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The Edison storage cell is represented as $Fe(s)/FeO(s)/KOH(aq)/Ni_2O_3(s)/Ni(s)$.The half cell reactions are $Ni_2O_3(s)+H_2O(l)+2e^-\rightarrow 2NiO(s)+2OH^-(aq)\Rightarrow E^0=0.4V$,$FeO(s)+H_2O(l)+2e^-\rightarrow Fe(s)+2OH^-(aq)\Rightarrow E^0=-0.87.V$.What will be the maximum amount of electrical energy that can be obtained from one mole of $Ni_2O_3$

$\begin{array}{1 1}(a)\;2.45\times 10^2KJmole^{-1}\\(b)\;2.45\times 10KJmole^{-1}\\(c)\;2.45\times 10^3KJmole^{-1}\\(d)\;None\end{array}$

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1 Answer

Maximum amount of electrical energy $\Delta G^0=-nFE^0_{cell}$
$\Rightarrow -2\times 96500\times 1.27$
$\Rightarrow -2.45\times 10^5Jmole$
$\Rightarrow -2.45\times 10^2KJmole^{-1}$
Hence (a) is the correct answer.
answered Feb 18, 2014 by sreemathi.v
 

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