Starting from the appropriate free energies of formation
(provided below) calculate the values of ∆G° (in kJ) and
E°cell (in V) at 298 K for the following
reaction,
2 Cu+(aq) -----> Cu2+(aq) + Cu(s)
DGf°(Cu+(aq)) = 49.98 kJ
mol-1
DfG°(Cu2+(aq)) = 65.49 kJ
mol-1
DfG°(Cu(s)) = 0.00 kJ mol-1
∆G° = 65.49-(2×49.98)
= -34.47 kJ
∆G° =- nFE°
E° = -34.47×1000/(-1×96500)
= 0.357 V
Starting from the appropriate free energies of formation (provided below) calculate the values of ∆G° (in...
Starting from the appropriate free energies of formation (provided below) calculate the values of ∆G° (in kJ) and E°cell (in V) at 298 K for the following reaction, 2 Na(s) + 2 H2O(l) <-----> 2 Na+(aq) + 2 OH-(aq) + H2(g) DfG°(Na(s)) = 0.00 kJ mol-1 DfG°(H2O(l)) = -237.13 kJ mol-1 DfG°(Na+(aq)) = -261.91 kJ mol-1 DfG°(OH-(aq)) = -157.24 kJ mol-1 DfG°(H2(g)) = 0.00 kJ mol-1
Starting with the standard free energies of formation from the
following table, calculate the values of ΔG° and E°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
1) FeO(s) +H2(g) ----Fe(s) +H2O(l)
change in G=18kj
E Cell= v
2) 2Pb(s)+O2(g)+2H2SO4(aq)--------2PbSO4(s)+2H2O(l)
change in G=-611.4
E°cell
= V
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