Q. $Ni\ /\ {Ni}^{2 +}\ \lbrack 1.0\ M\rbrack\ ||\ {Au}^{3 +}\ \lbrack 1.0\ M\rbrack\ /\ Au$ where $E^{{^\circ}}$ $for\ {Ni}^{2 +}\ /Ni\ is\ - 0.250\ V$; and $E^{{^\circ}}$ for ${Au}^{3 +}\ /\ Au$ is 0.150 V. The emf of the cell is
A) +1.25 V
B) -1.75 V
C) +1.75 V
D) +0.4 V
Explanation
(d)
$E_{cell}^{{^\circ}} = \ E_{cathode}^{{^\circ}}\ - \ E_{anode}^{{^\circ}}$
$Ni\ /\ {Ni}^{2 +}\ \lbrack 1.0\ M\rbrack\ ||\ {Au}^{3 +}\ \lbrack 1.0\ M\rbrack\ |\ Au$
$E_{cell\ }({Au}^{3 +}\ /\ Au)\ = \ 0.150\ V$
$E_{cell\ }({Ni}^{2 +}\ /\ Ni)\ = \ - \ \ 0.25\ \ V$
$E_{cell}^{{^\circ}}\ = \ E_{cathode}^{{^\circ}}\ - \ E_{anode}^{{^\circ}}$
= 0.150 – (- 0.25)
= 0.15 + 0.25
= + 0.4 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.