Q. Calculate the emf of the cell $Cu\ (s)\ |\ {Cu}^{2 +}\ (aq)\ ||\ {Ag}^{+}\ (aq)\ |\ Ag\ (s)$ Given, $\ E_{{Cu}^{2 +}/cu}^{{^\circ}}\ \ = \ + \ 0.34\ V\ ,\ E_{{Ag}^{+}/Ag}^{{^\circ}}\ \ = \ \ 0.80\ V\ ,$
A) +0.46 V
B) +1.14 V
C) +0.57 V
D) -0.46 V
Explanation
(a)
$E_{cell}^{{^\circ}}\ = \ E_{red\ (cathode)}^{{^\circ}}\ - \ E_{oxi\ (anode)}^{{^\circ}}$
$= E_{{Ag}^{+}/Ag}^{{^\circ}}\ - \ E_{Cu + \ /{Cu}^{2 +}}^{{^\circ}}$
= 0.80 – (+ 0.34) = + 0.46 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.