Q. Standard electrode potential of cell $\ H_{2}\left| H^{+} \right||{Ag}^{+}|Ag$ is $(Given,\ E{^\circ}_{{Ag}^{+}/Ag} = 0.80\ V)\$
A) 0.4 V
B) 0.8 V
C) 1.4 V
D) 1.8 V
Explanation
(b)
$\frac{1}{2}H_{2}|\ H^{+}\ ||\ {Ag}^{+}\ |\ Ag$
$E_{cell}^{{^\circ}} = E_{cathode}^{{^\circ}} - E_{anode}^{{^\circ}}$
$= E_{{Ag}^{+}/Ag}^{{^\circ}} - E_{H^{+}/\frac{1}{2}H_{2}}^{{^\circ}}$
$= (0.80) - (0.0) = 0.80\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.