Question

In an electrolysis experiment, a current was passed for $5$ hours through two cells connected in series. The first cell contains a 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 gold is $+3$, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of the current in ampere.

Answer 1

We know that,
$\frac{\text{Mass of Au deposited}}{\text{Mass of Cu deposite}}=\frac{\text{Eq. mass of Au}}{\text{Eq. mass of Cu}}$
Eq. mass of $Au=\frac{197}{3}$; Eq. mass of $Cu=\frac{63.5}{2}$
Mass of copper deposited
$=9.85\times \frac{63.5}{2}\times \frac{3}{197}g=4.7625g$
Let $Z$ be the electrochemical equivalent of $Cu$.
$E=Z\times 96500$
or $Z=\frac{E}{96500}=\frac{63.5}{2\times 96500}$
Applying $W=Z\times I\times t$
$t=5hour=5\times 3600second$
$4.7625=\frac{63.5}{2\times 96500}\times I\times 5\times 3600$
or $I=\frac{4.7625\times 2\times 96500}{63.5\times 5\times 3600}=0.804ampere$.