Question

A concentration cell based on the following half reaction at 274 K

Mg²⁺ + 2 e^- → Mg       SRP = -2.37 V

has initial concentrations of 1.31 M Mg²⁺, 0.239 M Mg²⁺, and a potential of 0.02008 V at these conditions. After 1.7 hours, the new potential of the cell is found to be 0.01032 V. What is the concentration of Mg²⁺ at the cathode at this new potential?

Answer 1

Eocell for concentration cell = -( RT /nF) ln K    where K = [Mg2+] anode /[Mg2+] cathode

     Initially Eo cell = - ( 8.314 x 274 /2 x 96485) ln ( 0.239/1.31)

                          = 0.02008 V

at anode we get oxidation hence we get more Mg2+ , at cathode Mg2+ decreases

hence at anode [Mg2+] = 0.239+X , at cathode [Mg2+] = 1.31-X

now Eocell =0.01032 = ( -8.314x274 /2 x 96485) ln ( 0.239+X/1.31-X)

0.4172 = ( 0.239+X/1.31-X)

0.54653 -0.4172X = 0.239+X

X = 0.217 ,

hence at cathode [Mg2+] = 1.31-0.217 = 1.093 M