In the figure below, which electrodes are negative terminals? I have worked out that E: Cu2+ -> Cu, F: OH- -> O2, G: H+ -> H2, H: OH- -> O2. However, I am still confused about the polarity of the electrodes and I have thought of two explanations

  1. By definition, electron flows out of the negative terminal, so E and F are negative terminals
  2. In an electrolytic cell, the negative terminal is the cathode. So E and G are negative terminals

Which one is correct?

enter image description here

  • 1
    $\begingroup$ At electrolysis, the cathode (absorbing electrons) is usually more negative electrode. But if for some reasons you reverse the voltage and current direction, the cathode is more positive electrode. $\endgroup$
    – Poutnik
    Oct 2 at 6:50
  • $\begingroup$ Note that for F and H case, it would be H2O -> O2 $\endgroup$
    – Poutnik
    Oct 2 at 7:39

2 Answers 2


If you remember a couple of simple things most of the confusions in electrochemistry will vanish.

  1. Positive and negative in electrochemistry mean that a particular electrode has an excess electrostatic charge.

  2. Anodes and cathodes are defined with respect to the chemical processes. Electrode where oxidation occurs is the anode, electrode where reduction occurs is the cathode.

  3. The bulk of the solution has to remain electrically neutral, i.e., the charge balance condition must hold all the time. If an electron enters the solution from one electrode, another electron from the solution must leave from another electrode.

Using the three points, we see that there is a battery, and electrode E is connected to its negative terminal.

(i) We can conclude that E is negatively charged because it has excess of electrons. If electrolysis is occurring, electronic current in the external circuit is flowing. If an electron left the electrode E, in order to balance the charge, another electron from electrode F must leave, so F is positive (cond. 3).

(ii) The electron which left the electrode F, has to enter the solution via electrode G. G has excess of electrons, and it will have a negative charge.

(iii) Electrode H is connected to the positive terminal of the battery. It must be positive.

  • 1
    $\begingroup$ I have heard from physics that positive terminal means a higher potential and a negative terminal means a lower potential. Does this apply to electrochemistry? If yes, could I also understand it as F having a higher potential than G? Edit: The potential refers to the electrical potential (as in Volts) $\endgroup$ Oct 2 at 5:32
  • $\begingroup$ (iii) It implies the external power source is harder voltage source than eventual galvanic cell. If - as extreme case - I attach 9V compact battery to 2 V cell of the car acid-lead battery to perform electrolysis, the lead-acid cell is much harder voltage source and would not care to which its contact is attached + contact of the 9V battery. $\endgroup$
    – Poutnik
    Oct 2 at 7:45
  • $\begingroup$ @JellyQwerty, Yes by convention, a positive positive (relative) is higher than negative potential. F is at higher potential than G. $\endgroup$
    – AChem
    Oct 2 at 13:51

Note that the potential of galvanic/electrolytic cell electrodes is relative to each other. They can be both positive or both negative wrt the potential of the standard hydrogen electrode (SHE), taken conventionally as 0.00 V potential reference. And the SHE has potential $\pu{+4.44 \pm 0.02 V}$ wrt the potential of a free electron.

Therefore, the "positive" electrode, contact or potential means being more positive than the "negative" one.

The electrolytic cell can be considered as a power source with the open circuit voltage $U_\text{cell}$ and internal resistance $R_\text{cell}$. The external source similarly is the combination of a voltage source $U_\text{src}$ with the internal resistance $R_\text{src}$.

Case 1 - The "Positive" contact of the voltage source is connected to the "positive" contact of the electrolytic cell. The source has higher voltage than the cell.

The current flows to the more positive electrode contact, electrons flow from it and this more positive electrode is the anode. The current $$I = \frac{U_\text{src} - U_\text{cell} }{ R_\text{src} + R_\text{cell}}$$

The voltage on the contacts is then

$$U_\text{cont} = \frac{ U_\text{src} R_\text{cell} + U_\text{cell} R_\text{src} } { R_\text{src} + R_\text{cell}} $$

The contact with the more positive potential is then obviously at the more positive contacts of both the source and the cell.

Case 2 - The "Positive" contact of the voltage source is connected to the "positive" contact of the electrolytic cell. The source has lower voltage than the cell.

The formulas above still stand, but there is the opposite direction of the current. The measured voltage on contacts is lower than the voltage of the cell at the open circuit, but higher than the voltage of the source at the open circuit. The cell works in the galvanic mode instead and the positive electrode is the cathode now.

Case 3 - The "Positive" contact of the voltage source is connected to the "negative" contact of the electrolytic cell.

In this case, we have two power sources, connected in a series, the cell working in the galvanic mode.

The current $$I = \frac{U_\text{src} + U_\text{cell} }{ R_\text{src} + R_\text{cell}}$$

The potential of the nominally "positive" electrode of the cell wrt to its nominally "negative" is: $$U_\text{cont} = \frac{ U_\text{cell} R_\text{src} - U_\text{src} R_\text{cell}} { R_\text{src} + R_\text{cell}} ,$$

and can be both positive and negative. As in the case 2, the cell works in the galvanic mode and the nominally "positive" electrode is the cathode now.

Note that this mode may lead to very high currents and potentially to damage or destruction of the source or the cell.


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