Question

Consider the following half reactions at 298K and pH 7.0

O2 + 4H+ + 4e- --> 2H2O   E°= +0.815V

Cystine + 2H+ + 2e- --> 2 Cysteine E°= -0.34

If you prepare a 0.010 M solution of cysteine at pH 7 and let it stand in contact with air at 298K, what will be the ratio of cysteine/cysteine at equilibrium? The partial Pressure of oxygen in the air is 0.20 bar. The activity coefficient mayu be taken as unity.

Answers: 1.61x10^18

Answer 1

E⁰cell = E⁰cathode -E⁰anode

E⁰cell = 0.815 - (-0.34) = 1.155 V

Using Nernst Equation

Ecell = E⁰cell - 2.303RT / nF log Q

R = 8.314 J / mol K

T = 298 K

F = 96485 C

Ecell = E⁰cell - 0.0596 / n log Q

n = 4

The overall equation will be

O2 + 4H+ + 4e- --> 2H2O

2 Cysteine ---> Cystine + 2H+ + 2e- ] X 2

........................................................

O2 + 4Cysteine ---> 2H2O + 2Cystine

Q = [Cystine]^2 / [Cysteine]^4 [pO2]

Putting values

at equilibrium Ecell = 0

E0cell = 0.0592/4 logQ

1.155 = 0.0148 logQ

logQ = 78.04

Q= 1.096 X 10^79 = [Cystine]^2 / [Cysteine]^4 [pO2]

[Cystine]^2 / [Cysteine]^4 = 0.2192 X 10&79