A) 0.33V
B) 1.69V
C) \[-\text{ }0.28\text{ }V\]
D) \[-\text{ }0.85\text{ }V\]
E) 0.85 V
Correct Answer: C
Solution :
Given: (i) \[M{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mn;E{}^\circ =-1.18V;\] \[\Delta G_{1}^{o}=-2\times F\times -1.18J\] (ii) \[M{{n}^{3+}}+{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}};{{E}^{o}}=1.51\,V;\] \[\Delta G_{2}^{o}=-1\times F\times 1.51\,J\] Aim: \[M{{n}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Mn;\,\,\,\,\,\,\Delta G_{3}^{o}=?\] (i) + (ii) gives the required results, i.e., \[\Delta G_{3}^{o}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\] \[=2.36\text{ }F+(-1.51\text{ }F)=0.85\text{ }F\] \[\therefore \] \[-nFE_{M{{n}^{3+}}/Mn}^{o}=0.85F\] Or \[E_{M{{n}^{3+}}/Mn}^{o}=-\frac{0.85}{3}V\] \[(\because n=3)\] \[=-0.28V\]You need to login to perform this action.
You will be redirected in
3 sec