A) \[\frac{2{{y}_{1}}-{{y}_{2}}}{3}\]
B) \[\frac{{{y}_{2}}-2{{y}_{1}}}{3}\]
C) \[2{{y}_{1}}-3{{y}_{2}}\]
D) \[{{y}_{2}}-2{{y}_{1}}\]
Correct Answer: B
Solution :
(i) \[{{A}^{3+}}+{{e}^{-}}\xrightarrow{{}}{{A}^{2+}},\,\,\Delta {{G}_{1}}=-1\,F\,{{y}_{2}}\] (ii) \[{{\operatorname{A}}^{2+}}+2{{e}^{-}}\,\xrightarrow{{}}\,\,A,\,\,\Delta {{G}_{2}}=\,\,-2F(-{{y}_{1}})=\,\,2\,F{{y}_{1}}\] Add, (i) and (ii) we get \[{{A}^{3+}}+3{{e}^{-}}\xrightarrow{{}}\,\,A;\] \[{{\operatorname{AG}}_{3}}=\,\,A{{G}_{1}}\,\,+\,\,A{{G}_{2}}\] \[-\,3FE{}^\circ =-F{{y}_{2}}+2F{{y}_{1}}\] \[-\,3FE{}^\circ =-F\left( {{y}_{2}}-2{{y}_{1}} \right)\] \[E{}^\circ =\frac{{{y}_{2}}-2{{y}_{1}}}{3}\]You need to login to perform this action.
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