Question
The rusting of iron takes place as follows 2H++2e−+12O2→H2O(l); E∘=+1.23 V Fe2++2e−→Fe(s); E∘=−0.44 V Calculate ΔG∘ for the net process
Answer: Option A
:
A
Fe(s)⟶Fe2++2e−;ΔG∘12H++2e−+12O2⟶H2O(I);ΔG∘2Fe(s)+2H++12O2⟶Fe2++H2O;ΔG∘3
Applying, ΔG∘1+ΔG∘2=ΔG∘3
ΔG∘3=(−2F×0.44)+(−2F×1.23)
ΔG∘3=−(2×96500×0.44+2×96500×1.23)
ΔG∘3=−322310J
∴ΔG∘3=−322kJ
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:
A
Fe(s)⟶Fe2++2e−;ΔG∘12H++2e−+12O2⟶H2O(I);ΔG∘2Fe(s)+2H++12O2⟶Fe2++H2O;ΔG∘3
Applying, ΔG∘1+ΔG∘2=ΔG∘3
ΔG∘3=(−2F×0.44)+(−2F×1.23)
ΔG∘3=−(2×96500×0.44+2×96500×1.23)
ΔG∘3=−322310J
∴ΔG∘3=−322kJ
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