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 and the magnitude of the current in ampere respectively.
Given:
$\text{Molar mass of gold}=197g/mol$
$\text{Molar mass of copper}=63.5g/mol$

• A. $4.67g\text{and}0.804A$
• B. $2.65g\text{and}2A$
• C. $70.5g\text{and}0.35A$
• D. $70.5g\text{and}0.26A$

Answer 1

The correct option is A $4.67g\text{and}0.804A$

Faradays second law of electrolysis:
If equal amount of charge (Q) is passed through two different solutions, the amount of substance deposited/liberated (W) is proportional to their chemical equivalent weights (E).

Thus,
$\frac{W_1}{W_2}=\frac{E_1}{E_2}=\frac{Z_1}{Z_2}$
$\therefore$
$\frac{\text{Mass of Au deposited}}{\text{Mass of Cu deposited}}=\frac{\text{Equivalent mass of Au}}{\text{Equivalent mass of Cu}}$
$\text{Equivalent mass of Au}=\frac{197}{3}$;
$\text{Equivalent mass of Cu}=\frac{63.5}{2}$
$\text{Mass of copper deposited}=9.85\times \frac{63.5}{2}\times \frac{3}{197}=4.76g$
Let 'Z' be the electrochemical equivalent of Cu.
$E=Z\times 96500$
$Z=\frac{E}{96500}=\frac{63.5}{2\times 96500}$

By Faradays first law,
$W=Z\times I\times t$

$t=5hour=5\times 3600s$

$4.76=\frac{63.5}{2\times 96500}\times I\times 5\times 3600$

$I=\frac{4.76\times 2\times 96500}{63.5\times 5\times 3600}$

$I=0.804A$