Q. The Edison storage cell is represented as : $Fe(s) + \ FeO(s)\left| \ KOH(aq) \right|{\ Ni}_{2}O_{3}(s)\left| \ {Ni}_{2}O_{3}(s) \right|\ Ni(s)$ The half reactions are ${Ni}_{2}O_{3}(s) + H_{2}O(l) + 2e^{-} \rightarrow 2NiO(s) + 2{OH}^{-};\ \ E{^\circ} = + 0.40\ V$ $FeO(s) + H_{2}O(l) + 2e^{-} \rightarrow Fe(s) + 2{OH}^{-};\ \ E{^\circ} = - 0.87\ V$ Choose the incorrect statement
A) $E_{anode}$ increases with increase in concentration of ${OH}^{-}$
B) $E_{cathode}$ decreases with increase in concentration of ${OH}^{-}$
C) $E_{cell}^{{^\circ}} = 1.27\ V$
D) $E_{cell}$ increases with increase in concentration of FeO
Explanation
(d)
$E_{cell}^{{^\circ}} = 0.87 + 0.40 = 1.27\ V$
Cell reaction $\ :Fe + {Ni}_{2}O_{3} \rightarrow FeO + 2NiO$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.