can someone explain the differences between electrolytic and galvanic cells

[mrh125]

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and what we should know about them?

I know that electrolytic ones are unfavorable and require an external power source (the potential "E" is negative) and that galvanic ones are favorable (positive E), but what else should we know and how does this affect how you solve problems?

I also know that the cathode is negative and the anode is positive in Electrolytic cells and the cathode is positive and anode is negative in Galvanic cells. What does that change in problems?

Also, the voltometer just conducts charge through both containers right?
Oh and where anions and cations go:
electrolytic: anions go to anode
cations go to cathode

galvanic: cations go to anode
anions go to cathode

 

[aesculapian]

It sounds like you pretty much know most of what you'd need to.. also know that having a negative E means a positive G (nonspontaneous, electrolytic cells) and a positive E = negative G (spontaneous, galvanic cells). Voltmeter tells you the potential, not the charge. Also keep in mind that there are two ways you can think about the equation for cell potential/solving problems. For Ecell = Ecat - Ean, you're using the reduction potentials (probably given in the passage or question stem); that is, the values of E. (And reduction = gaining electrons, so whatever element is having electrons added to it is being reduced.)

However, there's also the equation Ecell = Ered + Eox. In this case you'd want to use the reduction potential for whatever element is being reduced (larger, more positive reduction potential = greater propensity to be reduced and gain electrons) and the oxidation potential for the other species. So if you're given only the reduction potentials, you'll want to reverse the sign to get the oxidation potential.

In my opinion, using the first equation is easier, but it's up to you. I think all of this is right, but someone please correct me if I'm wrong!