Question

Given:

Cr3+(aq)+3e– ⇌Cr(s);E°=−0.74 V

Cl2(g)+2e– ⇌2Cl−(aq);E°=1.36  

What is the standard Gibbs free-energy change for the following reaction? (F = 96485 C per mole of electrons).

2Cr(s) +3Cl2(g) ⇔2Cr3+(aq) +6Cl−(aq)

Select one:

a. −1215 kJ

b. −231.3 kJ

c. −983 kJ

d. −322.4 kJ

e. 2309 kJ

Answer 1

2Cr(s) + 3Cl₂(g) ⇔2Cr³⁺(aq) +6Cl⁻(aq)

E⁰_rxn = E_r^o - E_l^o

E_r^o = standard electrode potential of reduction of half reaction (Cl2 undergoes reduction)

E_l^o = standard electrode potential of oxidation of half reaction (Cr undergoes oxidation)

E⁰_rxn = 1.36 - (-0.74) = 2.1V

Standard Gibbs free energy (ΔG^o) = - nF E⁰_rxn

n = no. of moles of electrons involved in the reaction = 6 in our case as each mole of Cr loses 3 moles of electrons.

ΔG^o = - 6 X96485 X 2.1V = - 1215.711 KJ