Q. The standard reduction potential $E^{{^\circ}}$ for half reactions are Zn = ${Zn}^{2 +}$ + Ze $E^{{^\circ}}$ = + 0.76 V Fe = ${Fe}^{2 +}$ + Ze $E^{{^\circ}}$ = + 0.41 V The emf of the cell reaction $Fe^{2 +}\ + \ Zn\ = \ Zn^{2 +}\ + \ Fe$ is
A) - 0.35 V
B) +0.35
C) +1.17 V
D) -1.`17 V
Explanation
(b)
$E_{cell} = E_{anode(op)}^{{^\circ}} - E_{cathode(op)}^{{^\circ}}$
= 0.76 – 0.41
= +0.35 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.