A) + 18.51 kJ \[mo{{l}^{-1}}\]
B) +11.87 kJ \[mo{{l}^{-1}}\]
C) - 8.10 kJ \[mo{{l}^{-1}}\]
D) - 10.41 kJ \[mo{{l}^{-1}}\]
Correct Answer: B
Solution :
\[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe\,;\,{{n}_{1}}=2,\,E_{1}^{o}=-0.447\,\,V\] \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}}\,;\,{{n}_{2}}=1,\,E_{2}^{o}=0.771\,\,V\] \[F{{e}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Fe\,\,;\,\,{{n}_{3}}=3,\,E_{3}^{o}=?\] \[\Delta {{G}_{3}}=\Delta {{G}_{1}}+\Delta {{G}_{2}}\] \[3E_{3}^{o}=-2E_{1}^{o}+E_{2}^{o}\] \[E_{3}^{o}=\frac{(-0.477\times 2)+0.771}{3}=-0.041\,\,V\] \[\Delta {{G}^{o}}=-nF{{E}^{o}}\] \[\Delta {{G}^{o}}=-3\times 96485\times (-0.041\,V)\] \[\Delta {{G}^{o}}=+11867.65\,\,mo{{l}^{-1}}\simeq +11.87\,\,kJ\,mo{{l}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
