logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...