Q. The cathodic reaction of a dry cell is represented by ${2MnO}_{2}(s) + {Zn}^{2} + {2e}^{-} \longrightarrow {ZnMn}_{2}O_{4}(s)$ If , there are 8 g of ${MnO}_{2}$ in the cathodic compartment then the time for which the dry cell will continue to give a current of 2 mA is
A) 25.675 day
B) 51.35 day
C) 12.8 day
D) 6.423 day
Explanation
(b)
$2MnO_{2}\ (s)\ + \ {Zn}^{2 +} + {2e}^{-} \longrightarrow {ZnMn}_{2}\ O_{4}\ (s)$
$2\ \times \ 87\ g\ MnO_{2}\ required\ 2\ \times \ 96500\ F\ charge$
$8\ g\ MnO_{2}\ = \ \frac{2\ \times \ 96500\ \times 8\ }{2\ \times 87}$
= 8873.56 F
$\mathcal{Q\ =}it$
$8873.56\ = \ 2\ \times \ 10^{- 3}\ \times t$
$t\ = \frac{8873.56\ \times \ 10^{3}}{2}\$
= 4436781.5 s
$= \ \frac{4436781.5}{24\ \times 60\ \times 60}\$
= 51.35 days
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.