Question

In an electrolysis experiment, current was passed for $5$ hours through two cells connected in series. The first cell contains a solution of gold at the oxidation state of $+3$ and second contain $CuS{O}_4$ solution. During electrolysis in the first cell $9.85gm$ of gold is deposited, determine the average gram of copper deposited per ampere of electricity passed in the nearest possible integers.
$Cu:63.5Au:197$

Answer 1

Answer: 6

Amount of $Au$ deposited $=\frac{9.85}{197}=0.05mol$
$\therefore$ Amount of electrons passed $=0.05\times 3$
$=0.15mol$
$\therefore$ Quantity of electricity passed $=0.15F$
$=96500\times 0.15C$
$\therefore$ Quantity of current passed
$=\frac{96500\times 0.15}{5\times 60\times 60}$
$=0.8042A$
Mass of $Cu$ deposited $=\frac{63.5}{2}\times \frac{0.15F}{F}$
$=4.7625g$
$\therefore \frac{\text{ mass of }Cu\text{ deposited }}{\text{ quantity of electricity passed }}=\frac{4.7625}{0.8042}$
$=5.92\simeq 6$