Question

A solution of copper sulphate electrolysed for 20 minutes with a current of $1.5$ Ampere. Calculate the mass of copper deposited at the cathode. (F = 96500C)

Answer 1

Hint: To answer this question, you should recall the concept of decomposition using Faraday’s law of electrolysis. Using the equation given, we can calculate the equivalent mass of copper by writing the electrochemical reactions and then use the current supplied to calculate the deposited mass.

Formula used:
${\text{m = Zit}}$ where ${\text{m}}$: mass deposited on the electrode, ${\text{Z}}$: Electrochemical equivalent for copper, ${\text{i}}$: current passed through the circuit and ${\text{t}}$: time (eq. 1)
${\text{Z = }}\dfrac{{\text{E}}}{{\text{F}}}$where ${\text{Z}}$: Electrochemical equivalent for copper, ${\text{E}}$: Equivalent mass and ${\text{F}}$: 1 Faraday = 96500C (eq. 2)
Complete Step by step solution:
We know that during electrolysis of copper sulphate, the electrolyte copper(II) sulfate provides a high concentration of copper(II) ions ${\text{C}}{{\text{u}}^{{\text{2 + }}}}$ and sulfate ions ${\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$ to carry the current during the electrolysis process. Also, there is the production of tiny concentrations of hydrogen ions ${{\text{H}}^{\text{ + }}}$ and hydroxide ions ${\text{O}}{{\text{H}}^{\text{ - }}}$ from the self-ionization of water itself, but these can be ignored. The cathode attracts ${\text{C}}{{\text{u}}^{{\text{2 + }}}}$ ions and ${{\text{H}}^{\text{ + }}}$ ions. But we know that the less reactive a metal, the more readily its ion is reduced on the electrode surface. Copper is below hydrogen in the reactivity series, hence, copper ion is discharged, and reduced to copper metal. The reaction can be written as: $C{u^{2 + }}\left( {aq} \right)\;{\text{ }} + \;{\text{ }}2{e^-} \to \;{\text{ }}Cu\left( s \right)$.
This results in a brown copper deposit. Now let us find the answer to this question:
Substituting equation (i) in (ii), the formula for weight of deposit can be calculated by ${\text{W = }}\dfrac{{{\text{Eit}}}}{{\text{F}}}$. Now substituting the desired values into the formula, we have,
W=$\dfrac{{63.52 \times 1.5 \times 20 \times 60}}{{2 \times 96500}} = 0.529{\text{ g}}$​
Hence, the mass of copper deposited at the cathode is $0.529{\text{ g}}$.

Note: We should know that at the electrodes, electrons are absorbed or released by the atoms and ions. There is gain or loss of electrons which are released and passed into the electrolyte. The uncharged atoms separate from the electrolyte upon exchanging electrons. This is called discharging and this concept helps determine the release of ions on different electrodes.
The increasing order of discharge of few cations is: ${{\text{K}}^{\text{ + }}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}{\text{, N}}{{\text{a}}^{\text{ + }}}{\text{, M}}{{\text{g}}^{{\text{2 + }}}}{\text{, A}}{{\text{l}}^{{\text{3 + }}}}{\text{, Z}}{{\text{n}}^{{\text{2 + }}}}{\text{, F}}{{\text{e}}^{{\text{2 + }}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, C}}{{\text{u}}^{{\text{2 + }}}}{\text{, A}}{{\text{g}}^{\text{ + }}}{\text{, A}}{{\text{u}}^{{\text{3 + }}}}$
Increasing Order of discharge of few anions is: ${\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}$