Q. If in a voltaic cell, 5 g of zinc is consumed, we will get how many ampere hour (given that ECE of zinc is $3.38 \times 10^{- 7}kgC^{- 1}$)
A) 2.05
B) 8.2
C) 4.1
D) $5 \times 3.338$ $\times 10^{- 7}$
Explanation
(c) $It = \frac{m}{z} = \frac{5 \times 10^{- 3}}{3.387 \times 10^{- 7}}$
$= \frac{5 \times 10^{4}}{3.387 \times 60 \times 60}$ Ah = 4.1 Ah
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.