Electrolysis and Predicting Products for the ESAT

The Fundamentals of Electrolysis

Electrolysis is a chemical process that uses electricity to break down an ionic compound, known as the electrolyte. For electrolysis to occur, the ions must be free to move, meaning the compound must be in a molten state or dissolved in water. The apparatus consists of two electrodes: the cathode (the negative electrode) and the anode (the positive electrode).

During the process, positively charged ions (cations) are attracted to the negative cathode, where they gain electrons in a process called reduction. Negatively charged ions (anions) are attracted to the positive anode, where they lose electrons in a process called oxidation. This movement of ions constitutes the flow of current through the electrolyte.

Electrolysis of Molten Binary Compounds

A binary compound consists of only two elements, typically a metal and a non-metal. When these compounds are molten, there are only two types of ions present. Predicting the products is straightforward: the metal will form at the cathode and the non-metal will form at the anode.

Worked Example: Electrolysis of molten lead(II) bromide

When lead(II) bromide ($PbBr_2$) is heated until it melts, the $Pb^{2+}$ and $Br^-$ ions become mobile.

1. At the cathode: Lead ions move to the cathode and gain electrons to form lead atoms. The half-equation is: $Pb^{2+} + 2e^- \rightarrow Pb(l)$. This is reduction.

2. At the anode: Bromide ions move to the anode and lose electrons to form bromine molecules. The half-equation is: $2Br^- \rightarrow Br_2(g) + 2e^-$. This is oxidation.

Electrolysis of Aqueous Solutions

In aqueous solutions, the situation is more complex because water ($H_2O$) itself partially dissociates into hydrogen ions ($H^+$) and hydroxide ions ($OH^-$). This means there is more than one type of ion attracted to each electrode. Competitive discharge rules determine which ion is actually reacted.

Predicting Products at the Cathode

At the cathode, both the metal cation from the salt and the $H^+$ ions from the water are attracted. The rule is that the less reactive element is discharged.

1. If the metal is higher than hydrogen in the reactivity series (such as sodium, potassium, or magnesium), then hydrogen gas is produced at the cathode. The half-equation is: $2H^+ + 2e^- \rightarrow H_2(g)$.
2. If the metal is lower than hydrogen in the reactivity series (such as copper or silver), then the metal is produced at the cathode. For example: $Cu^{2+} + 2e^- \rightarrow Cu(s)$.

Predicting Products at the Anode

At the anode, both the non-metal anion from the salt and the $OH^-$ ions from the water are attracted. The product depends on whether the anion is a halide.

1. If the anion is a halide (chloride, bromide, or iodide) and the solution is concentrated, the corresponding halogen ($Cl_2$, $Br_2$, or $I_2$) is produced. For example: $2Cl^- \rightarrow Cl_2(g) + 2e^-$.
2. If the anion is not a halide (such as sulfate $SO_4^{2-}$ or nitrate $NO_3^-$), then oxygen gas is produced from the discharge of hydroxide ions. The half-equation is: $4OH^- \rightarrow O_2(g) + 2H_2O(l) + 4e^-$.

Worked Example: Electrolysis of aqueous sodium chloride (Brine)

In a concentrated solution of $NaCl$, the ions present are $Na^+$, $Cl^-$, $H^+$, and $OH^-$.

• At the cathode: $Na^+$ and $H^+$ are present. Sodium is more reactive than hydrogen, so hydrogen gas is discharged.
• At the anode: $Cl^-$ and $OH^-$ are present. Since $Cl^-$ is a halide, chlorine gas is discharged.
• Result: The $Na^+$ and $OH^-$ ions remain in the solution, forming sodium hydroxide ($NaOH$).