[Sis290025]

If the voltage of the following cell:
Zn(s)|Zn+2(aq)||Ag+(aq)|Ag
is 1.41 V when the concentration of Zn+2 is 0.0447, what is the concentration of Ag+ ion?.

Zn --> Zn(2+) + 2e- oxidized
2(e- + Ag+ -->Ag)     reduced

use E = E* - (0.0592/n)*log Q

E_cell = 1.41 V
[Zn (2+)] = 0.0447 M
[Ag+] = ?

E* = E_Zn + E_Ag
= 0.80 V + 0.76 V
= 1.56 V

so logQ = n(E - E*)/(0.0592 V)

log [Zn (2+)]/[Ag+]^2] = -2(1.41 V - 1.56 V)/0.0592 V

log [Zn (2+)]/[Ag+]^2] = +5.067567568
log[Zn (2+)] - 2log[Ag+] = +5.067567568
- 2log[Ag+] = +5.067567568 - log[0.0447 M]
- 2log[Ag+] = 6.424113004

log [Ag+] = (-3.21205)
[Ag+] = antilog(-3.2 . .) = 6.1368E-4

Thank you very much.

[Albert]

I got 6.185E-4. However, they are almost the same.