The rusting of iron takes place as: 2H++2e−+12O2→H2O(l);Eo=+1.23V Fe2++2e−→Fe(s);Eo=−0.44V Thus, ΔGo for the net process is:
-322 kJ/mol
-161 kJ/mol
-1522 kJ/mol
-76 kJ/mol
A
-1522 kJ/mol
B
-322 kJ/mol
C
-161 kJ/mol
D
-76 kJ/mol
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Solution
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2H++2e+12O2→H2O(l);Eo=+1.23V Fe2++2e→Fe(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−+12O2→H2O(l);Eo=+1.23V Fe2++2e−→Fe(s);Eo=−0.44V Thus, ΔGo for the net process is:
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Q2
The rusting of iron takes place as follows : 2H++2e+12O2→H2O(l);Eo=+1.23V
Fe2++2e→Fe(s);Eo=−0.44V
ΔGo for the net process is:
View Solution
Q3
The half cell reactions for rusting of iron are :
2H++12O2+2e−→H2O;Eo=1.23V Fe2++2e−→Fe;Eo=−0.44V
ΔGo (in kJ) for the complete cell reaction is :
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Q4
The half cell reactions for the corrosion are 2H++1/2O2+2e−→H2O;Eo=1.23V Fe2++2e−→Fe(s);Eo=−0.44V. Find teh ΔGo (in kJ) for the overall reaction:
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Q5
The rusting of iron takes place as follows: 2H⊕+2e−+12O2→H2O(l);E⊖=+1.23V Fe2+(aq)+2e−→Fe(s);E⊖=−0.44V Calculate ΔG⊖ for the net process.