Question

a)You have a galvanic cell set up like a Daniell cell but using a silver electrode in a silver nitrate solution and an iron electrode in an FeSO4 solution. Instead of a semipermeable membrane, you use a salt bridge. What is a salt bridge? Where is cathode and anode? What is the voltage under standard contitions? If the silver nitrate solution is 0.2 mmol/l and the FeSO4 solution is 3 mol/l, what is the voltage you measure? Compare to the previous result.

Answer 1

Fe2+(aq) + 2e-      Fe(s) Eo = - 0.44V

Ag+(aq) + e-   Ag(s) Eo = +0.80V

The cathode is where the reduction takes place. But from the two half reactions, it can be seen that Ag is reduced.

So, cathode is Ag and anode is Fe.

To combine these half-reactions to get a positive cell potential, we reverse the Fe half-reaction.

Fe(s) Fe2+(aq) + 2e- Eo = +0.44 V

2x [Ag+(aq) + e-     Ag(s)] Eo = +0.80 V

The overall reaction is:

Fe(s) + 2Ag+(aq)    Fe2+(aq) + 2 Ag(s)

Eo = 0.44 + 0.80

Eo = 1.24 V

[Ag+] = 0.2 mmol/l = 2x10^-4 mol/l

[Fe2+] = 3 mol/l

E = Eo + 0.0591 / 2 log [3 / 2x10^-4 ]

= 1.24 + 0.0591/ 2 * 4.17

= 1.24 + 0.123

E = 1.36 V