Q. The standard reduction potential, $E{^\circ}$ for the half-reactions are as $Zn \rightleftharpoons {Zn}^{2 +} + 2e^{-},\ E{^\circ} = \oplus 0.76\ V$ $Fe\ \rightleftharpoons {Fe}^{2 +} + 2e^{-},\ E{^\circ} = + 0.41\ V$ The $E{^\circ}_{cell}$ for the cell formed by these two electrodes is
A) $- 0.35\ V$
B) $- 1.17\ V$
C) +0.35 V
D) +1.17 V
Explanation
(c)
In this reaction,
[image:3_image2.emf]
$EMF = E_{cathod} - E_{anode}$
$= - 0.41 - ( - 0.76)$
$= + \ 0.35\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.