1.59C-1.60C - Electrolysis half-equations and practical

1.59C-1.60C - Electrolysis half-equations and practical

Electrolysis questions become much easier when you separate the cell into two electrode reactions. A half-equation shows the particles and electrons involved at one electrode only. In this lesson, you will write those half-equations, classify them as oxidation or reduction, and connect them to the practical investigation of aqueous solutions.

Half-equations and electron transfer

A normal symbol equation describes the overall reaction. An ionic half-equation zooms in on one electrode: it shows which ion or molecule changes there, and where the electrons go.

Ionic half-equation

An ionic half-equation is an equation for the reaction at one electrode, showing electrons gained or lost.

In electrolysis:

ElectrodeCharge in electrolysisWhat happens to electronsRedox word
Cathodenegativeions gain electronsreduction
Anodepositiveions lose electronsoxidation

The quickest exam check is the position of the electrons.

Electron positionMeaningExample
Electrons on the leftelectrons are gainedCu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu
Electrons on the rightelectrons are lost2Cl−→Cl2+2e−2Cl^- \rightarrow Cl_2 + 2e^-

Balance half-equations in this order: balance the atoms, then balance the charge by adding electrons. The total charge on the left must equal the total charge on the right.

For example, Cu2+Cu^{2+} has a charge of +2. It needs two electrons, each with a -1 charge, to become neutral copper:

Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu

This is reduction because copper(II) ions gain electrons.

Cathode half-equations

Positive ions move to the cathode because the cathode is negative. At the cathode, the positive ion or water gains electrons. That is always reduction.

For a metal ion that is discharged, add enough electrons to make a neutral atom:

Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu Ag++e−→AgAg^+ + e^- \rightarrow Ag

In aqueous solutions, hydrogen may form instead of a metal. For a strong acid, the hydrogen half-equation can be written as:

2H++2e−→H22H^+ + 2e^- \rightarrow H_2

For many aqueous solutions, the Pearson core practical guide gives the more general water form:

2H2O+2e−→H2+2OH−2H_2O + 2e^- \rightarrow H_2 + 2OH^-

Both equations show reduction because electrons are on the left. The water equation is useful when the solution is not strongly acidic, such as aqueous sodium chloride.

Common cathode observations are a gas bubbling if hydrogen forms, or a solid coating on the electrode if a metal forms. In copper(II) sulfate solution with inert electrodes, copper forms as a pink-brown deposit at the negative electrode.

Anode half-equations

Negative ions move to the anode because the anode is positive. At the anode, ions or water lose electrons. That is always oxidation.

Halide ions can form halogen molecules. The half-equation for chlorine from chloride ions is:

2Cl−→Cl2+2e−2Cl^- \rightarrow Cl_2 + 2e^-

This is oxidation because chloride ions lose electrons. In an exam answer, name the ion that loses electrons. "Chloride ions lose electrons" is precise; "chlorine loses electrons" is not.

When oxygen forms at the positive electrode in an aqueous solution, a common half-equation is:

2H2O→O2+4H++4e−2H_2O \rightarrow O_2 + 4H^+ + 4e^-

In alkaline conditions, the same oxygen formation may be represented as:

4OH−→O2+2H2O+4e−4OH^- \rightarrow O_2 + 2H_2O + 4e^-

Both versions show oxidation because electrons are produced on the right. In the copper(II) sulfate and dilute sulfuric acid examples, sulfate ions remain in the solution, so oxygen forms at the anode instead.

In electrolysis, reduction happens at the cathode and oxidation happens at the anode. The electrons in the half-equation prove it.

Investigating aqueous solutions

The practical uses a direct current supply, two inert electrodes, and an aqueous electrolyte. The positive terminal is connected to the anode, and the negative terminal is connected to the cathode.

Continue this lesson.

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