via the external potential. And this external potential doesn't get altered in the course of the reaction because the "sucked in" electrons are transported away by the voltage source. Thus, those electrons can't build-up a negative charge at the electrode and the electrode potential stays the same. $\endgroup$ – Philipp Sep 27 '14 at 4:54
  • 2
    $\begingroup$ @user2612743 In a galvanic cell the reaction proceeds spontaneously according to the difference in Gibb's free energy $\Delta G$. At the anode you have the oxidation and electrons enter the electrode leading to a build-up of neg. charge. At the cathode you have the reduction and electrons pass out of the electrode leading to a build-up of pos. charge. The electrodes are often made of metal and the electrons originate from those metal atoms by leaving them thus creating positive metal ions. $\endgroup$ – Philipp Sep 27 '14 at 5:17
  • 1
    $\begingroup$ @user2612743 Concerning electrolysis: The electrodes are charged according to the potential you enforce. This charge is not altered during the reaction because any electrons that go in or out are immediately transported away or replaced by the external voltage source. The voltage source is the force behind the pull that electrons feel at the anode and it is also the force behind the push the electrons feel in the cathode. The electrons don't flow directly from the anode to the cathode. They are rather collected and redistributed by the voltage source, so to speak. $\endgroup$ – Philipp Sep 27 '14 at 5:26
14
$\begingroup$

The electrode at which oxidation takes place is known as the anode, while the electrode at which reduction take place is called the cathode.

Reduction -> cathode  
Oxidation -> anode

If you see galvanic cell reduction take place at the left electrode, so the left one is the cathode. Oxidation takes place at the right electrode, so the right one is the anode.

While in electrolytic cell reduction takes place at the right electrode, so right one is the cathode. Oxidation takes place at the left electrode, so the left one is the anode.

enter image description here

$\endgroup$
  • $\begingroup$ Yes, I got that, but why does is the anode considered negative in a galvanic cell but positive in a electrolytic cell? $\endgroup$ – 1110101001 Sep 27 '14 at 4:26
  • $\begingroup$ Unless I'm missing something, I don't quite yet understand why the anode is considered negative in a galvanic cell but positive in a electrolytic cell... $\endgroup$ – 1110101001 Sep 27 '14 at 4:36
  • $\begingroup$ @user2612743 I have tried to make it clear, see if you get it. $\endgroup$ – Freddy Sep 27 '14 at 4:45
  • $\begingroup$ My question was about the sign convention for the cathode and anode though, not about the definition of anode and cathode. I understand that oxidation occurs at the anode and reduction occurs at the cathode. I am looking to understand why the cathode is considered positive in a galvanic cell but negative in an electrolytic cell. $\endgroup$ – 1110101001 Sep 27 '14 at 4:49
4
$\begingroup$

I'm no expert nor scholar, but from what I am reading in all of these explanations, and what I notice from the illustration, it becomes obvious...at least to me...which I feel may clarify the polarity change between the Galvanic cell and electrolytic cell for this user.

As established and understood, the source of electrons and transfer of ions flows from the negative pole, (Anode) and is received by the positive pole (Cathode) (intentionally using most basic terms) the anode is negative here because the the flow originates FROM the electrolyte, into the light bulb, for which, if the terminals of the bulb were labeled, they would match the electrolyte in the other cell as it is the force coming from the bulb pushing the flow to the cell's cathode, and the cell's cathode is pulling from the bulb.

In the electrolytic cell, the "electrolyte" is taking the role of the light bulb of the Galvanic cell, since the electrons are being SENT TO it from the power source, and is not in itself the SOURCE of flow, but is SUBJECT TO the force from the source of flow.

SO just as the Galvanic cell's anode sends to the light bulb, and the electrolyte is labeled like the load of the galvanic cell, and transferring its incoming negative force from the current source, and this pushes through the electrolyte like the flow FROM the light bulb.

It may be easier if you note that the SOURCE of power is NOT the electrolyte and technically, the black terminal of the power supply is the TRUE anode (Sending), and the red side the TRUE Cathode, (Receiving) but when identifying the reactive substance submerged/surrounded by the electrolytic substance, the anode is giving up its ions, which then add to the Cathode which is receiving them.

Therefore the tags in the electrolytic cell are not naming the "source of flow", but the reaction of the substances involved, due TO the force/flow imposed on them from the power source, but is not THE source of power, and therefore should not be labeled AS one...and there are only two options for labeling them, and since it cannot be changed at the power source it can only b changed at the point of contact with the electrolyte!

At least this is what I have come to understand by reviewing the comments and illustrations.

I sincerely hope it helps clarify the rationale for the reversal of labels for this user and any others struggling with the concept of being due to the source of current having to be labeled as - Anode and + Cathode... forcing the object the current plays upon to be the opposite despite their poles and due to direction of flow.

$\endgroup$
2
$\begingroup$

I mean this answer as a complement to previous answers.

As alreadey disccused, in the ANODE you will always have an oxidation reaction $\mathrm{ \;Red\; \longrightarrow \; Ox + e^-} $, while in the CATHODE you will observe the reduction reaction $\mathrm{\;Red\; + e^- \longrightarrow \; Ox }$.

The reduction an oxidation reactions are always coupled so one electrode acts as a source of electrons and the other as a sink. In the galvanic cell the overall reaction is spontaneous and the current flows from the anode to the cathode. On the other hand, in an electrolitic cell we drive the reaction in a non-spontaneous sense applying an external potential (for example, using a power supply).

I think this image should make clear the operation of both kinds of cells,the processes that occur at each electrode and the sign convention.

galvanic versus electrolytic cell

Although it illustrates a specific reaction, you may generalised it to other systems.

The source of the image is Electrolysis I at Chemistry.LibreTexts.

$\endgroup$
  • 1
    $\begingroup$ This does constitute enough meat to be an answer to the question, but a few words of your own could get my upvote, making it a very good answer in my opinion. $\endgroup$ – M.A.R. Dec 17 '18 at 12:54
1
$\begingroup$

The anode is the electrode at which the oxidation half-reaction takes place.

In a galvanic cell, the reaction is spontaneous, there is no external potential applied, and when the anode material is oxidized that makes the anode the negative electrode. In an electrolytic cell, it is the external potential that drives the reaction, the anode is the electrode where the oxidation reaction happens, consequently this time it is the electrode with the positive potential.

$\endgroup$
  • $\begingroup$ The anode is the electrode where the oxidation reaction happens, consequently this time it is the electrode with the positive potential. I don't see the link between these two statements. Doesn't oxidation always occur at the anode, regardless of whether it is a galvanic or electrolytic cell? So why is it considered positive for an electrolytic cell but negative for a galvanic one? $\endgroup$ – 1110101001 Sep 27 '14 at 4:53

protected by Community Mar 8 '15 at 21:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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