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?
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.
Got it, thank you!
To find the standard cell potential (E°_cell), you can use the standard reduction potentials of the Mn/Mn^2+ and Fe/Fe^2+ half-reactions and apply the following equation:
E°_cell = E°_cathode - E°_anode
In this case, the Mn/Mn^2+ electrode is the cathode, and the Fe/Fe^2+ electrode is the anode. Since reduction potentials are always for reduction half-reactions, you need to flip the sign of the reduction potential for the anode half-reaction:
E°_cell = E°_cathode - E°_anode
= E°_Mn/Mn^2+ - E°_Fe/Fe^2+
= -1.18 V - (-0.44 V)
= -1.18 V + 0.44 V
= -0.74 V
Therefore, the standard cell potential (E°_cell) for this galvanic cell is -0.74 V.