A) \[-\,\,0.057\,\,V\]
B) \[+\,\,0.057\,\,V\]
C) \[+\,\,0.30\,\,V\]
D) \[-\,\,0.30\,\,V\]
Correct Answer: A
Solution :
[a] \[\Delta G{}^\circ =-nFE{}^\circ \] (i) \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe;\]\[E{}^\circ \,\,=-0.47V;\] (ii) \[\underline{F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}};}\]\[E{}^\circ =+0.77\,\,V;\] (iii) \[F{{e}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Fe\] (i) \[\Delta G{}^\circ =-nF{{E}^{0}}=-2\,\,(-0.47)F=0.94\,F\]) (ii) \[\underline{\Delta {{G}^{0}}=-nF{{E}^{0}}=-1\,(+0.77)F=-0.77\,\,F}\] (iii) On adding: \[\Delta {{G}^{0}}=+\,0.17\,F\] \[\Delta {{G}^{0}}=-nF{{E}^{0}}\] \[{{E}^{0}}\,\text{for}\,\,(F{{e}^{3+}}\xrightarrow{{}}Fe)\] \[=\frac{\Delta {{G}^{0}}}{-nF}=\frac{0.17F}{-3F}=-0.057\,\,V\]You need to login to perform this action.
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