For what concentration of Ag+ (aq) will the EMF of the given cell be zero at 298 K if the conc. of Cu^2+ (aq) is 0.1M
Cu(s)/Cu 2+ (aq)(0.1M) //Ag(aq)+ /Ag (s)
Need help in the calc. part of nernst eq.THANK YOU
Ecell = Eocell - (0.0592/n) log K
When a cell is at equilibrium, E = 0.00 and the expression becomes an equilibrium constant K, which bears the following relationship:
n E°log K = --------................10.0592where E° is the difference of standard potentials of the half cells involved.
2 Ag+(aq) + 2 e-  2 Ag(s) | Eoreduction = + 0.80 V |
| Cu(s) Cu2+(aq) + 2 e- | Eooxidation = - 0.34 V |
| 2 Ag+(aq) + Cu(s) 2 Ag(s) + Cu2+(aq) | Eocell = + 0.46 V |
n = 2 moles of electrons
log K = log Cu 2+/[ Ag+]2
According to eq. 1
log Cu 2+/[ Ag+]2 = 2 x 0.46V / 0.0592 = 15.6345
log [0.01]/ [ Ag+]2 = 15.6345
log [0.01]- log [ Ag+]2 = 15.6345
log [0.01]-15.6345 = log [ Ag+]2
log [0.01]-15.6345 = 2 log [ Ag+]
-2-15.6345/2 = log [ Ag+]= -8.8173
[ Ag+] = antilog of -8.8173 = 1.532 x 10 -9