Construct ionic half-equations for reactions at the anode (to show oxidation) and at the cathode (to show reduction)

⚡️ Electrochemistry – Electrolysis

Imagine a battery that is turned on by electricity instead of a chemical reaction. That’s what electrolysis does – it uses an external electric current to drive a chemical reaction that would otherwise be impossible or very slow. In the classroom we often see it with simple solutions like copper(II) sulfate or sodium chloride. Let’s learn how to write the ionic half‑equations that describe what happens at each electrode.

🔋 What is Electrolysis?

Electrolysis is the process of using an electric current to split a compound into its constituent ions. Think of it as a “chemical blender” that forces the ions to move to the electrodes where they either lose or gain electrons.

🧪 Key Concepts

• Electrodes – the anode (positive) and cathode (negative).
• Oxidation – loss of electrons, happens at the anode.
• Reduction – gain of electrons, happens at the cathode.
• Half‑equations – separate equations for oxidation and reduction.
• Overall reaction – sum of the two half‑equations.

⚙️ Writing Ionic Half‑Equations

1. Identify the ions present in the electrolyte solution.

2. Decide which ion will go to which electrode based on charge and the applied voltage.

3. Write the half‑equation showing the ion gaining or losing electrons.

4. Balance atoms and charge using electrons, H⁺, H₂O, and OH⁻ as needed.

Example of an oxidation half‑equation:

\$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}\$

Example of a reduction half‑equation:

\$2\text{H}2\text{O} + 2e^- \rightarrow \text{H}2 + 2\text{OH}^-\$

🧪 Example: Electrolysis of Copper(II) Sulfate

Setup: A dilute solution of \$\ce{CuSO4}\$ is placed in a beaker with two inert electrodes (e.g., platinum). When a DC power supply is connected, copper metal plates onto the cathode and sulfuric acid forms at the anode.

Step 1 – Identify ions: \$\ce{Cu^{2+}}\$ and \$\ce{SO4^{2-}}\$ are the main ions in solution. Water molecules are also present.

Step 2 – Anode (oxidation): The sulfate ion is oxidised to sulfur dioxide and oxygen gas.

\$\ce{2SO4^{2-} -> 2SO3 + 2e^-}\$

(In practice, \$\ce{SO4^{2-}}\$ can also form \$\ce{O2}\$ directly: \$\ce{2H2O -> O2 + 4H^+ + 4e^-}\$)

Step 3 – Cathode (reduction): Copper ions gain electrons to form solid copper.

\$\ce{Cu^{2+} + 2e^- -> Cu(s)}\$

Overall reaction (simplified):

\$\ce{Cu^{2+} + 2SO4^{2-} + 2H2O -> Cu(s) + 2SO3 + 2H2O}\$

(The water cancels out, leaving the net production of copper metal and sulfur dioxide.)

📊 Common Electrolytes and Their Products

🧠 Quick Tips for Remembering Half‑Equations

1. Remember the phrase “Anode = Oxidation, Cathode = Reduction”.
2. Use the electron balance rule – electrons always appear on the side of the reaction that loses or gains them.
3. Check the charge balance after adding electrons.
4. Don’t forget to include water (H₂O) or hydroxide (OH⁻) when balancing in aqueous solutions.

💡 Real‑World Connection

Electrolysis is used in metal plating (e.g., nickel plating on coins), in the production of chlorine and sodium hydroxide from brine, and even in the manufacturing of hydrogen fuel. Understanding the half‑equations helps engineers design efficient processes and troubleshoot problems.

📝 Practice Problem

Write the ionic half‑equations for the electrolysis of a dilute solution of \$\ce{FeCl3}\$.

Hint: Iron(III) ions will be reduced at the cathode, while chloride ions may be oxidised at the anode. Try balancing the equations yourself before checking the answer in your textbook!