Q. Consider the following $E^{{^\circ}}$ values ${E{^\circ}}_{{Fe}^{3 +}/\ {Fe}^{2 +}} = \ + \ 0.77\ V$ $\ {E{^\circ}}_{{Sn}^{2 +}/\ Sn}\ = \ - \ 0.14\ V\$ Under standard conditions the potential for the reaction $Sn\ (s)\ + \ {2Fe}^{3 +}(aq)\ \rightarrow \ {2Fe}^{2 +}(aq)\ + \ {Sn}^{2 +}(aq)\ is$
A) 1.68 V
B) 1.40 V
C) 0.91 V
D) 0.63 V
Explanation
(c)
$Sn\ (aq)\ + \ {2Fe}^{3 +}(aq)\ \rightarrow \ {2Fe}^{2 +}(aq)\ + \ {Sn}^{2 +}(aq)$
$E_{cell}^{{^\circ}}\ = \ E_{oxi}^{{^\circ}}\ + \ E_{red}^{{^\circ}}\$
$\ \ \ \ \ \ \ \ \ \ = \ E_{{Sn\ /Sn}^{2 +}}^{{^\circ}}\ + \ E_{{Fe}^{3}/\ {Fe}^{2 +}}^{{^\circ}}$
Given , $\ {E{^\circ}}_{{Sn}^{2 +}/\ Sn}\ \ = \ - \ 0.14\ V\$
∴ $E_{{Sn\ /Sn}^{2 +}}^{{^\circ}}\ = \ + \ 0.14\ V$
$E_{{Fe}^{3 +}/\ {Fe}^{2 +}}^{{^\circ}}\ = \ \ 0.77\ V$
$E_{cell}^{{^\circ}}$ = 0.14 + 0.77 = 0.91 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.