Redox Reactions And Electrochemistry Question 49
Question: The rusting of iron takes place as follows 2H+ + 2e- + ½O2 $ \xrightarrow{{}} $ H2O(l) ; E° = +1.23 V Fe2+ + 2e- $ \xrightarrow{{}} $ Fe(s) ; E° = -0.44 V Calculate DG° for the net process [IIT 2005]
Options:
A) -322 kJ mol-1
B) -161 kJ mol-1
C) -152 kJ mol-1
D) -76 kJ mol-1
Show Answer
Answer:
Correct Answer: A
Solution:
Fe(s) $ \xrightarrow{{}} $ Fe2+ + 2e- ; $ \Delta G_1^{o} $ 2H+ + 2e- + ½O2 $ \xrightarrow{{}} $ H2O(l) ; $ \Delta G_2^{o} $ Fe(s) + 2H+ + ½O2 $ \xrightarrow{{}} $ Fe2+ + H2O ; $ \Delta G_3^{o} $
Applying, $ \Delta G_1^{o}+\Delta G_2^{o}=\Delta G_3^{o} $
$ \Delta G_3^{o} $ = (-2F ´ 0.44) + (-2F ´ 1.23)
$ \Delta G_3^{o} $ = -(2 ´ 96500 ´ 0.44+ 2 ´ 96500 ´ 1.23)
$ \Delta G_3^{o} $ = -322310 J
$ \Delta G_3^{o} $ = -322 kJ
BETA
