Br2 + 2 Fe2+(aq) = 2 Br-(aq) + 2 Fe3+(aq) Electrochemistry?

Question

For the reaction above, the following data are available:

2 Br-(aq) =Br2(l) + 2e- Eo = -1.07 volts

Fe2+(aq) = Fe3+(aq) + e- Eo = -0.77 volts

______________________________________...

So, cal/mole K

Br2(l) 58.6 Fe2+(aq) -27.1

Br-(aq) 19.6 Fe3+(aq) -70.1

______________________________________...

(a) Determine delta Srxn

(b) Determine delta Grxn

(c) Determine delta H rxn

Answer 1

Br-(aq) =Br2(l) + 2e- Eo = -1.07 volts

Fe2+(aq) = Fe3+(aq) + e- Eo = -0.77 volts

so  

Br2(l) + 2e- =>  Br-(aq)     Eo = + 1.07 volts

Fe2+(aq) = Fe3+(aq) + e- Eo = -0.77 volts

Eo = +0.30

(b) Determine delta Grxn:

d Grxn = -nFEo

dG = - 2 e- (96.5kJ/mol-ev)(0.30 volts)

your first answer: dG = - 57.9 kJ

(but it really should be rounded off to -58kJ (2sigfigs in the "0.30" volts)

===========================

(a) Determine delta Srxn

Br2 + 2 Fe2+(aq) = 2 Br-(aq) + 2 Fe3+(aq)

dS = S prod - S reactants

dS = [2(19.6) & 2 (-70.1)] -[58.6 & 2(-27.1)]

dS = (-101) - (4.4)

dS = -105.4 joules/K

===============================

(c) Determine delta H rxn , @ 298K since you have standard conditions:

dG = dH -TdS

-57,900 Joules = dH  - (298K)(-105.4J/K)

-57,900 Joules = dH  + 31,409 joules

dH = -57,900 Joules - 31,409 joules

dH = -89309 Joules

your answer (3 sigfigs) : dH = -89.3 kJ

but I still hold that it is a 2 sigfig problem,,, so I'd round it off to dH = -89kJ