Asked by Jackie
A voltaic cell consists of Ag/Ag+2 electrode E= 0.80 and a Fe+2/Fe+3 electrode E=0.77 with the following initial molar concentrations: [Fe+2}= 0.30 M [Fe+3]= 0.10 M [Ag+]=0.30 M. What is the equilibrium concentration of Fe+3? (Assume the anode and the cathode solutions are of equal volume, and at the temperature 25 degreees Celcius)
Answers
Answered by
DrBob222
Did you intend Ag/Ag^+ or Ag/Ag^2+. I will assume you intended Ag/Ag^+
Ag^+ + e ==> Ag Eo = 0.80
Fe^2+ ==> Fe^3+ Eo = -0.77
==================
Ag^+ + Fe^2+ ==> Ag + Fe^3+ Eocell = 0.03
Then Ecell = EoCell - (0.0592/n)log Q
where Q = (Fe^3+)/(Fe^2+)
Ag^+ + e ==> Ag Eo = 0.80
Fe^2+ ==> Fe^3+ Eo = -0.77
==================
Ag^+ + Fe^2+ ==> Ag + Fe^3+ Eocell = 0.03
Then Ecell = EoCell - (0.0592/n)log Q
where Q = (Fe^3+)/(Fe^2+)
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