I am very much confused: in a simple voltaic cell ($\ce{Zn}$, $\ce{Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ is positively charged and is the anode, where in a Galvanic cell ($\ce{Zn}$ in $\ce{ZnSO4}$ and $\ce{Cu}$ in $\ce{CuSO4}$), $\ce{Cu}$ is positively charged but is the cathode. How can I define it, where both are electric or voltaic cell?


3 Answers 3


Cathode is the electrode where the Reduction reaction takes place.

Anode is the electrode where the Oxidation reaction takes place


Voltaic cell is a synonym for a Galvanic cell. In the mentioned system ($\ce{Zn}$, $\ce{Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ can be charged both positively and negatively. It depends how you connect it to the power source. In a Galvanic/Voltaic cell (as Babounet mentioned):

The cathode is a negatively charged electrode where reduction takes place, and the anode is a positively charged electrode where oxidation takes place.

Let's analyse the second cell. Without any additional information, I assume that no external power source is connected. This system is known as Daniell cell:

Wikipedia Daniell cell

\begin{align} \ce{Zn | Zn^2+ &|| Cu^2+ | Cu}\\ \ce{Zn - 2e- &-> Zn^2+} &\therefore\text{oxidation} &= \text{anode}\\ \ce{Cu^2+ + 2e- &-> Cu} &\therefore\text{reduction} &= \text{cathode} \end{align}

The situation is nearly the same for the first cell, but this situation is used and known as Sacrificial Anode. $\ce{Zn}$ acts in the system as an anode and it is dissolved.

You can have two electrodes that are both positively charged such as $\ce{Ag}$ and $\ce{Cu}$. Standard electrode potential of both metals are positive: $\ce{Ag+|Ag (s)} = \pu{+0.80 V}$ and $\ce{Cu^2+ | Cu (s)} = \pu{+0.34 V}$, but if you connect them in a galvanic cell, one will act as cathode ($\ce{Au}$) and the other as anode ($\ce{Cu}$).

As far as I can tell (based on provided informations), the copper electrode will be in both cases positively charged and therefore be a cathode. There must be a mistake in the description or in the statement.


We can take this definition - Cathode is the electrode which ejects electrons into the site of interest/study. This holds for both electrolytic and galvanic cells.

In an electrolytic cell, the solution in which the electrodes are immersed is the area of interest and cathode would be that electrode which would eject electrons into the solution.

In a galvanic cell, the half cells (the electrodes dipped in their metal salt solutions) where chemical reactions happen responsible for electric current generation is the area of interest and not the wire. In a typical galvanic cell, the Daniell cell, we can see that Cu electrode would be the cathode as it ejects electrons into the CuSO4 solution reducing the copper(II) ions to copper.

In a cathode ray tube experiment, the CRT would be the area of interest and electrons are ejected from the cathode into the tube and are incident on the glass behind the anode.

The statement that cathode is the electrode where reduction takes place is slightly misleading. This is because oxidation and reduction take place simultaneously(have to). In Daniel cell, Cu electrode is the cathode as Cu(II) ions get reduced to Cu. In electrolysis of molten KCl, the electrode where K(I) ions get reduced to K metal is the cathode. However, note that K(I) ions in the second case come from the solution and the electrodes need not be made of K whereas in the Daniel cell, Cu metal is the electrode itself. In one case we talk about the reduction of electrode while in the other we talk about reduction of something present in the vicinity of electrode. If this is kept in mind, then we could remember cathode as that electrode where reduction takes place.

So, the bottomline is, it is important to know the site of interest/study which is different for a galvanic and an electrolytic cell as the cathode and anode are terms based on that.

Edit: In a galvanic cell, the cathode is positive because it ejects electrons and becomes positive (the electrode being neutral in the beginning and attains an equilibrium). In an electrolytic cell, we are maintaining a potential difference across the electrodes and forcing the reaction to take place and electrons move from negative to positive terminal (being negatively charged, they feel a force opposite to the direction of electric field) and cathode has negative polarity.

  • $\begingroup$ Welcome to Chemistry.se! Here mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. Please don't use edit statements in you post, always think of them as final versions targeted at users who read them for the very first time. If someone is interested in the timeline of a post, an edit-history is linked below the post. $\endgroup$ Commented Sep 9, 2020 at 14:55

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.