Oxidation and reduction in terms of electrons (HT only)

AQA·Chemistry (Combined Science)·5.4.1.4·Trilogy — Higher

Oxidation and reduction as electron transfer

During some reactions, electrons move from one species to another. A species is a substance or particle, such as an atom or ion.

  • Oxidation is the loss of electrons. A species that gives electrons away is oxidised.
  • Reduction is the gain of electrons. A species that accepts electrons is reduced.

These processes occur together when electrons lost by one species are gained by another.

Tracking electrons in a reaction

When magnesium reacts with copper ions, magnesium atoms form magnesium ions and copper ions form copper atoms:

Mg+CuX2+→MgX2++Cu\ce{Mg + Cu^2+ -> Mg^2+ + Cu}

The electrons do not appear in the full equation, but the changes in charge show what has happened. Each magnesium atom becomes a $$\ce{Mg^2+}$$ ion by losing two electrons, so magnesium is oxidised. Each $$\ce{Cu^2+}$$ ion becomes a neutral copper atom by gaining two electrons, so the copper ion is reduced.

Electron transfer in half-equations

A half-equation shows the electron loss or gain for one species. For the reaction above, the oxidation half-equation is:

Mg→MgX2++2 eX−\ce{Mg -> Mg^2+ + 2e-}

The electrons are on the product side because magnesium releases them. The reduction half-equation is:

CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}

The electrons are on the reactant side because the copper ion accepts them. The two electrons lost by magnesium are the two gained by the copper ion.

To identify oxidation or reduction in a half-equation, look for the electrons: electrons produced show oxidation; electrons taken in show reduction. In a full equation, compare the charges of the species before and after the reaction to work out which has lost electrons and which has gained them.

Ionic equations for displacement reactions

In a displacement reaction, one element replaces another in a compound. An ionic equation shows the particles that undergo chemical change, leaving out ions that remain unchanged. These unchanged ions are called spectator ions.

To write an ionic equation, first identify the reactants and products. Write each dissolved ionic substance as its separate aqueous ions. Solids and substances present as molecules remain in their chemical forms. State symbols show whether a species is aqueous, $$(aq)$$, or solid, $$(s)$$.

Writing an ionic equation for a displacement

Zinc displaces silver from aqueous silver nitrate. The balanced equation for the reaction is:

Zn(s)+2 AgNOX3(aq)→Zn(NOX3)X2(aq)+2 Ag(s)\ce{Zn(s) + 2AgNO3(aq) -> Zn(NO3)2(aq) + 2Ag(s)}

Silver nitrate and zinc nitrate are dissolved ionic substances, so write them as their aqueous ions. Zinc and silver are solids, so they remain as $$\ce{Zn(s)}$$ and $$\ce{Ag(s)}$$:

Zn(s)+2 AgX+(aq)+2 NOX3X−(aq)→ZnX2+(aq)+2 NOX3X−(aq)+2 Ag(s)\ce{Zn(s) + 2Ag+(aq) + 2NO3^-(aq) -> Zn^{2+}(aq) + 2NO3^-(aq) + 2Ag(s)}

The nitrate ions, $$\ce{NO3^-}$$, appear unchanged on both sides. They are spectator ions, so remove them from both sides. The resulting overall ionic equation is:

Zn(s)+2 AgX+(aq)→ZnX2+(aq)+2 Ag(s)\ce{Zn(s) + 2Ag+(aq) -> Zn^{2+}(aq) + 2Ag(s)}

This equation shows that zinc forms zinc ions while silver ions form solid silver.

Balancing atoms and charge

An ionic equation must have the same number of atoms of each element and the same total electric charge on both sides. In the equation above, there is one zinc atom and two silver atoms on each side. The total charge is $$+2$$ on the left, from two $$\ce{Ag+}$$ ions, and $$+2$$ on the right, from one $$\ce{Zn^{2+}}$$ ion.

Use coefficients in front of formulas to balance an equation: they change the quantities of species. Do not change subscripts within formulas, because that would change the substances or ions being represented.

Oxidation and reduction in terms of electrons (HT only) | AQA GCSE Combined Science: Trilogy (8464) — Chemistry — Higher Revision Notes | KnowledgeSet