Question

What is E for a concentration cell where [Ag+]=0.00012 M in the diluted half-cell and 1.50 M in the more concentrated one at SATP? -0.56 V 0.56 V 0.24 V -0.24 V

Answer 1

For concentration cell, cathode and anode are same electrode
So, Eo = 0
Number of electron being transferred in balanced reaction is 1
So, n = 1
If E is positive anode will be the one with lower concentration

use:
E = Eo - (2.303*RT/nF) log {[Ag+] at anode/[Ag+]at cathode}

Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591

So, above expression becomes:
E = Eo - (0.0591/n) log {[Ag+] at anode/[Ag+]at cathode}
E = 0 - (0.0591/1) log (1.2E-4/1.5)
E = 0.2422 V
Answer: 0.24 V