Question

A professor adds to water to facilitate its electrolysis in a lecture demonstration.

(a) What is the purpose of the ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ ?

(b)Why is the water electrolyzed instead of the salt?

Short answer

a) To improve conductivity and speed up the electrolysis reaction, is utilized.

b) Due to its lower reduction potential, water is more easily electrolyzed than other ions in the solution.

Step-by-step solution

Definition

It's an electrochemical process that involves passing a current between two electrodes through an ionized solution (the electrolyte) to deposit positive ions (anions) on the negative electrode (cathode) and negative ions (cations) on the positive electrode (cathode) (anode).

Determining the purpose

(a)

To boost the water conductivity in this scenario, sodium sulfate is utilized.

Pure water is said to contain only ${\mathrm{H}}^{+}$and ${\mathrm{OH}}^{-}$ions in low concentrations $-{10}^{-7}\mathrm{M}$at $\mathrm{pH}$$7$ , limiting its conductivity. More ions would be present in the solution as more salt was added, and the current would more quickly travel through the water to complete the circuit, resulting in much quicker electrolysis.

Determining the water electrolyzed instead of the salt

(b)

In the case of ${\mathrm{Na}}_2{\mathrm{SO}}_4$addition to water, several reactions are plausible.

For oxidation (on the anode), the following reactions are considered:

- oxidation of water to ${\mathrm{O}}_2$:

$\begin{array}{l}2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+4{\mathrm{e}}^{-}\\ {\mathrm{E}}_{\mathrm{ox},{\mathrm{H}}_2\mathrm{O}}=-(1.40\mathrm{V}\left)\right(\mathrm{withovervoltage})\end{array}$

- oxidation of ${\mathrm{SO}}_4^{2-}$ ions is not possible because $\mathrm{S}$ is in their maximum oxidation state $\left(+6\right)$already, so they cannot be further oxidized.

The oxidation with the less positive (more negative) electrode potential happens in aqueous solution with mixed salts. Therefore, the oxidation of water to ${\mathrm{O}}_2$gas and ${\mathrm{H}}_3{\mathrm{O}}^{+}$ions is expected on anode.

Consider the following reduction processes on the cathode:

Determining the water electrolyzed

- reduction of water to ${\mathrm{H}}_2$:

$\begin{array}{l}2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right)+2\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{E}}_{\mathrm{red},{\mathrm{H}}_2\mathrm{O}}=-1.00\mathrm{V}\left(\mathrm{withovervoltage}\right)\end{array}$

- reduction of ${\mathrm{Na}}^{+}$ions to $\mathrm{Na}$:

$\begin{array}{l}{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Na}\left(\mathrm{s}\right)\\ {\mathrm{E}}_{\mathrm{red},,{\mathrm{K}}^{+}}=-2.71\mathrm{V}\end{array}$

The reduction with the less negative (more positive) electrode potential happens in an aqueous solution containing mixed salts. On the cathode, water is predicted to be reduced to ${\mathrm{H}}_2$gas and (hydroxide) ions.

Overall, the water is still more readily electrolyzed with the addition of ${\mathrm{Na}}_2{\mathrm{SO}}_4$owing to the half-reaction potentials difference, as seen above. As a result, the anode reaction produces ${\mathrm{O}}_2$gas and${\mathrm{H}}_3{\mathrm{O}}^{+}$ ions from water, whereas the cathode reaction produces ${\mathrm{H}}_2$gas and ${\mathrm{OH}}^{-}$ions from water.