4.0 Electrolysis and electrode Reactions

Electrolysis is a process in which chemical reactions take place at the electrodes which are dipped into the electrolytes when the voltage is applied across these electrodes. Positively charged electrode is termed as anode and negatively charged electrode is termed as cathode.

The electrodes which transfer electrons to and from the solutions are termed as inert electrodes. There are some reactive electrodes also which take part in the electrode reaction. During electrolysis, reduction takes place at cathode and oxidation takes place at anode.

There are some general features of electrode reactions:

$(i)$ Ions which carry current are discharged at the electrodes.

$(ii)$ For a positively charged ion , it is difficult to get discharged at the cathode, results into the decomposition of water and consequently form ${H_2},O{H^ - }$ and absorption of electron.

$(iii)$ For a negatively charged ion , it is difficult to get discharged at the anode, results into the decomposition of water and consequently form ${O_2}$, ${H^ + }$ and electron.

An ideal cell for the electrolysis of molten $NaCl$ is shown below.

Direct current source is connected to a pair of inert electrodes which are immersed in the molten sodium chloride. The salt has been heated until it starts to melt. Sodium ions ($N{a^ + }$) moves towards the negative electrode and the chloride ions ($C{l^ - }$) move toward the positive electrode. When $N{a^ + }$ ions collide to the negative electrode, the battery carries large potential to force these ions and pick up the electrons to form sodium metal.

Negative Electrode (cathode): $N{a^ + } + {e^ - } \to Na$

$C{l^ - }$ ions collide to the positive electrode and oxidized to $C{l_2}$ which bubbles off at this electrode.

Positive Electrode (anode): $2C{l^ - } \to C{l_2} + 2{e^ - }$

The net effect of passing an electric current is to decompose the sodium chloride into its elements i.e., sodium metal and chloride gas.

(At cathode): $N{a^ + } + {e^ - } \to Na$

(At anode): $2C{l^ - } \to C{l_2} + 2{e^ - }$

The potential required to oxidize $C{l^ - }$ to $C{l_2}$ is $-1.36$ volts and potential to reduce $N{a^ + }$ ions to sodium metal is $-2.71$ volts. So the battery required to drive this reaction must have potential of $4.07$ (will discuss in the next topics)

This illustration best explain the difference between the voltaic cell as well as the electrolytic cell.

In voltaic cell, it uses energy released in the spontaneous reaction to perform electrical work whereas electrical work is utilized to drive the reaction in the opposite direction.

There are $N{a^ + }$ and $S{O_4}^{2 - }$ ions in the aqueous solution for the conduction of electric current. When the current is passed, then $S{O_4}^{2 - }$ ions moves towards anode and $N{a^ + }$ ions moves towards cathode but they are not discharged. Following reactions take place at the electrodes.

Cathode: $2{H_2}O + 2{e^ - } \to {H_2} + 2O{H^ - }or2{H^ + } + 2{e^ - } \to {H_2}$

Anode: $2{H_2}O \to {O_2} + 4{H^ + } + 2{e^ - }or4O{H^ - } \to {O_2} + 2{H_2}O + 4{e^ - }$

Overall Reaction: $2{H_2}O \to {O_2} + 2{H_2}$

In this case, ${H_2}$ is liberated at cathode and ${O_2}$ is at anode.

The possible species involved in the half-reactions are $N{a^ + }$, $C{l^ - }$ and ${H_2}O$. The possible cathode reactions are: $$N{a^ + }_{(aq)} + {e^ - } \to N{a_{(s)}}\ \ \ \ \ \ \ \ \ \ {E^0} = - 2.71V$$

$$2{H_2}{O_{(l)}} + 2{e^ - } \to {H_2}_{(g)} + 2O{H^ - }_{(aq)}\ \ \ \ \ \ \ \ {E^0} = - 0.83V$$

Under standard conditions, ${H_2}O$ should reduced in preference to $N{a^ + }$. Hydrogen is evolved at the cathode.

$$2{H_2}{O_{(l)}} \to {O_2}_{(g)} + 4{H^ + }_{(aq)} + 4{e^ - }\ \ \ \ \ \ \ \ \ {E^0} = 1.23V$$

$$2C{l^ - }_{(aq)} \to C{l_2}_{(g)} + 2{e^ - }\ \ \ \ \ \ \ {E^0} = 1.36V$$

Under standard conditions,${H_2}O$ is to be oxidized in preference to $C{l^ - }$ . As potentials are so close and over voltages at the electrodes could alter this conclusion. So it is difficult to give a genera statement about the product formed at the anode.Electrode Potential depend on the concentration. If solution is concentrated enough in $C{l^ - }$ then $C{l_2}$ will be the product but if it is dilute then ${O_2}$ will form. To see this we can simply apply Nernst Equation in the half cell reaction $$C{l^ - }/C{l_2}$$ half reaction.

$$C{l^ - }_{(aq)} \to C{l_2}_{(g)} + 2{e^ - }$$

Lets start with the dilute solution of $NaCl$, we found that oxidation potential of $C{l^ - }$ is negative so ${H_2}O$ is to be reduced in preference to $C{l^ - }$.Now start increasing the concentration of $NaCl$ and make it concentrate thenoxidation potential of $C{l^ - }$ increases . Now, $C{l^ - }$is to be oxidized in preference to ${H_2}O$ . And product changes from ${O_2}$ to $C{l_2}$.

Following are the half cell and overall reactions for the electrolysis of aqueous sodium chloride to chlorine and hydroxide ion

Anode: $2C{l^ - }_{(aq)} \to C{l_2}_{(g)} + 2{e^ - }$

Cathode: $2{H_2}{O_{(l)}} + 2{e^ - } \to {H_2}_{(g)} + 2O{H^ - }_{(aq)}$

Overall Reaction: $$2{H_2}{O_{(l)}} + 2C{l^ - }_{(aq)} \to {H_2}_{(g)} + 2O{H^ - }_{(aq)} + C{l_2}_{(g)}$$

We can generalize it as:

$(i)$ If two cations are present, then the cation with the higher reduction potential will be liberated at the cathode. And if two anions are present then anion with lower reduction potential will be discharged at the anode.

$(ii)$ If there is an active electrode at the cathode, metal goes on depositing on cathode and metal is dissolved at the anode.

$(iii)$ SRP does not decide which cation will discharge, it is the reduction potential which is a deciding factor. SRP's used only when the concentration are $1M$.

Qualitatively, it is impossible to predict the reduction and oxidation extent, which may enhance and which get hamper. But quantitatively, we can predict it by the use of standard hydrogen electrode (SHE).