A) 0.33 V ; the reaction will not occur
B) −0.33 V ; the reaction will occur
C) −2.69 V ; the reaction will not occur
Correct Answer: C
Solution :
\[M{{n}^{+2}}+2{{e}^{-}}\to Mn\]\[E_{1}^{0}=-1.18V,\Delta G_{1}^{0}=-2FE_{1}^{0}\] \[\frac{2\left( M{{n}^{+3}}+{{e}^{-}}\to M{{n}^{+2}} \right)E_{2}^{0}=1.51V,}{3M{{n}^{+2}}\to Mn+2M{{n}^{+3}}E_{3}^{0}=?},\begin{matrix} \Delta G_{2}^{0}=-2FE_{2}^{0} \\ \Delta G_{3}^{0}=-2FE_{3}^{0} \\ \end{matrix}\]\[\Delta G_{3}^{0}=\Delta G_{1}^{0}-\Delta G_{2}^{0}\] \[E_{3}^{0}=E_{1}^{0}-E_{2}^{2}=-1.18-1.51=-2.69V\] −2.69 V ; the reaction will occurYou need to login to perform this action.
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