Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: For a galvanic cell, when can you add half-cell reaction potentials?  (Read 1141 times)

0 Members and 1 Guest are viewing this topic.


  • Very New Member
  • *
  • Mole Snacks: +0/-0
  • Offline Offline
  • Posts: 1

In one problem I have (#3), you're given a net redox reaction and you break it up into an oxidation reaction and a reduction reaction. Then you look at a "standard potentials" table in the back of the textbook, add the two values (with one of their signs switched, of course) and you get the potential for the net reaction.

In another problem I have (#4), you're given two half reactions and their potentials. You have to take those values, use them to find deltaG for each reaction, add the values for deltaG, and then use deltaG=-vFEo to get the potential of the net reaction.

Here's an imgur album containing the two problems and my solutions:

I don't see what's different about these two problems and why for the first one I can just add the potentials to get the net potential but for the second one I have to use this other method.



  • Full Member
  • ****
  • Mole Snacks: +19/-3
  • Offline Offline
  • Posts: 129
  • USNCO High Honors 2012, 2013
Re: For a galvanic cell, when can you add half-cell reaction potentials?
« Reply #1 on: November 12, 2012, 02:58:25 PM »

For the first problem (#3) you are finding E° for a complete reaction given its component half reactions, whereas in #4 you are finding E° for another half-reaction given the E° for two related half reactions.

To calculate E° of the new half reaction, you need to convert to ΔG because the half reactions have different numbers of electrons involved (which you denoted as v). If the number of electrons in each half reaction were the same i.e. Cu2+ to Cu+ and then Cu+ to Cu, then you could sum the E° values up normally.

However for #4, you can't subtract them because adding E° Eu3+/Eu and Eu2+/Eu involves different numbers of electrons and thus the v factor in ΔG=vFE° is different. Thus ΔG represents the free energy change in the reduction reaction for different numbers of electrons. The free energy change ΔG is the only true measure of the energetics of the reaction; the voltage difference is a result of the energetics, and is dependent on other factors as well including the number of electrons in the reaction.

Also, for #3 I believe the reaction is Cd(OH)2 :rarrow: Cd2+ + 2OH-, there is no extra Cd2+ on the reactants side. :)
"Opportunity is missed by most people because it is dressed in overalls and looks like work." - Thomas Edison

Pages: [1]   Go Up

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.172 seconds with 19 queries.