What quantity of current is required to deposit \(2.4 g\) of copper in a period of 750 seconds during an electrolytic deposition process? \(\left[ Cu =64,1 F =96500 coulmol ^{-1}\right] \)
A. \(9.65\) A B. \(10.81\) A C. \(12.33\) A D. \(15.54 A\)
Correct Answer: A
Explanation
\begin{aligned} &1 Cu ^{2+}=2 F \\ &64 g =193000 C \end{aligned} Therefore, \(2.4 g\) of copper will be produced by \begin{aligned} &\left(\frac{193000}{64} \times 2.4\right) \text { C of electricity } \\ &Q=I t, I=\frac{Q}{t}=\left(\frac{7237.5}{750}\right) A \\ &=9.65 A \end{aligned}
