Question

Calculate the time to deposit $1.27g$ of copper at, cathode when a current of $2A$ was passed through the solution of $CuS{O}_4$. (Molar mass of $Cu=63.5gmo{l}^{-1},1F=96500Cmo{l}^{-1}$)

Answer 1

${Cu}^{2+}+{2e}^{-}\to Cu$

moles of $Cu$ deposited $=\frac{1.27}{63.5}=0.02mol$

2 mol Faraday gives $63.5g$ of $Cu$

$\therefore e^{-}$ required for deposition $=0.02\times 2=0.04$

$0.04=\frac{charge}{96500}=\frac{2\times t}{96500}$

$t=19306$