If the cost of electricity required to deposit \(1 g\) of aluminum is \(\# 4.00\), how much would it cost to deposit \(24 g\) of copper' \([ A ]=27, Cu =64]\)
A. #27.02 B. #37.02 C. #47.02 D. #57.02
Correct Answer: A
Explanation
No of mole of \(\frac{1 g }{27 g / mol }\). \(=0.037\) motl. deposited The discharge of \(Al l^{3+}\) is: \(Al _{(\text {oq })}^{3+}+3 e ^{-} \rightarrow Al _{(s)}\) 1 mole of \(Al\) is deposited by \(3 F\) \(0.037\) mole of \(A /\) would be deposited by. \((0.037 \times 3) F =0.111 F\)lso The no of mole of Cato beideposited \(=\frac{24 g }{64 g / mol }=0.375 mol\) The discharge of \(Cu ^{2+}\) is: \(Cu _{(a+)}^{2+}+2 e ^{-} \rightarrow Cu _{(s)}\) 1 mole of \(Cu\) is deposited by \(2 F\) \(0.375 mole\) of \(Cu\) would be deposited by \((0.375 \times 2) F =0.75 F\)ut the cost of \(0.111 F\) of electricity is \(\# 4.00\) \(0.75 F\) would cost \(\left(\frac{0.75 F \times \# 4: 00}{0.111 F }\right)\) \(=\# 27.02\)
