Question

Electrolysis involves electronation and de-electronation at the respective electrodes. Anode of an electrolytic cell is the electrode at which de-electronation takes place whereas at cathode electronation is noticed. If two or more ions of the same charge are to be electronated or de-electronated, the ion having lesser discharge potential is discharged. Discharge potential of an ion refers for \[{\text{E}}_{{\text{OP}}}^{\text{o}}\,{\text{or E}}_{{\text{RP}}}^{\text{o}}\] as the case may be.
The products formed at either electrode is given in terms of Faraday's laws of electrolysis i.e. \[{\text{w}} = \dfrac{{{\text{Eit}}}}{{96500}}\]
1.During electrolysis of acidulated ${{\text{H}}_{\text{2}}}{\text{O}}$ the molar ratio of gases formed at cathode and anode is:
A.$1:2$
B.$2:1$
C.$3:1$
D.$1:3$
2.During electrolysis of ${\text{CuS}}{{\text{O}}_{\text{4}}}(aq.)$ the pH of the solution becomes:
A.$ < 7$
B.$ > 7$
C.$ = 7$
D.$ \geqslant 7$

Answer 1

Hint: To answer this question, you should recall the concept of decomposition using faraday law. Calculate the equivalent mass of water by writing the electrochemical reactions and then use it to calculate the gas released at anode and cathode.

Complete step by step answer:
1. The reaction for the dissociation of water is:
 \[2{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right) \to 2{{\text{H}}_{\text{2}}}\left( g \right) + {{\text{O}}_2}\left( g \right)\]
 As shown in the equation, the number of hydrogen molecules produced is twice the number of oxygen molecules. Therefore, the mole ratio of hydrogen and oxygen gas liberated is \[2:1\].

Hence option B is correct.

2.We know that during electrolysis of copper sulphate the electrolyte copper(II) sulfate, provides a high concentration of copper(II) ions ${\text{C}}{{\text{u}}^{2 + }}$ and sulfate ions ${\text{S}}{{\text{O}}_4}^{2 - }$ to carry the current during the electrolysis process. Also, there is the production of tiny concentrations of hydrogen ions ${{\text{H}}^ + }$ and hydroxide ions ${\text{O}}{{\text{H}}^ - }$ from the self-ionization of water itself, but these can be ignored. Cathode attracts ${\text{C}}{{\text{u}}^{2 + }}$ ions and ${{\text{H}}^ + }$ 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: \[{\text{C}}{{\text{u}}^{2 + }}\left( {aq} \right)\;{\text{ }} + \;{\text{ }}2{{\text{e}}^-} \to \;{\text{ Cu}}\left( s \right)\]. This results in a brown copper deposit. The copper sulphate solution will break down into ${\text{C}}{{\text{u}}^{2 + }}$ cation and ${\text{S}}{{\text{O}}_4}^{2 - }$ ​ anion. As a result, the solution of copper sulphate which is blue will no longer retain. This ${\text{S}}{{\text{O}}_4}^{2 - }$​ combines with \[{{\text{H}}_2}\] ​ and increases the acidity.

Hence, option A is correct.

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: \[{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}\]