— Physical Chemistry — Electrochemistry Chemistry Question
Chlorine electrochemistry
State the Nernst equation.
You are given the following set of standard electrode potentials and half cell reactions for chlorine.
Calculate the following quantities
i) The ionic product of water, Kw.
ii) The equilibrium constants for the disproportionation reaction of chlorine to oxidation states +1 and –1 under both acidic and alkaline conditions.
iii) The pKa value for HOCl.
iv) The concentrations at pH 7.5 of HOCl and ClO– in a solution where the total concentration of hypochlorite (chlorate (I)) is 0.20 mmol dm–3, and the electrode potential for the reduction of this system to chlorine at this pH with unit activity of chlorine. These conditions are typical of a swimming pool.
Model Answer
i) The difference between the alkaline and acidic perchlorate half cells is:
E° / V G° / kJ mol –1
0.37 –71.4
ClO4 − + H2O + 2 –e → ClO3− + 2 OH−
ClO4 − + 2 H+ + 2 –e → ClO3− + H2O 1.20 –231.6
2 H2O → 2 H+ + 2 OH− –80.1
Hence Kw = 9.2×10–15
ii) Alkaline conditions
E° / V G° / kJ mol –1
1.36 –131.2
½ Cl2 + –e → Cl−
ClO− + H2O + –e → ½ Cl2 + 2 OH− 0.42 –40.5
Cl2 + 2 OH− → ClO- + Cl- + H2O –90.7
Hence Kc = 7.9×1015.
Acidic conditions
E° / V G° / kJ mol –1
1.36 –131.2
½ Cl2 + –e → Cl−
HOCl + H+ + –e → ½ Cl2 + H2O 1.63 –157.3
Cl2 + H2O → HOCl + Cl- + H+ +26.1
Hence Kc = 2.7×10–5
iii) The pKa value for HOCl.
G° / kJ mol –1
HOCl + H+ + –e → ½ Cl2 + H2O –157.3
ClO− + H2O + –e → ½ Cl2 + 2 OH− –40.5
HOCl + H+ + 2 OH− → ClO− + 2 H2O –116.8
2 H2O → 2 H+ + 2 OH− +160.2
HOCl → H+ + ClO− +43.4
Ka = 2.4×10–8 and pKa = 7.61
iv) With this value of Ka, at pH 7.5 we have the ratio [HOCl] / [OCl − ] = 1.29, thus [HOCl] = 0.113×10–3 and [OCl − ] = 0.087×10–3
In either case, taking the activity of chlorine as 1, the result is 1.13 V.