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Given below are the half - cell reactions\begin{array}{1 1}Mn^{2+}+2e^-\rightarrow Mn;E^{\large\circ}=-1.18V\\2(MN^{3+}+e^-\rightarrow Mn^{2+});E^{\large\circ}=-1.18V\end{array}The $E^{\large\circ}$ for $3Mn^{2+}\rightarrow Mn+2Mn^{3+}$ will be

$\begin{array}{1 1}(A)\;-0.33V;\text{the reaction will not occur}\\(B)\;-0.33V;\text{the reaction will occur}\\(C)\;-2.69V;\text{the reaction will not occur}\\(D)\;-2.69V;\text{the reaction will occur}\end{array} $

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1 Answer

The $E^{\large\circ}$ for $3Mn^{2+}\rightarrow Mn+2Mn^{3+}$ will be $-2.69V$ the reaction will not occur
Hence (C) is the correct answer.
answered Apr 7, 2014 by sreemathi.v
 

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