14.21 - Galvanic Cells

Moderators: Chem_Mod, Chem_Admin

Raman Ishwar 1J
Posts: 11
Joined: Fri Sep 25, 2015 3:00 am

14.21 - Galvanic Cells

Postby Raman Ishwar 1J » Tue Feb 02, 2016 7:21 pm

In all of the cell diagrams in problem 14.21, the answer key shows two reduction half-reactions instead of one reduction half-reaction and one oxidation half-reaction.

I always thought that you start out with two reductions, but then you flip the one you got from the left side of the cell diagram because the cell diagram always has oxidation on the left and reduction on the right. When you do this flip, the E value you got from the reduction changes signs.

In the answer key, however, this flip does not occur, and both half-reactions are left as reductions and there is no oxidation taking place. Why is that? Is it only for galvanic cells?

Ronald Yang 2F
Posts: 86
Joined: Fri Sep 25, 2015 3:00 am

Re: 14.21 - Galvanic Cells

Postby Ronald Yang 2F » Tue Feb 02, 2016 10:25 pm

No, you are correct. You must flip the half-reaction resembling reduction that would occur at the anode to yield the half-reaction resembling oxidation. The reason why they leave the "oxidation" half-reaction in its reduction form is because they want to emphasize that the standard cell potential of the anode is a reduction potential, according to the equation Eo(cell)=Eo(cathode) - Eo(anode) [to better explain, view it like this: Eo(cell)=Eo(cathode) + (-Eo(anode))]. When they include that "minus sign," they are essentially flipping the reduction potential, symbolized by Eo(anode), to become an oxidation potential, symbolized by -Eo(anode).


Return to “Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams”

Who is online

Users browsing this forum: No registered users and 3 guests