Q. Agalvanic cell with electrode potential of ‘A’ = +2.23 V and ‘B’ = -1.43 V. The value of $E_{cell}^{{^\circ}}$ is
A) 3.66 V
B) 0.80 V
C) -0.80 V
D) -3.66 V
Explanation
(a)
As $‘A’$ has more $E_{red}^{{^\circ}}$ value than $B,\ A$ will act as cathode in the galvanic cell.
Hence,
$E_{cell}^{{^\circ}}$ = $E_{cathode}^{{^\circ}}$ - $E_{anode}^{{^\circ}}$
= (2.23) – (-1.43)
= 2.23 + 1.43 = 3.66 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.