Question

Ascorbate and Cu ions have the following electrochemical potentials;

Dehydroascorbate + 2H+ + 2e- --> Ascorbate Eo' = +0.08V

Cu2+ + e- --> Cu+ Eo' = 0.159 V

10 mL of a 0.02M solution of ascorbate in a buffered solution at pH 7.0; 25 degrees Celsius is mixed with 10mL of 0.02 M of Cu2+. Write a balanced chemical equation for the reaction, and figure out the equilibrium concentration of dehydroascorbate, ascorbate, Cu2+, and Cu+.

Answer 1

for the above shown data

Balanced chemical equation,

2Cu2+ + ascorbate ---> 2Cu+ + dehydroascrobate + 2H+

Eo = 0.159 - 0.08 = 0.079 V

using,

dGo = RTlnK = nFEo

8.314 x (25 + 273) lnK = 2 x 96500 x 0.079

equiilibrium constant K = 470.6

initial [Cu2+] = 0.02 M x 10 ml/20 = 0.01 M

initial [ascorbate] = 0.02 M x 10 ml/20 = 0.01 M

pH = -log[H+] = 7

[H+] = 1 x 10^-7 M

ICE chart

                 2Cu2+ + ascorbate ---> 2Cu+ + dehydroascrobate + 2H+

I                  0.01            0.01             -                    -                    1 x 10^-7

C                -2x               -x              +x                  +x                    +2x

E             0.01-2x          0.01-x             x                    x                  1 x 10^-7-x

So,

K = [dehydroascorbate][Cu+]^2.[H+]^2/[Cu2+]^2.[ascorbate]

let x be a small change

470.6 = (x)(2x)^2(1 x 10^-7)^2/(0.01-2x)^2.(0.01-x)

1 x 10^-4 - 0.04x + 4x^2

1 x 10^-5 - 4 x 10^-4x + 0.04x^2 - 1 x 10^-4x + 0.04x^2 - 4x^3

1882.4x^3 + 37.65x^2 - 0.24x + 4.71 x 10^-3 = 0

solving,

x = 0.003 M

So concentration of,

[Cu2+] = 0.01 - 2 x 0.003 = 0.004 M

[ascorbate] = 0.01 - 0.003 = 0.007 M

[Cu+] = 2 x 0.003 = 0.006 M

[dehdyroascrobate] = 0.003 M