Question

1. If the reaction below was performed in a lab within a 1 L vessel and showed to produce 1.550 mol of HI while leaving 0.225 mol of H₂ and 0.225 mol of I₂ remaining.

H₂(g) + I₂(g) ⇌2HI(g)         

a.Calculate the standard free-energy change for the reaction at 458°C.

2. Use the chemical equation below to answer the following questions.

Ca(s) + Cl₂(g) Ca²⁺ (aq) + 2 Cl^-(aq)

a.Write the cell notation for this equation.

b.What is the half-cell reaction for the anode and for the cathode?

c.What is the standard cell potential of the cell? Use the table from your textbooks for standard reduction potentials.

Answer 1

1a) The equilibrium reaction is

H₂ (g) + I₂ (g) <=====> 2 HI (g)

The equilibrium constant for the reaction is given as

K = [HI]²/[H₂][I₂] where the terms in square braces indicate molar concentrations at equilibrium.

It is given that the system has 1.550 mole of HI and 0.225 mole each of H₂ and I₂ at equilibrium. Given the volume of the container, we can easily calculate the molar concentrations of the gaseous species at equilibrium as

[HI]= (moles of HI at equilibrium)/(volume of container) = (1.550 mole)/(1.0 L) = 1.550 mol/L

[H₂] = (0.225 mole)/(1.0 L) = 0.225 mol/L

[I₂] = (0.225 mole)/(1.0 L) = 0.225 mol/L

Plug in values to find K.

K = (1.550 mol/L)²/(0.225 mol/L).(0.225 mol/L) = (1.550)²/(0.225)² = 47.4568

Use the relation ΔG⁰ = -R*T*ln K to find the standard free energy change of the reaction. It is given that the reaction takes place at 458⁰C so that T = (273 + 458) K = 731 K. Plug in values.

ΔG⁰ = -(8.314 J/mol.K)*(731 K)*ln (47.4568) = -(6077.534 J/mol)*(3.8598) = -23458.0657 J/mol = -(23458.0657 J/mol)*(1 kJ/1000 J) = -23.45806 kJ/mol ≈ -23.458 kJ/mol (ans).

2a) The cell reaction is given as

Ca (s) + Cl₂ (g) --------> Ca²⁺ (aq) + 2 Cl^- (aq)

It is quite evident that Ca is oxidized and Cl₂ is reduced. Oxidation takes place at the anode and reduction at the cathode. Therefore, we can write the cell notation as

Ca (s), Ca²⁺ (aq) ││ Cl₂ (g), Cl^- (aq)│Pt

Here Pt acts as the solid support to support the Cl₂/Cl^- electrode.

b) The half cell reactions and the standard reduction potential values are listed below:

Reduction: Cl₂ (g) + 2 e^- --------> 2 Cl^- (aq); E⁰_red = +1.358 V

Oxidation: Ca (s) -------> Ca²⁺ (aq) + 2 e^-; E⁰_ox = + 2.87 V (tables in text books give the standard reduction potential for the reduction process, i.e, the reverse process; we need to reverse the sign of the standard reduction potential to obtain the standard oxidation potential)

c) The standard cell potential is given as E⁰_cell = E⁰_red + E⁰_ox = (+1.358 V) + (+2.87 V) = +4.228 V (ans).