Asked by ciaociao

A galvanic cell consists of a Mn/Mn^2+ electrode (E° = -1.18v) and a Fe/Fe^2+ electrode (E° = -0.44v). Calculate the ratio [Mn2+]/[Fe2+] if E_cell = .78v at 25°C.

I know to use the nernst eq.
E_cell = E°_cell - RT/nF * ln(Q)
where Q is the ratio I'm looking for and I a know all of the other #s except for how to find E°_cell?
Answered by DrBob222
Mn==> Mn^2+ Eo = 1.18 (as written)
Fe^2+ ==> Fe Eo = -0.44 as written
--------------------------
Mn + Fe^2+ ==> Mn^2+ + Fe
Eocell = 1.18-0.44 = +0.74.
Answered by ciaociao
Got it, thank you!
There are no AI answers yet. The ability to request AI answers is coming soon!