consider the cell: Fe(s)/Fe2+(.325 M)//Ag+(5.50*10^-4M)/Ag(s) a)determine anode

Question

consider the cell: Fe(s)/Fe2+(.325 M)//Ag+(5.50*10^-4M)/Ag(s)

a)determine anode and cathode Fe2+/Fe =>Eo=-0.44, Ag+/Ag=>Eo=0.7993

b)calculate Eocell and express balanced reaction

c)calculate the actual cell potential

d)which direction to the electrons flow

e)NaCl is addded to the silver half-cell to preciptate most of the Ag+ as AgCl(s), establishing an equilibrium concentration of .250 M Cl-(NOTE: this won't affect the Fe/Fe2+ cell) if the new cell potential (Ecell) is 0.7055 V, what is the Ksp pf AgCl?

any help is greatly appreciated!

Explanation / Answer

1) anode is Fe+2 ?Fe

Cathode is Ag+/Ag

Eo cell = Eo cathode - Eo anode

= 0.7993 - (-0.44)

= 1.2393

Eo cell is 1.2393 V

Balanced reaction is

2Ag+ + Fe ------> 2 Ag + Fe +2

2) according to nernst equation

E cell = Eo cell - 0.059/n logk

here n = 2 as 2 moles of electrosn are transferred

E cell = 1.2393 - 0.059/2 log [ Fe +2] / [ Ag+ ] 2

E cell = 1.2393 - 0.059/2 log 0.325 / (5.5 x 10-4 )2

E cell = 1.0614 V

3) the eclectrons flow from Ag + ( cathode ) to Fe ( anode)

4) 0.7055 = 1.2393 - 0.059 /2 log 0.325 / [Ag+]2

[Ag+] = 5.11 x 10-10 M

Ksp = [Ag+] [Cl-]

= 5.11 x 10-10 x .25

Ksp = 1.277 x 10-10   

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