Q. The emf of the cell, $Ag\ |\ {Ag}^{+}\ (0.1\ M)\ ||\ {Ag}^{+}\ (1\ M)\ |\ Ag$ at 298 K is
A) 0.0059 V
B) 0.059 V
C) 5.9 V
D) 0.59 V
Explanation
(b)
Ag / ${Ag}^{+}$ (0.1 M) || ${Ag}^{+}$ (1 M) / Ag
anode cathode
$E_{cell}$ = $\frac{2.303RT}{nF}\log\frac{C_{1}}{C_{2}}$
Here, $n$ = number of electrons in cell reaction = 1
$C_{1}$ = concentration of cathodic electrolyte = 1 M
$C_{2}$ = concentration of anodic electrolyte = 0.1 M
∴ $E_{cell}$ = $\frac{0.059}{1}\log\frac{1}{0.1}$
$E_{cell}$ = 0.059 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.