During electrolysis, two cells each containing molten \(Al _{2} O _{3}\) and fuse \(CaCl _{2}\) were connected in series. A current of \(15 Amp\) was passed through the cells for a given period of time. At the end of the electrolysis \(9 g\) of calcium was found to have been deposited at the cathode. What mass of aluminum would be deposited in the second cell. [A = \(27, Ca =40]\)
A. \(8.82 g\) B. \(4.44 g\) C. \(17.60 g\) D. \(4.05 g\)
Correct Answer: D
Explanation
From Faraday's second law of electrolysis, \(\eta \alpha \frac{1}{c}\) \(n=\) no of mole, \(c=\) charge on the ion \(\Rightarrow n_{1} c_{1}=n_{2} c_{2}\) \(\eta=\frac{\text { Mass }}{\text { Molar Mass }}\) \(\Rightarrow \frac{M_{A 1} \times \text { Charge on } Al ^{3+}}{\text { Molar Mass of } Al }\) \(=\frac{M_{C a} \times \text { Charge on } Ca ^{3+}}{M M_{C a}}\) \(M_{A l}=\frac{9 \times 2 \times 27}{40 \times 3}=4.05 g\)
