2
$\begingroup$

Pure water does not conduct electricity. When you dissolve a salt like $\ce{Na2SO4}$ in water the solution becomes a better conductor.

If electrolysis is done on this solution, oxygen and hydrogen gases form at the electrodes. The ions $\ce{Na+}$ and $\ce{SO^{2-}4}$ don't seem to be doing anything further. The electrons from the cathode don't involve the ions:

$$ \begin{align} \ce{4 H+ + 4 e- &-> 2 H2} \\ \ce{4 OH- &-> 2 H2O + O2 + 4 e-} \end{align} $$

How are the salt ions helping here or what are they doing?

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ Does this answer your question? Electrolysis with two similar metals $\endgroup$
    – M. Farooq
    Jan 11 '21 at 3:58
  • 1
    $\begingroup$ An inert salt here is basically reducing the resistance of the electrolytic cell. Anything which is thermodynamically feasible gets reduced/oxidized at the electrode, in this case it is water. Note that sulfate ion is not always inert. It can be oxidized to peroxydisulfate ion under certain conditions. $\endgroup$
    – M. Farooq
    Jan 11 '21 at 4:02
  • 1
    $\begingroup$ This is most certainly not a duplicate of the linked question. Did you even read the two questions? $\endgroup$
    – Jan
    Jan 19 '21 at 9:01
7
$\begingroup$

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's see what's going on at each electrode.

At the cathode (negative pole), water is decomposed according to : $\ce{2 H2O + 2 e- -> H2 + 2 OH-}$. This is only possible if there are some positive ions available near the cathode to compensate the negative charges appearing in the solution with these newly created $\ce{OH-}$ ions. Without positive ions like $\ce{Na+}$ near the cathode, these new $\ce{OH-}$ ions would repel further electrons coming from the outer circuit and prevent them from arriving to the electrode and from reacting with water. This would stop the electrolysis.

At the anode (positive pole), water is decomposed according to : $\ce{4 H2O -> O2 + 4 H+ + 4 e-}$. This is only possible if there are negative ions arriving near the anode to compensate the positive charges appearing in solution because of these newly created $\ce{H+}$ ions. Without negative ions like $\ce{SO4^{2-}}$ in the solution near the anode, these new $\ce{H+}$ ions would make a positively charged solution. This would prevent further electrons to be ejected from water. This would stop the electrolysis.

At the end, the composition of the solution is modified around the electrodes. Near the cathode, the solution contains more and more $\ce{NaOH}$. Near the anode, the solution contains more and more $\ce{H2SO_{4}}$ Of ourse if you stir and mix anodic and cathodic solutions, the acid and the base neutralize each other, which regenerate the solution of the original salt.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.