Q. The $E^{{^\circ}}$of ${Fe}^{2 +}\ /\ Fe$ and ${Sn}^{2 +}\ /Sn$ are -0.44 V and -0.14 V respectively. If cell reaction is $Fe\ + \ {Sn}^{2 +}\ \rightarrow \ {Fe}^{2 +}\ + \ Sn$ then emf of the cell is
A) +0.30 V
B) -0.58 V
C) +0.58 V
D) -0.30 V
Explanation
(a)
Here Fe acts as anode while Sn act as cathode. We know that,
$E_{cell}^{{^\circ}}$ = $E_{cathode}^{{^\circ}}$ - $E_{anode}^{{^\circ}}$
= (-0.14) – (-0.44)
= - 0.14 + 0.44
= 0.30 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.