# Galvanic cell vs. electrolytic cell

#### laczlacylaci

2+ Year Member
Is this picture incorrect or is it just me?

Here is info that I have, pls correct if I am wrong:
Galvanic cells: anode-; cathode+. electrons will flow towards cathode, because it is being reduced (gain electrons) For cell potential calculations, it is [email protected] (ox)[email protected](red).

Electrolytic cells (ie. gel electrophoresis): anode+; cathode-. electrons will flow toward cathode, because it is being reduced (gain electrons) For cell potential calculations it is the same as above. [email protected] (ox)[email protected](red).

I am always confused with the cell potential calculation, because different systems use different methods +/-

#### theonlytycrane

5+ Year Member
I think you have the correct general idea. The pictures can get confusing. Here's my take:

electrons always flow from anode to cathode (in both cells) and the random mnemonic helps "an ox, red cat"

galvanic (also called voltaic) cells are spontaneous (no battery). anode is (-) and cathode is (+). Usually reduction potentials are given for both half-reactions so one of them has to be flipped. The more positive the reduction potential, the higher affinity for electrons (to be reduced). So I always find the one with the highest 'E' value first and mark that as the cathode (red cat from above). The other half-reaction has to be the anode so you flip the reaction and sign on the E. The net E is now just the sum of the two.

electrolytic cells are non-spontaneous and battery driven. The potential will just be that from the battery. Cathode is now (-) and anode is (+) but electrons still flow from anode to cathode.

For a double-check, using "an ox, red cat", whichever half-reaction we mark as the anode should have something getting oxidized. The e- term will be on the product side. Conversely, the half-reaction of the cathode will have the e- term on the reactant side.

Sarahka74
OP

#### laczlacylaci

2+ Year Member
I think you have the correct general idea. The pictures can get confusing. Here's my take:

electrons always flow from anode to cathode (in both cells) and the random mnemonic helps "an ox, red cat"

galvanic (also called voltaic) cells are spontaneous (no battery). anode is (-) and cathode is (+). Usually reduction potentials are given for both half-reactions so one of them has to be flipped. The more positive the reduction potential, the higher affinity for electrons (to be reduced). So I always find the one with the highest 'E' value first and mark that as the cathode (red cat from above). The other half-reaction has to be the anode so you flip the reaction and sign on the E. The net E is now just the sum of the two.

electrolytic cells are non-spontaneous and battery driven. The potential will just be that from the battery. Cathode is now (-) and anode is (+) but electrons still flow from anode to cathode.

For a double-check, using "an ox, red cat", whichever half-reaction we mark as the anode should have something getting oxidized. The e- term will be on the product side. Conversely, the half-reaction of the cathode will have the e- term on the reactant side.
When you say the sum of the two, don't you have to be careful in if the cathode comes first or the anode, in case one of them is negative? I think it is cathode+anode for both galvanic & electrolytic, correct?

#### theonlytycrane

5+ Year Member
The questions are usually given like:

In this case, I figure out which half-reaction corresponds to the cathode (the one getting reduced will be the one with the more positive reduction potential). Then I flip the other reaction and potential and sum them.

OP

#### laczlacylaci

2+ Year Member
The questions are usually given like:

View attachment 207820

In this case, I figure out which half-reaction corresponds to the cathode (the one getting reduced will be the one with the more positive reduction potential). Then I flip the other reaction and potential and sum them.
So this would be ~+0.32V?

#### theonlytycrane

5+ Year Member
yeah and that positive deltaE works into the equation deltaG = -n * F * deltaE, making deltaG negative and the electron transfers spontaneous. (n is the # of electrons transferred and F is the Faraday constant = 96500 C / 1 mol e-)

laczlacylaci
OP

#### laczlacylaci

2+ Year Member
yeah and that positive deltaE works into the equation deltaG = -n * F * deltaE, making deltaG negative and the electron transfers spontaneous. (n is the # of electrons transferred and F is the Faraday constant = 96500 C / 1 mol e-)

#### betterfuture

2+ Year Member
Just do a quick check with the answer.

If they provide reduction potentials, you should know that the more negative a reduction potential is, the more likely it is going to be oxidized, while the more positive a reduction potential is, the more likely it is going to be reduced. Now, if they give you any reaction and you figure out which species is being oxidized and which one is being reduced, ask yourself, with the table they provided, does molecule X usually become oxidized or reduced according to their reduction potentials? If the species is becoming reduced and you know that it usually will be oxidized, you know that the E standard of cell is going to be (-) and the reaction is going to be nonspontaneous, standard G = (+)

Example

Zn+2(aq) + 2e--->Zn(s) E standard= (-0.76)
Cu+2(aq) + 2e---->Cu(s) E standard = (+0.34)

The reduction potential of Zn 2+ is -0.76 and that of Cu+2 is 0.34. You know that Zn likes to be oxidized in this case and Cu likes to be reduced. If however, the reaction you are given, provided in the question stem, Zn is reduced and Cu is oxidized, this is a nonspontaneous reaction and powered by an outside source - i.e. battery, so that the reaction may occur, meaning E standard of cell should come out (-)

Remember, E standard of cell is always opposite of standard gibbs free energy G.

Sarahka74