Question

The standard redox potentials for the reactions
$ M{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mn $ and $ M{{n}^{3+}}+{{e}^{-}}\xrightarrow{{}}M{{n}^{2+}} $
are $ -1.18\text{ }V $ and $ 1.51\text{ }V $ respectively. What is the redox potential for the reaction $ M{{n}^{3+}}+3{{e}^{-}}\to Mn $

• A. $0.33\,V$
• B. $1.69\,V$
• C. $ -\text{ }0.28\text{ }V $
• D. $ -\text{ }0.85\text{ }V $
• E. $0.85\, V$

Answer 1

Answer: (C)

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 $