Two related questions:

(1) It's my understanding that for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons. If I have a sodium chloride electrolyte, the sodium ions will move toward the cathode. However, does that mean chloride will not be reduced at the cathode?

(2) Is the same true for an electrolytic cell? The cathode is (-), so I assume it stays negative. But, what if I wanted to force the reduction of an anion like perchlorate - is it possible or will there be too great of an electrostatic barrier?

I see people reporting on the electrocatalytic reduction of anions, but I don't understand how this is possible if the ionic current is the opposite direction.

  • 1
    $\begingroup$ If the anion is the species that gets reduced, it will do so at the cathode. $\endgroup$
    – Jan
    Jan 11, 2017 at 19:54
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    $\begingroup$ Reduction should happen exclusively at the cathode. Oxidation at the anode. By definition. $\endgroup$
    – Zhe
    Jan 11, 2017 at 22:22
  • $\begingroup$ Yes, of course anode is oxidation, cathode is reduction. I'm asking about movement of ions toward electrodes. So, ionic movement is driven by potential difference of the ionic species? $\endgroup$ Jan 12, 2017 at 15:21

1 Answer 1


Theoretically, it is possible for anions to get reduced more. According to wikipedia's reduction potential table page, this reaction occurs at +0.08V vs SHE:

$\ce{S4O6^2− + 2 e− ⇌ 2 S2O3^2−}$

In your chloride example, though, you're trying to reduce $\ce{Cl^-}$ to (at least) $\ce{Cl^2-}$ and the potential for that reaction would be extremely negative. In most real-world scenarios, you would start reducing other things way before you hit that potential. Most likely you'd start reducing water at -0.8V.

$\ce{2H2O + 2 e− ⇌ H2(g) + 2 OH−}$

As an aside, be careful with your terminology. Reduction always happens at the cathode, that's what "cathode" means. Sometimes we get sloppy and don't switch the terms when we go from galvanic to electrolytic mode, but that doesn't change the chemistry. If electrons are flowing into an electrode, it's the cathode. I find the (+)/(-) labels on terminals more confusing than anything else.

Also, you said that

for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons

This is backwards. The cathode does not become negatively charged because electrons have flowed into it. Rather, reduction occurs spontaneously at the cathode, leaving it electron-deficient. Electrons then flow from the anode to the cathode to compensate this charge imbalance and make both sides neutral again.

  • $\begingroup$ Okay, I think we are referring to two different states of a galvanic cell. I'm referring to the point just before oxidants diffuse to the electrode surface. At that point, the cathode would be negatively charged as no reaction has occurred, not positive as you have stated, which would be at equilibrium. However, if the oxidant never reaches the electrode due to electrostatics then no reduction will occur. That's what I'm really asking here: what's the state of electrostatics? Will anions be able to diffuse (effectively) toward a negatively charged cathode? $\endgroup$ Jan 12, 2017 at 15:25
  • $\begingroup$ Wait, or are the ions already adsorbed to the electrode before the electrons arrive? This is what caused the flow of electrons in the first place...this created potential difference? So, I can ask this another way, if assuming the cathode is net positively charged throughout the experiment - will cationic oxidants effectively diffuse to the electrode surface or will there be electrostatic repulsion preventing this? $\endgroup$ Jan 12, 2017 at 15:29
  • $\begingroup$ Ok, let's say the cell is not connected to start with. Electrodes are submerged in the electrolyte. The oxidant diffuses through the solid-electrolyte interface and adsorbs to the surface. An electron transfers from the electrode's conduction band to the oxidant, which gets reduced, leaving the electrode with a net positive charge. The quickly stops as electrostatics repels more oxidant. Now we connect the external circuit and electrons flow from the anode to the cathode to balance the built up charge, which relieves the electrostatic repulsion and let's redox continue. $\endgroup$
    – m3wolf
    Jan 13, 2017 at 0:32
  • $\begingroup$ Okay, so that brings me back to my original thought - can anions or cations be reduced at the cathode, irrespective of its charge? ie there is no electrostatic barrier? $\endgroup$ Jan 13, 2017 at 23:14
  • $\begingroup$ Sure. It's just really hard to do for most anions and neutral metals. $\endgroup$
    – m3wolf
    Jan 14, 2017 at 3:31

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