Question

What is the minimum amount of [Ag+] that a cell is able to function (EMF=0). Use Fe2+/Fe3+ as the anode. They are 0.10M and 0.20 M respectively.

Answer 1

Solution:

The anode is the electrode at which oxidation take place and at cathode reduction takes place.

Thus,

Oxidation half cell : Fe3+ / Fe2+

Reduction half cell : Ag+ / Ag

Thus, E°cell = E°Ag+/Ag - E°Fe3+/Fe2+

E ° = 0.80 V - 0.77 V = 0.03 V

Emf of the cell is calculated as,

Ecell = E° - (0.0591/n) log [Fe3+/Fe2+] / [Ag+]

0 = 0.03 - 0.0591 x log (0.20/0.10) /[ Ag+]

0.03 = 0.0591 { log 2 - log [Ag+]}

0.03 = 0.0591{0.3010 - log [Ag+]}

0.3010 - log [Ag+] = 0.03/0.0591 = 0.508

log [Ag+] = - 0.207

[Ag+] = antilog - 0.207

[Ag+] = 0.62 V

Thus, [Ag+] = 0.62 M