Calculate the quantity of electricity required to deposit \(64.0 \mathrm{~g}\) of copper from copper (II) chloride \((\mathrm{Cu}=64=96500 \mathrm{C}) \)
A. \(3.09 \times 10^6 \mathrm{C} \) B. \(1.93 \times 10^5 \mathrm{C}\) C. \(9.65 \times 10^4 \mathrm{C} \) D. \(6.18 \times 10^6 \mathrm{C}\)
Correct Answer: B
Explanation
\(\mathrm{Cu}^{2+}+\mathrm{e}^{-} \rightarrow \mathrm{Cu}\) Since 2 moles of electrons are involved in the deposition of 1 mole of \(\mathrm{Cu}\), the deposition of 1 mole of \(\mathrm{Cu}(64 \mathrm{~g})\) requires \(2 \mathrm{~F}(2 \times 96500 \mathrm{C})\) of electricity \(\therefore \mathrm{Q}=2 \times 96500 \mathrm{C}=193000 \mathrm{C}=1.93 \times 10^5 \mathrm{C}\)
