Question

A voltaic cell is constructed that uses the following reactions and operates at 298K:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

If the cell generates an emf = +0.493 V and the [Ag+] = 0.795 M, what is the [Cu2+] ?

Answer 1

Lets find Eo 1st

from data table:

Eo(Cu2+/Cu(s)) = 0.337 V

Eo(Ag+/Ag(s)) = 0.7996 V

As per given reaction/cell notation,

cathode is (Ag+/Ag(s))

anode is (Cu2+/Cu(s))

Eocell = Eocathode - Eoanode

= (0.7996) - (0.337)

= 0.4626 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

use:

E = Eo - (0.0592/n) log {[Cu2+]^1/[Ag+]^2}

0.493 = 0.4626 - (0.0592/2) log ([Cu2+]^1/0.795^2)

log ([Cu2+]^1/0.795^2) = -1.027

([Cu2+]^1/0.795^2) = 9.397*10^-2

[Cu2+] = 5.94*10^-2 M

Answer: 5.94*10^-2 M