Redox Reactions And Electrochemistry Question 183
Question: The rusting of iron takes place as follows $ 2{H^{+}}+2{e^{-}}+1/2O_2\to H_2O(l); $ $ E^{o}=+1.23V $ $ F{e^{2+}}+2{e^{-}}\to Fe(s);E^{o}=-0.44V $ Calculate $ \Delta G{}^\circ $ for the net process.
Options:
A) $ -152 $ kJ $ mo{l^{-1}} $
B) $ -161 $ kJ $ mo{l^{-1}} $
C) $ -322 $ kJ $ mo{l^{-1}} $
D) $ -76 $ kJ $ mo{l^{-1}} $
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Answer:
Correct Answer: C
Solution:
[c] $ \Delta G^{o}=-nFE^{o} $
Applying. $ \Delta G_1^{o}+\Delta G_2^{o}=\Delta G_3^{o} $
$ \Delta G_3^{o}=(-2F\times 0.44)+(-2F\times 1.23) $ $ \Delta G_3^{o}=-(2\times 96500\times 0.44+2\times 96500\times 1.23) $ $ \Delta G_3^{o}=-322310J $
$ \therefore \Delta G_3^{o}=-322KJ $
BETA
