Question

Calculate ΔG∘rxn and K for each of the following reactions. Part A The reaction of Cr3+(aq) and Cr(s) to form Cr2+(aq). [The reduction potential of Cr2+(aq) to Cr(s) is -0.91 V.] Calculate ΔG∘rxn.

Answer 1

reaction: 2Cr+3(aq) + Cr(s) ----> 3Cr+2(aq)

standard data:

Cr3+ + 3 e− ----> Cr(s)    E0 = −0.74 v

Cr2+ + 2 e− ----> Cr(s)    E0 = −0.91 v

calculation of standard reduction potential of Cr3+ + 1e- ----> Cr2+ E0 = ?
Cr3+ + 3 e− ----> Cr(s)     DG01 = - 3F*-0.74

Cr(s) ------> Cr2+ + 2 e− DG02 = -2F*0.91
------------------------------------------------------------------------
Cr3+ + 1e- ----> Cr2+   DG0 = DG01+DG02
------------------------------------------------------------------------

   -FE0 = - 3F*-0.74+-2F*0.91

     E0 = 2*0.91-3*0.74 = - 0.4 V

Cr3+ + 1e- ----> Cr2+ , E0 = -0.4 V

cell half reaction,

oxidation half: anode   =  Cr(s) ----> Cr3+ + 3 e−

reduction half:cathode = Cr3+ + 1e- ----> Cr2+

E0cell = E0cathode - E0anode

        = -0.4-(-0.91)

        = 0.51 v

DG0rxn = -nFE0cell

n= no of electrons transfered in the reaction = 2

F = faraday constant = 96500 C

E0cell = 0.51 V

    = -2*96500*0.51

    = -98430 joule

    = -98.43 kj

DG0 = - RTlnK

-98430 = -8.314*298lnK

K = equilibrium constant = 1.8*10^17