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Redox reactions

A redox reaction is one in which both oxidation and reduction take place.

Equations for redox reactions can be produced by adding together the two ion-electron equations representing each half-step (either reduction or oxidation).

The ion-electron equations must be balanced and added together.

Displacement reactions are a good example of redox reactions. Metals higher in the electrochemical series will displace lower metals from a solution of their ions.

For example, if magnesium metal is added to a solution of blue copper sulfate, the solution decolourises and copper metal forms on the surface of the magnesium.

Magnesium metal is oxidised (loses electrons) to form magnesium ions. The ion-electron equation for the oxidation step is:

\(Mg(s)\rightarrow Mg^{2+} + 2e^{-}\)

The reduction reaction involves copper ions in the solution being reduced (gaining electrons) to form copper metal, and is shown by the following ion-electron equation:

\(Cu^{2+} (aq) + 2e^{-} \rightarrow Cu(s)\)

The sulfate ion is a spectator and doesn't participate in the reaction. Some ion-electron equations for common elements can be found in the data booklet.

Adding the two half equations so that the electrons cancel out gives the equation for the redox reaction.

\(\scriptsize{Mg(s)\rightarrow Mg^{2+}}(aq)+2e^{-}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\scriptsize{OX}\)

\(\scriptsize{Cu^{2+} (aq) + 2e^{-} \rightarrow Cu(s)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\scriptsize{RED}\)

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\(\scriptsize{Mg(s)+ Cu^{2+} (aq)\rightarrow Mg^{2+}(aq)+CU(s)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\scriptsize{REDOX}\)

When the equations do not add together to cancel the electrons on the left and right hand sides, the equations must be multiplied to balance out when they are added together.

Example

Write the ion-electron equation for the displacement reaction between silver nitrate and zinc.

Firstly, write both ion-electron equations.

\(Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OX\)

\(Ag^{+}(aq)+e^{-}\rightarrow Ag(s)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,RED\)

The imbalance in the number of electrons means that the ion-electron equation involving silver (the reduction step) must be multiplied by two before the equations are added together.

\(\scriptsize{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\scriptsize{OX}\)

\(\scriptsize{Ag^{+}(aq)+e^{-}\rightarrow Ag(s)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\scriptsize{RED}\)

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\(\scriptsize{Zn(s)+ 2Ag^{+} (aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\scriptsize{REDOX}\)

Oxidsing and reducing agents can be identified in redox reactions, e.g

\(Li(s)+{Ag}^{+}(aq)\rightarrow{{Li}^{+}}(aq)+Ag(s)\)

Step 1: Split in to two equations

\(Li(s)\rightarrow{{Li}^{+}}(aq)\)

\({Ag}^{+}(aq)\rightarrow{Ag(s)}\)

Step 2: Balance each half equation

\(Li(s)\rightarrow{{Li}^{+}}(aq){+e}^{-}\)

\({Ag}^{+}(aq)+{e}^{-}\rightarrow{Ag(s)}\)

Step 3: Identify reduction and oxidation equations

\(\scriptsize{Li(s)\rightarrow{{Li}^{+}}(aq)+{e}^{-}~OXIDATION~=~REDUCING~AGENT}\)

\(\scriptsize{Ag^{+}(aq)+e^{-}\rightarrow{Ag(s)}~REDUCTION~=~OXIDISING~AGENT}\)

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