Question

In an electrolysis experiment, a current was passed for 5 hours through 2 cells connected in series. The first cell contains the solution of gold salt and the second cell contains copper sulphate solution. 9.85g of gold was deposited in the first cell. If the oxidation number of golds is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of current in ampere.

Answer 1

Hint: Faraday's laws of electrolysis states that amount of element deposited on electrode is directly proportional to the amount of current passed and secondly if same amount of charge is passed through different electrolytes then quantity of charge deposited at different electrodes is directly proportional to the ratio of their equivalent masses.


Complete answer:

Faradays gave two laws:
Faraday's first law of electricity states that the amount of electrolyte deposited is equal to the amount of current passed through it.
Faraday's second law of electrolysis states that when the same quantity of charge is passed through different electrolytes the masses of different substances deposited at respective electrodes will be in ratio of their equivalent masses.
So, as per second law of faraday

\[\dfrac{Mass~of~Cu~deposited}{Mass~of~Au~deposited}=\dfrac{Eq.~mass~of~Cu}{Eq.~mass~of~Au}\]
Equivalent mass is equal to molar mass divided by valency factor.
For acids, the valency factor is equal to their basicity.
And for bases, the valency factor is equal to their acidity.
Valency factor of gold is 3
Therefore,
\[Eq.~mass~of~Au=\dfrac{197}{3}\]
Valency factor of copper is 2
Therefore,
 \[Eq.\text{ }mass\text{ }of~Cu=\dfrac{63.5}{2}\]

\[Mass\text{ }of\text{ }copper\text{ }deposited=9.85\times \dfrac{63.5}{2}\times \dfrac{3}{197}g=4.7625~g\]
Mass of copper deposited on electrodes is 4.7625g.
Now as per the first law of faraday, the amount of copper deposited on electrodes is directly proportional to the amount of current passed through it.
Let Z be the equivalent of Cu.
\[\begin{align}
  & E=Z\times 96500 \\
 & Z=\dfrac{96500}{E} \\
 & Z=\dfrac{96500\times 2}{63.5} \\
\end{align}\]
Now,
 \[W=Z\times I\times t\]
A current is passed through the cell for 5 hours therefore time is equal to 5 hours.
$\begin{align}
  & I=\dfrac{W}{Z\times t} \\
 & I=\dfrac{4.6725\times 2\times 96500}{63.5\times 5\times 3600} \\
\end{align}$
I=0.804 ampere

Note: Faraday gave two laws of electrolysis:
First law of electrolysis states that amount of element deposited on electrode is directly proportional to the amount of current passed
 \[W=Z\times I\times t\]
Second law of electrolysis states that if the same amount of charge is passed through different electrolytes then quantity of charge deposited at different electrodes is directly proportional to the ratio of their equivalent masses.
\[\dfrac{Mass~of~firstelectrolyte~deposited}{Mass~of~\sec ondelectrolyte~deposited}=\dfrac{Eq.~mass~of~firstelectrolyte}{\begin{align}
  & Eq.~mass~of~\sec ondelectrolyte \\
 & \\
\end{align}}\]