Question

\[{\text{2}}{\text{.5F}}\] of electricity is passed through \[CuS{O_4}\] solution. The number of gm equivalent of \[Cu\] deposited on the anode is:
A) Zero
B) $2.5$
C) $1.25$
D) $5.0$

Answer 1

Hint:
To answer this question, you should recall the concept of Faraday’s law of electrolysis. Write the electrochemical reaction and then use it to calculate the deposited mass at the anode.

Complete step by step solution:
You should know that during electrolysis of copper sulphate, the electrolyte provides a high concentration of copper(II) ions $C{u^{2 + }}$ and sulfate ions $SO_4^{2 - }$ to carry the current during the electrolysis process. Also, there is the production of tiny concentrations of hydrogen ions ${H^ + }$ and hydroxide ions $O{H^ - }$ from the self-ionization of water itself, but these can be ignored.
Cathode attracts $C{u^{2 + }}$ ions and ${H^ + }$ 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:
\[C{u^{2 + }}\left( {aq} \right)\;{\text{ }} + \;{\text{ }}2{e^-} \to \;{\text{ }}Cu\left( s \right)\].
One Faraday of electricity will deposit one equivalent of copper.
Hence, \[{\text{2}}{\text{.5F}}\] of electricity will deposit $2.5$ equivalents of copper on the cathode.
But in the question, it is asked how much \[Cu\] is deposited on the anode which is zero as copper is deposited on the cathode.

Hence, the correct option is option A.

Note:
The question can also be solved by a trick. We know that for copper to deposit from its solution, it must undergo reduction. Reduction always occurs at the cathode in an electrochemical cell. So, the amount deposited at anode will be zero. 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: \[{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}\]