Question

A concentration cell based on the following half reaction at 312 k ag+ + e- --------> ag srp = 0.80 v has initial concentrations of 1.25 m ag+, 0.221 m ag+, and a potential of 0.04865 v at these conditions. after 8.3 hours, the new potential of the cell is found to be 0.01323 v. what is the concentration of ag+ at the cathode at this new potential? Please explain all the steps used to find the answer.

Answer 1

The quantitative relationship between electrolyte concentration and cell potential is given by the following Nernst Equation:                        E_cell = E^o_cell - (RT/nF) lnQ;                

E_cell is cell potential under non-standard conditions, while E^o_cell is cell potential under standard conditions (calculated from standard reduction potentials), R = 8.314 J/(mol.K), F = 96,485 C/mol, is the Faraday’s constant; Q is the reaction quotient.

For the given half cell Ag⁺ + e^- = Ag,

Q = 1/ [Ag⁺] , E^o_cell = 0.80V, T= 312K, n=1

E_cell = E^o_cell - (RT/nF) lnQ;

        = 0.8 - (8.314)(312)/96500 [ln(1/ [Ag⁺] )]

0.01323 v = 0.8v - 0.0268[ln(1/[X])

    [log(1/[X]) = 29.4

        1/x = 10^-29.4

          X = 1/10^-29.4