Cliffs AP Chemistry, 3rd Edition

10. For the reaction

Pb^^shh++ +PbO 2442 s 42 H+__aqiiSO2-aq " 2 PbSO ^^shh+ 2 H O ,

which is the overall reaction in a lead storage battery, ∆H°= −315.9 kJ/mole and ∆S°=

263.5 J/K ⋅mole. What is the proper setup to find E°at 75°C?

A.

.

..

2 96 487

315 9 348 0 2635

• 
  

--

^

^

h

h

B.

.

..

2 96 487

-+348 315 9 0 2635

^

^

h

h

C..

..

96 487

-+348 315 9 0 2635^h

D. ..

.

96 487 315 9

2 348 263 5

+

-- +^h

E.

.

..

96 487 348

2 315 9 -263 5

^^

^

hh

h

Answer: A

Use the relationships

∆G°= −nFE°= ∆H°−T∆S°

to derive the formula

E°= ∆∆HTS-n-F

% %

Next, take the given equation and break it down into the oxidation and reduction half-reactions
so that you can discover the value for n, the number of moles of electrons either lost or gained.

Anode reaction (oxidation)

Pb(s) + SO 42 - →PbSO 4 (s) + 2 e–

Cathode reaction (reduction)

s s aq aq s

saqaq s

Pb PbO H SO PbSO H O

PbO SO H e PbSO H O

42 2 2

42 2

2442

24 42

"

"

,

,

++ + +

+++ +

• 
  


• 
  
  
  
  • 
    
  
  
  
  

• 
  

+

2

2

^^__ ^^

^ __ ^^

hhiihh

h ii hh

Now substitute all the known information into the derived equation.

E°=H°−T∆S°/−nF= (−315.9 kJ/mole−348 K⋅0.2635 kJ/K⋅mole/ −2(96.487 kJ/V⋅mole)

Part II: Specific Topics