Starting with the standard free energies of formation from the following table, calculate the values of ΔG° and E°cell of the following reactions. SubstanceΔΔG°f (kJ/mol)FeO(s)-255.2H2(g)0Fe(s)0H2O(l)-237.2Pb(s)0O2(g)0H2SO4(aq)-744.5PbSO4(s)-813.0 FeO(s)+H2​(g)Fe(s)+H2​O(l) ΔΔG° =     kJ     FeO(s)+H2​(g)Fe(s)+H2​O(l) E°cell =     V     2Pb(s)+O2​(g)+2H2​SO4​(aq)2PbSO4​(s)+2H2​O(l)  ΔΔG° =     kJ      Pb(s)+O2​(g)+2H2​SO4​(aq)2PbSO4​(s)+2H2​O(l)  E°cell =     V

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Asked Nov 19, 2019
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Starting with the standard free energies of formation from the following table, calculate the values of ΔG° and cell of the following reactions.
 
Substance ΔΔG°f (kJ/mol)
FeO(s) -255.2
H2(g) 0
Fe(s) 0
H2O(l) -237.2
Pb(s) 0
O2(g) 0
H2SO4(aq) -744.5
PbSO4(s) -813.0
 
FeO(s)+H2​(g)Fe(s)+H2​O(l)
 ΔΔG° =     kJ   

 
 
FeO(s)+H2​(g)Fe(s)+H2​O(l)
 E°cell =     V   

 
 
2Pb(s)+O2​(g)+2H2​SO4​(aq)2PbSO4​(s)+2H2​O(l)
  ΔΔG° =     kJ   
 

 

 
Pb(s)+O2​(g)+2H2​SO4​(aq)2PbSO4​(s)+2H2​O(l)
  cell =     V   
 
 
 
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Expert Answer

Step 1

For reaction (1),

The enthalpy change is calculated for the given reaction as follows,

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Step 2

The standard cell potential is calculated using,

FeO(s) H (g) » Fe(s) + H,0(1)
Н.
AGO nFEO
cell
-(2) (96.485 kJ/volt) EQ
15 kJ/mol
cell
15kJ/mol
ЕO
cell
=-0.0777V
=
-(2 mole e(96.485 kJ/volt)
help_outline

Image Transcriptionclose

FeO(s) H (g) » Fe(s) + H,0(1) Н. AGO nFEO cell -(2) (96.485 kJ/volt) EQ 15 kJ/mol cell 15kJ/mol ЕO cell =-0.0777V = -(2 mole e(96.485 kJ/volt)

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Step 3

For reaction (2),

The enthalpy change is calculat...

2Pb (s) O2(g) + 2H2SO4(aq)- 2P6SO4(s) 2H20(1)
AG (kJ/mol
0
0
-744.5
-813.0
-237.2
nAGproducts - n AGReactants
AGo
=[2(-237.2)+2813.0)]-[2(-744.5)] = -611.4kJ /mol
help_outline

Image Transcriptionclose

2Pb (s) O2(g) + 2H2SO4(aq)- 2P6SO4(s) 2H20(1) AG (kJ/mol 0 0 -744.5 -813.0 -237.2 nAGproducts - n AGReactants AGo =[2(-237.2)+2813.0)]-[2(-744.5)] = -611.4kJ /mol

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