A) 1.653V
B) 0.111 V
C) 0.330V
D) 1.212V
Correct Answer: D
Solution :
\[Fe\xrightarrow{{}}F{{e}^{2+}}+2{{e}^{-}}\] \[\Delta G_{1}^{o}=-2Fx.441\] \[2F{{e}^{3+}}+2{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}}\]\[\Delta G_{2}^{o}=-2Fx.771\] \[Fe+2F{{e}^{3+}}\xrightarrow{{}}3F{{e}^{2+}}\]\[\Delta G_{3}^{o}=-2FE_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\]\[=-2Fx.441-2Fx.771\] \[E_{3}^{o}=1.212\,V\]You need to login to perform this action.
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