Explanation:
The given data is as follows.
m = 19.2 g, I = 15 A
The given reaction equation will be as follows.
[tex]CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}[/tex]
[tex]Ca(s) \rightarrow Ca^{2+} + 2e^{-}[/tex]
As, 1 mole will give [tex]2 \times 96500 C[/tex] of charge. So, calculate the amount of charge deposited by 19.2 g as follows.
[tex]2 \times 96500 C \times \frac{19.6 g}{40.07 g/mol}[/tex]
= 94404.791 C
Hence, 94404.791 C are needed to produce 19.2 g of Ca metal.
Therefore for 15 A, minutes required to produce 94404.791 C and 19.2 g of Ca(s) will be calculated as follows.
= [tex]94404.791 C \times \frac{1 Amp sec}{1 C} \times \frac{1}{15 Amp} \times \frac{1 min}{60 sec}[/tex]
= 104.9 mins
Thus, we can conclude that 104.9 minutes are required if a cell runs at 15 A, how many minutes will it take to produce 19.2 g of Ca(s).
