Electrolysis

Q: Write the ionic half-equation for the reaction occurring at the anode during the electrolysis of dilute sulfuric acid.

$4OH^- rightarrow 2H_2O + O_2 + 4e^-$

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Definition and Basic Setup

Electrolysis is defined as the decomposition of an ionic compound, when molten or in aqueous solution, by the passage of an electric current. This process breaks down the compound into its constituent elements.

To perform electrolysis, you need a simple electrolytic cell consisting of:

Electrolyte: The molten or aqueous ionic substance that undergoes electrolysis. It contains free-moving ions.
Electrodes: Two conductors dipped into the electrolyte.
- The anode is the positive electrode. Negative ions (anions) are attracted here.
- The cathode is the negative electrode. Positive ions (cations) are attracted here.
Power Supply: Provides the electric current to drive the non-spontaneous reaction.

Charge Transfer Mechanism
Electrolysis involves the movement of charge through two different media:

In the external circuit (wires): Charge is carried by the movement of electrons. Electrons flow from the negative terminal of the power supply to the cathode, and from the anode back to the positive terminal.
In the electrolyte: Charge is carried by the movement of ions. Positive ions (cations) move toward the cathode; negative ions (anions) move toward the anode.

Note: Electrons do not flow through the electrolyte. Ions do not flow through the wires.

Electrolysis

Electrolysis: The decomposition of an ionic compound, when molten or in aqueous solution, by the passage of an electric current.

Key terms to remember:

Decomposition: Breaking down a complex substance into simpler substances.
Ionic compound: A substance made of positive and negative ions held together by electrostatic forces.
Molten: Melted (liquid state due to heat).
Aqueous solution: Dissolved in water.

Predicting Products: Molten Binary Compounds

General Rule for Molten Binary Ionic Compounds
When an ionic compound is molten (melted), it contains only its own positive and negative ions. There is no water present to interfere.

At the cathode (negative electrode): The metal cation gains electrons (reduction) to form the metal.
At the anode (positive electrode): The non-metal anion loses electrons (oxidation) to form the non-metal.

This applies to any molten binary ionic compound. For example, in molten lead(II) bromide (PbBr_2):

Ions present: Pb^{2+} and Br^-
Cathode product: Lead metal (Pb)
Anode product: Bromine gas (Br_2)

Electrolysis of Molten Lead(II) Bromide

Setup: Molten PbBr_2 with inert carbon electrodes.

Ions Present: Pb^{2+} (positive) and Br^- (negative).

At the Cathode (-):

Pb^{2+} ions are attracted to the negative electrode.
They gain 2 electrons each: Pb^{2+} + 2e^- \rightarrow Pb.
Observation: A silvery-grey solid (lead metal) forms at the bottom of the crucible.

At the Anode (+):

Br^- ions are attracted to the positive electrode.
They lose electrons each: 2Br^- \rightarrow Br_2 + 2e^-.
Observation: Red-brown fumes of bromine gas are seen.

⚠︎ Confusing Molten vs. Aqueous Products

Mistake: Assuming that water always participates in the reaction or that the products are the same for molten and aqueous compounds.

Correction: In molten compounds, only the ions from the salt are present. Therefore, the metal forms at the cathode and the non-metal at the anode. In aqueous solutions, water molecules also provide H^+ and OH^- ions, which compete with the salt's ions for discharge. You must consider concentration and reactivity series to predict products in aqueous solutions.

Describing Observations Accurately

When to use: When asked to describe observations during electrolysis.

Why examiners accept this: Examiners look for specific physical states and colors. Vague terms like 'gas' or 'solid' are often insufficient without color or state details.

Correct Usage: Instead of saying 'bromine is formed', write 'red-brown fumes are observed'. Instead of 'lead forms', write 'silvery-grey solid collects at the bottom**. For hydrogen, say 'colorless gas bubbles'. For chlorine, say 'pale green gas' or 'bleaches damp litmus paper'.

Predicting Products: Aqueous Halides

General Rule for Aqueous Halide Solutions (Cl⁻, Br⁻, I⁻)
When a halide compound is dissolved in water, the solution contains halide ions (X^-), hydrogen ions (H^+) from water, and hydroxide ions (OH^-) from water.

At the Cathode (-): Hydrogen gas (H_2) is always formed because H^+ ions are preferentially discharged over metal ions (unless the metal is very unreactive like copper or silver, but even then, hydrogen is often cited in this syllabus context for dilute solutions). Note: For IGCSE/O Level, hydrogen is typically formed at the cathode for Group 1 and 2 metals.
At the Anode (+):
- If the solution is concentrated: The halogen (Cl_2, Br_2, or I_2) is formed. The high concentration of halide ions favors their discharge over hydroxide ions.
- If the solution is dilute: Oxygen gas (O_2) is formed. The low concentration of halide ions means hydroxide ions are preferentially discharged.

Electrolysis of Concentrated Aqueous Sodium Chloride (Brine)

Setup: Concentrated NaCl(aq) with inert carbon electrodes.

Ions Present: Na^+, Cl^- from salt; H^+, OH^- from water.

At the Cathode (-):

H^+ ions are discharged in preference to Na^+ ions.
Half-equation: 2H^+ + 2e^- \rightarrow H_2.
Observation: Colorless gas bubbles (hydrogen).

At the Anode (+):

Because the solution is concentrated, Cl^- ions are discharged in preference to OH^- ions.
Half-equation: 2Cl^- \rightarrow Cl_2 + 2e^-.
Observation: Pale green gas (chlorine) or bleaching of damp litmus paper.

Electrolysis of Dilute Sulfuric Acid

Setup: Dilute H_2SO_4(aq) with inert carbon electrodes.

Ions Present: H^+, SO_4^{2-} from acid; H^+, OH^- from water.

At the Cathode (-):

H^+ ions are discharged.
Half-equation: 2H^+ + 2e^- \rightarrow H_2.
Observation: Colorless gas bubbles (hydrogen).

At the Anode (+):

The solution is dilute, so there are few sulfate ions relative to water volume. OH^- ions from water are discharged.
Half-equation: 4OH^- \rightarrow 2H_2O + O_2 + 4e^-.
Observation: Colorless gas bubbles (oxygen).

Note: The overall process is the electrolysis of water, producing hydrogen and oxygen in a 2:1 volume ratio.

⚠︎ Incorrect Half-Equations for Oxygen

Mistake: Writing the half-equation for oxygen production using water molecules directly (2H_2O \rightarrow O_2 + 4H^+ + 4e^-) or confusing it with oxide ions.

Correction: In this syllabus, the standard accepted half-equation for oxygen formation at the anode in aqueous solutions is derived from hydroxide ions:
4OH^- \rightarrow 2H_2O + O_2 + 4e^-
Ensure you balance both atoms and charge. The charge on the left is -4 and on the right is -4 (from 4e^-).

Writing Ionic Half-Equations

When to use: When asked to write the ionic half-equation for a reaction at an electrode.

Why examiners accept this: Examiners require correct species, states (if asked), and balanced charge. The markscheme specifically looks for the reactant ion on the left and the product molecule/atom plus electrons on the right.

Correct Usage: For chlorine formation: 2Cl^- \rightarrow Cl_2 + 2e^-. Do not write Cl^- \rightarrow Cl + e^- as chlorine exists as diatomic molecules (Cl_2). Ensure the number of electrons lost equals the change in oxidation state.

Electroplating

Purpose: Metal objects are electroplated to improve their appearance (e.g., make them look shiny or gold-colored) and to increase resistance to corrosion (rusting).

How it Works:
Electroplating uses electrolysis to coat a conductive object with a thin layer of metal.

The object to be plated is made the cathode (negative electrode). It attracts positive ions of the plating metal.
The plating metal is made the anode (positive electrode). It dissolves to replenish the metal ions in the solution.
The electrolyte must be a solution containing ions of the plating metal (e.g., silver nitrate for silver plating).

Process: Metal atoms at the anode lose electrons and enter the solution as positive ions (M \rightarrow M^+ + e^-). These ions move to the cathode, gain electrons, and deposit as solid metal (M^+ + e^- \rightarrow M).

Electroplating a Silver Spoon with Silver

Setup:

Cathode (-): The silver spoon (object to be plated).
Anode (+): A piece of pure silver metal.
Electrolyte: Silver nitrate solution (AgNO_3).

Reactions:

At the Anode (+): Silver atoms lose electrons and dissolve into the solution.
Ag \rightarrow Ag^+ + e^-
At the Cathode (-): Silver ions from the solution gain electrons and deposit onto the spoon.
Ag^+ + e^- \rightarrow Ag

Result: The spoon is coated with a layer of silver. The concentration of the electrolyte remains constant because for every ion leaving the solution at the cathode, one enters from the anode.

⚠︎ Electroplating Electrode Connections

Mistake: Connecting the object to be plated to the positive electrode (anode).

Correction: The object to be plated must be the cathode (negative). Positive metal ions are attracted to the negative electrode. If connected to the anode, the object would dissolve or not receive the plating layer.

Mnemonic: Cathode is Negative (Cat-N). The Plating Object is Negative.

Electrolysis of Copper(II) Sulfate with Copper Electrodes

When to use: When asked about the electrolysis of aqueous copper(II) sulfate using active (copper) electrodes versus inert (carbon/platinum) electrodes.

Why examiners accept this: This is a key distinction. With inert electrodes, the electrolyte changes color/concentration. With copper electrodes, it does not. Examiners test if you understand that the anode material participates in the reaction.

Correct Usage: State clearly: 'With copper electrodes, the anode dissolves (Cu \rightarrow Cu^{2+} + 2e^-) and the color of the electrolyte remains constant because the concentration of Cu^{2+} ions does not change. With inert electrodes, oxygen is produced at the anode, and the blue color fades as Cu^{2+} ions are removed.'

Predicting Products in Molten vs. Aqueous Solutions

Predict the products formed at the electrodes during the electrolysis of molten calcium chloride using inert electrodes.

At the cathode: Calcium metal (Ca). At the anode: Chlorine gas (Cl₂).

State the difference in products formed at the anode when concentrated aqueous sodium chloride is electrolysed compared to dilute aqueous sodium chloride.

In concentrated solution, chlorine gas is formed. In dilute solution, oxygen gas is formed.

Writing Half-Equations

Write the ionic half-equation for the reaction occurring at the anode during the electrolysis of dilute sulfuric acid.

4OH^- \rightarrow 2H_2O + O_2 + 4e^-

Write the ionic half-equation for the formation of hydrogen at the cathode during the electrolysis of aqueous sodium chloride.

2H^+ + 2e^- \rightarrow H_2

Electroplating Setup and Observations

Describe how you would electroplate a steel key with copper. Include the materials used for the electrodes and the electrolyte.

The steel key is connected to the negative terminal (cathode). A piece of copper metal is connected to the positive terminal (anode). The electrolyte is an aqueous solution containing copper(II) ions, such as copper(II) sulfate.

During the electrolysis of aqueous copper(II) sulfate using copper electrodes, what happens to the mass of the anode and the color of the solution?

The mass of the anode decreases (it dissolves). The color of the solution remains blue (no change).