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Question

The rusting of iron takes place as:
2H++2e+12O2H2O(l);Eo=+1.23V
Fe2++2eFe(s);Eo=0.44V
Thus, ΔGo for the net process is:
  1. -322 kJ/mol
  2. -161 kJ/mol
  3. -1522 kJ/mol
  4. -76 kJ/mol

A
-1522 kJ/mol
B
-322 kJ/mol
C
-161 kJ/mol
D
-76 kJ/mol
Solution
Verified by Toppr

2H++2e+12O2H2O(l);Eo=+1.23V
Fe2++2eFe(s);Eo=0.44V
For net cell reaction,
Eo=EoOPFe+EoRPH2O=0.44+1.23=1.67V
As we know,
ΔG=nFEo
ΔGo=n×Eo×F=2×1.67×96500
=322.31kJ/mole

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Similar Questions
Q1
The rusting of iron takes place as:
2H++2e+12O2H2O(l);Eo=+1.23V
Fe2++2eFe(s);Eo=0.44V
Thus, ΔGo for the net process is:
View Solution
Q2

The rusting of iron takes place as follows : 2H++2e+12O2H2O(l);Eo=+1.23V

Fe2++2eFe(s);Eo=0.44V

ΔGo for the net process is:
View Solution
Q3
The half cell reactions for rusting of iron are :

2H++12O2+2eH2O ; Eo=1.23 V
Fe2++2eFe ; Eo=0.44 V

ΔGo (in kJ) for the complete cell reaction is :

View Solution
Q4
The half cell reactions for the corrosion are
2H++1/2O2+2eH2O;Eo=1.23V
Fe2++2eFe(s);Eo=0.44V. Find teh ΔGo (in kJ) for the overall reaction:
View Solution
Q5
The rusting of iron takes place as follows:
2H+2e+12O2H2O(l);E=+1.23V
Fe2+(aq)+2eFe(s);E=0.44V
Calculate ΔG for the net process.
View Solution