Question

What is the OCP (open circuit potential) of the H2(g)/Cl2(g) electrochemical cell [Pt(s)|H2(g)|HCl(aq)|Cl2(g)|Pt(s)] at 25 oC, if partial pressure of hydrogen and chlorine are 0.5 bar and the concentration of HCl(aq) is 5 mol/kg?

Answer 1

The electrochemical reactions involved are:

Oxidation:

(Reference electrode)

Reduction:

Thus,

.......................... (i)

But, C should be in mol/L and P in bar

Concn of HCl has been given as 5 mol/kg

5 mol of HCl is 5*36.5 g of HCl= 182.5 g of HCl where 36.5 g is the molecular weight of HCl

Thus, in a 1 kg solution we have 182.5 g of HCl which implies that the mass % of HCl is 18.25%

This, translates to a density of 1.089 g/mL as can be seen from literature for HCl

Thus, 1 kg of this solution, will occupy a volume of 918.3 mL =0.9183 L

Hence, there are 5 moles in 0.9183 L which implies by unitary method that there will be (5/0.9183) moles in 1 L

Thus, C_HCl = 5.445 mol/L

P_H2 = 0.5 bar

P_Cl2 = 0.5 bar

T= 298.16 K, R=8.314, F=96485 Coulumb/Volt, n= number of electrons exchanged=1

E⁰_Cell =0.679 V

Thus, substituting these values in equation (i), we get: