A) \[-\,0.476\]volt
B) \[-\,0.404\]volt
C) 0.440 volt
D) 0.772 volt
Correct Answer: D
Solution :
[d] \[F{{e}^{3+}}3e\to Fe,\]\[{{E}^{o}}_{1}=-0.036\,volt;\Delta {{G}^{o}}_{1}\] \[Fe\to F{{e}^{2+}}+2{{e}^{-}},\] \[{{E}^{o}}_{2}=0.44\,volt;\Delta {{G}^{o}}_{2}\] \[F{{e}^{3+}}+{{e}^{-}}\to \Delta {{G}^{o}}=\Delta {{G}^{o}}_{1}+\Delta {{G}^{o}}_{2}\] \[-nF{{E}^{0}}=-F{{E}^{0}}_{1}+(-2F{{E}^{0}}_{2})\] \[\Rightarrow -F{{E}^{0}}=-F(3{{E}^{0}}_{1}+2{{E}^{0}}_{2})\] \[{{E}^{o}}=(3\times (0.036)+(2\times 0.44)\] \[{{E}^{o}}=0.772volt\]
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