Redox as electron transfer

Oxidation and reduction always happen together. If one species loses electrons, another species must gain those electrons. That paired process is called redox.

A useful memory aid is OIL RIG:

• Oxidation Is Loss of electrons.
• Reduction Is Gain of electrons.

The definition is about electron transfer, not just oxygen. A metal atom forming a positive ion has lost electrons, so the metal has been oxidised. A positive ion forming a neutral atom has gained electrons, so the ion has been reduced.

Worked example: electron transfer in a simple redox reaction

Magnesium reacts with copper(II) ions:

Magnesium changes from Mg to Mg^2+. It has lost two electrons, so magnesium is oxidised.

Copper(II) ions change from Cu^2+ to Cu. Each copper(II) ion has gained two electrons, so copper(II) ions are reduced.

The two changes fit together:

• Mg loses 2 electrons.
• Cu^2+ gains 2 electrons.

The electrons lost and gained must balance overall.

Oxidation-number changes

Electron transfer is often invisible in a full equation. Oxidation numbers are a bookkeeping method that lets you see whether electrons have effectively been lost or gained.

For this lesson, use the oxidation-number rules from the previous redox work as tools. The most common quick checks are:

• an uncombined element has oxidation number 0
• a simple ion has oxidation number equal to its charge
• hydrogen is usually +1 in acids
• oxygen is usually -2 in ordinary oxides and many compounds

Do not confuse oxidation number with the whole formula charge. For example, in FeCl3, each chloride ion is -1, so iron is +3. The compound as a whole is neutral, but the iron atom has oxidation number +3.

Worked example: tracking oxidation-number change

In the reaction below, zinc reduces copper(II) ions:

Track only the elements that change.

Zinc is a d-block metal example. The same idea works for s-block examples such as magnesium and p-block examples such as aluminium: the block of the element does not change the redox rule.

Worked example: p-block aluminium

Aluminium reacts with chlorine to form aluminium chloride:

Aluminium changes from 0 in Al to +3 in AlCl3. Its oxidation number increases, so aluminium is oxidised.

Chlorine changes from 0 in Cl2 to -1 in AlCl3. Its oxidation number decreases, so chlorine is reduced.

Full equations for metals with acids

Reactive metals can react with dilute acids to form a salt and hydrogen gas.

For this specification boundary, write full equations for these reactions. Ionic equations are not required here.

The acid controls the salt name:

• hydrochloric acid forms chlorides
• sulfuric acid forms sulfates

The metal controls the positive ion in the salt. Use its usual ion charge or the formula of the salt to balance the equation.

Worked example 1: s-block metal with hydrochloric acid

Magnesium forms Mg^2+ ions and hydrochloric acid forms chloride salts.

Products: magnesium chloride, MgCl2, and hydrogen, H2.

This is a full equation. It includes the acid formula, the salt formula and hydrogen gas.

Worked example 2: p-block metal with hydrochloric acid

Aluminium forms Al^3+ ions, so the chloride salt is AlCl3.

The coefficients are in the ratio 2 : 6 : 2 : 3. Ratio checking matters because H atoms must balance into H2 molecules.

Worked example 3: d-block metals with sulfuric acid

Iron can form iron(II) sulfate with dilute sulfuric acid:

Zinc forms zinc sulfate:

These examples keep to the routine pattern: salt plus hydrogen. Nitric acid and concentrated sulfuric acid can behave as oxidising acids, so do not automatically use the salt plus hydrogen pattern for them unless the reaction information tells you to.

Interpreting metal-acid redox

A metal-acid equation is not just an acid reaction. It is also a redox reaction.

Look at magnesium with hydrochloric acid:

In this equation:

• Mg has oxidation number 0 in Mg(s).
• Mg has oxidation number +2 in MgCl2.
• H has oxidation number +1 in HCl.
• H has oxidation number 0 in H2.
• Cl is -1 in HCl and -1 in MgCl2, so chlorine has not been oxidised or reduced.

Magnesium's oxidation number increases from 0 to +2. Magnesium has been oxidised. In electron-transfer language, each magnesium atom has lost two electrons.

Hydrogen's oxidation number decreases from +1 to 0. Hydrogen has been reduced. In electron-transfer language, hydrogen ions from the acid have gained electrons to form hydrogen gas.

Worked example: interpreting aluminium with acid

Use the full equation:

Aluminium:

• before: Al(s), oxidation number 0
• after: AlCl3, aluminium is +3
• change: 0 to +3, so aluminium is oxidised

Hydrogen:

• before: HCl, hydrogen is +1
• after: H2, hydrogen is 0
• change: +1 to 0, so hydrogen is reduced

Now check the electron ratio. Each aluminium atom loses 3 electrons. There are 2 aluminium atoms, so 6 electrons are lost in total. Each hydrogen atom gains 1 electron. There are 6 hydrogen atoms in 6HCl, forming 3H2, so 6 electrons are gained in total.

That ratio check is why the full equation has 2Al and 6HCl.

Predicting unfamiliar redox changes

For an unfamiliar redox equation, do not try to memorise the reaction. Use a repeatable oxidation-number method.

1. Assign oxidation numbers to the elements that might have changed.
2. Identify any oxidation-number increase: that species has been oxidised and has lost electrons.
3. Identify any oxidation-number decrease: that species has been reduced and has gained electrons.
4. Compare the size of the changes to check the electron ratio.
5. Ignore species whose oxidation numbers have not changed.

This method lets you make predictions from the equation, even if the reaction context is new. The prediction might be which species is oxidised, which is reduced, or what coefficient ratio is needed for electron loss and gain to balance.

Worked example: unfamiliar full equation

Tin(II) chloride reacts with iron(III) chloride:

Use chloride as -1 throughout.

For tin:

• SnCl2: tin is +2
• SnCl4: tin is +4
• change: +2 to +4, so tin is oxidised
• electron language: tin has lost 2 electrons per tin atom

For iron:

• FeCl3: iron is +3
• FeCl2: iron is +2
• change: +3 to +2, so iron is reduced
• electron language: each iron has gained 1 electron

The equation needs 2FeCl3 because two Fe^3+ centres each gain 1 electron, matching the 2 electrons lost by one Sn^2+ centre. This is M0.2 ratio thinking in a redox context.

Boundary caution

Some unfamiliar reactions may involve nitric acid or concentrated sulfuric acid. In those cases, do not assume the simple metal + acid -> salt + hydrogen pattern. Use the equation or information given, then track oxidation numbers and electron loss or gain.