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 $CuSO _{4}$ solution. During electrolysis in the first cell $9.85\, gm$ of gold is deposited, determine the average gram of copper deposited per ampere of electricity passed in the nearest possible integers.
$Cu: 63.5 Au: 197$

Answer 1

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