| Calculate \[\Delta G{}^\circ \] for the reaction: |
| \[C{{u}^{2+}}(aq)+Fe(s)F{{e}^{2+}}(aq)+cu(s)\] |
| Given that: \[E_{C{{u}^{2+}}/Cu}^{o}=+0.34V,\] |
| \[E_{F{{e}^{2+}}/Fe}^{o}=-0.44V\] |
A) \[180.55\operatorname{kJ}\]
B) \[140.35kJ\]
C) \[-130.151\operatorname{kJ}\]
D) \[-150.54\operatorname{kJ}\]
Correct Answer: D
Solution :
| the cell reactions are: |
| At anode: \[\operatorname{Fe}(s)\xrightarrow{{}}{{\operatorname{Fe}}^{2+}}(aq)+2{{e}^{-}}\] |
| At cathode:\[{{\operatorname{Cu}}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{{}}\operatorname{Cu}(s)\] |
| We know that: \[\Delta G{}^\circ =-nFE_{cell}^{o};n=2mol\] |
| \[E_{cell}^{o}=[E_{(C{{u}^{2+}}/Cu)}^{o}-E_{(F{{e}^{2+}}/Fe)}^{o}]\]\[=(+0.34\operatorname{V})-(-0.44\operatorname{V})=+0.78\operatorname{V}\] |
| \[F=96500\,\,C\,mo{{l}^{-1}}\] |
| \[\therefore \Delta G{}^\circ =-nFE_{cell}^{o}\] |
| \[=-(2\,mol)\times (96500C\,mo{{l}^{-1}})\times (+0.78V)\] |
| \[=-150540CV=-150540J(\because \,1CV=1J)\] |
| \[=-150.54\operatorname{kJ}\] |
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