Question

Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of $CuS{O}_4$.

(Molar mass of $Cu=63.5$ g mol${}^{-1}$, $1$ F $=96500$ C mol${}^{-1}$)

Answer 1

From faraday's law of electrolysis, we get

$W=Zit=\frac{E}{F}it$

Where, $W=$ Deposited amount

$E=$ Equivalent weight

$F=$Faraday

$i=$Electricity passed

$t=$ Time

$\Rightarrow 1.27=\frac{63.5}{96500}\times 2\times t$

$\Rightarrow t=965sec$