Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 is cell reaction?

$\begin{array}{1 1}(a)\;Ni_2O_3+H_2O\rightarrow 2NiO+2OH^-\\(b)\;Fe+Ni_2O_3\rightarrow FeO+2NiO\\(c)\;FeO+H_2O+2e^-\rightarrow Fe+2OH^-\\(d)\;None\end{array}$

Can you answer this question?

1 Answer

0 votes
Given cell is $Fe(s)/Ni_2O_3(s)/FeO/Ni_2O_3(s)/Ni(s)$
Cell reaction
LHS:$Fe^++2OH^-\rightarrow FeO+H_2O+2e^-$
RHS:$Ni_2O_3+H_2O+2e^-\rightarrow 2NiO+2OH^-$
$Fe+Ni_2O_3\rightarrow FeO+2NiO$
Hence (b) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App