Question

Determine the concentration of Cd2+ ions in the following electrochemical cell:
Fe|Fe2+ (0.1M) ||Cd2+ (x M)|Cd, Given that Ecell = -0.02 V and Eo cell = 0.04 V at 298K

Answer 1

$E_{cell}=E_{0cell}+\frac{0.059}{2}lg\frac{a_{Cd^{2+}}}{a_{Fe^{2+}}}$

$a_{Cd^{+2}}=a_{Fe^{+2}}·10^{\frac{2·(E_{cell}-E_{0cell})}{0.059}}=0.1·10^{\frac{2·(-0.02-0.04)}{0.059}}=9.2·10^{-4}M$

Answer $9.2·10^{-4}M$