Question

a) Choose all of the statements from below which correctly describe issues pertaining to the relationship between the emf of a reaction and its Gibbs free energy change.

a)The Gibbs free energy change and emf values of a reaction have opposite algebraic signs (+ vs. -).
b)Faraday's constant must be used in order to interconvert a Gibbs free energy change and a reaction emf.
c)Faraday's constant is used to interconvert coulombs of electrons and moles of electrons.
d)A Gibbs free energy change involves coulombs of substances whereas an emf involves moles of electrons.
e)A Gibbs free energy change involves moles of substances whereas an emf involves coulombs of electrons.

Calculate the standard Gibbs free energy changes at 25 ^oC for each of the reactions shown below using the E^o values given. Select whether each of these reactions is nonspontaneous, at equilibrium, or spontaneous under standard conditions.

(a) 1 Fe(CN6)3-(aq) + 1 Co2+(aq) 1 Fe(CN6)4-(aq) + 1 Co3+(aq)

Eo = -1.482 V

Go =  kJ/mol

nonspontaneous

at equilibrium

spontaneous

(b) 1 Mn2+(aq) + 2 Br-(aq) 1 Mn(s) + 1 Br2(l)

Eo = -2.245 V

Go =  kJ/mol

nonspontaneous

at equilibrium   

spontaneous

(c) 1 F2(g) + 2 Li(s) 2 F-(aq) + 2 Li+(aq)

Eo = 5.920 V

Go =  kJ/mol

nonspontaneous

at equilibrium

spontaneous

(d) 2 Cr3+(aq) + 2 Cl-(aq) 2 Cr2+(aq) + 1 Cl2(g)

Eo = -1.769 V

Go =  kJ/mol

nonspontaneous

at equilibrium

spontaneous

Answer 1

Answen. we know that (AGO = - nf El Where, as - standrad free energy change na no. of election change fa farraday's constant = 96500 C. mol f' = Standard reduction potential. when AG Do the ran is non-spontaneous. AG=0 - ar equilibrium. | AGoTo the moon at is spontaneous. a fe (CN) ( + 0 2t (24) ► Fe (CN) & (aq) + co 3+ (aq) Hene n=1, € = -1482 >> AG°= - 1 x 90500 C.mom) x (-1,482 v) =) AGO = + 143.01 *J/mol (*&v=3) AS, AG Xo is so the reaction nonspontaneous

6 1m2 + (aq) + 2Br (an) ² MM (3) + Brz (4) Here, n=2 and € = -2.245V Standard '. The free energy change, 14Gº = - NAE Cou => AG = -2 x (96500 Cimo7')(-2.2454) => AG° = 216.64 kJ/moll (3= v) AG yo, so the road is nonspontaneous. AB 9 fz(y) + 2 Wi (3) — 2F (09) + 2 i (p) Hero, n=2 and € = 5.920v 1.490 - - 2 x 66500 Ci mot') x (5.9201) y AG = - 1182560 C.v. mot => AGO =-1142.56 kJ/mole L.v>]] AS AGO TO So the rx is spontaneous. d) 2623+ (aq) + 2 61 (aq) 260 2+1aq) + Cl2 (7) Here, n=2 and € = -1.769 v 1.160 = -2x (96,500 Cimoi') x (-1769V) = 46° = + 341417 Civimoiy

=> 44° = 341.42 kJ/moll (as civ= 1) As the 46 70, so the reaction is nonspontaneous.