Question

If in a voltaic cell, $5\, g$ of zinc is consumed, we will get how many ampere hour? $(ECE$ of zinc is $3.387 \times 10^{-7} kg\, C^{-1})$

• A. $2.05$
• B. $8.2$
• C. $4.1$
• D. $5.2$

Answer 1

Answer: (C)

$4.1$

Answer 2

According to Faraday's first law of electrolysis $m = ZIt$ $It=\frac{m}{Z}=\frac{5 \times 10^{-3}kg}{3.387 \times 10^{-7}kg\,C^{-1}}$ $=\frac{5\times10^{4}}{3.387}A\,s$ $=\frac{5\times10^{4}}{3.387\times60\times60}A\,h$ $=4.1\,A\,h$