Electrochemistry

Oxidation-Reduction Reactions

Defining Oxidation-Reduction Reactions

During an oxidation-reduction reaction, electrons move from one element to another. The element that loses electrons is oxidized. The element that gains electrons is reduced.

Electrochemistry is the branch of science that studies chemical reactions that cause electrons to move, resulting in the transfer of electric charge. Reactions in which electrons move from one element to another are called oxidation-reduction reactions, commonly referred to as redox reactions. In many chemical reactions, the enthalpy change involves a transfer of heat into or out of the system. Redox reactions can be set up so that the enthalpy change in the reaction involves a flow of electrons instead of an exchange of heat.

In redox reactions, an element that loses one or more electrons is said to be oxidized, while an element that gains one or more electrons is said to be reduced. A helpful way to remember the relationship is through this phrase: LEO (Loses Electrons—Oxidation) goes GER (Gains Electrons—Reduction). The species that is oxidized is also called the reducing agent, or reductant, while the species that is reduced is also called the oxidizing agent, or oxidant.

For example, the Haber-Bosch process can be shown by the following redox reaction.
N2+3H22NH3{\rm N}_2+3{\rm{H}}_2\rightarrow2{\rm{NH}}_3
Nitrogen gas, N2, is the oxidizing agent because it gains electrons and is reduced during the reaction. Hydrogen gas (H2) is the reducing agent because it loses electrons and is oxidized during the reaction. The oxidation number of each species is the number of electrons gained or lost by the species.

Oxidation Number Change in a Redox Reaction

In this redox reaction, N2 is the oxidizing agent because it accepts an electron from H2, the reducing agent. This can be seen by the oxidation numbers of each species.
A redox reaction can be broken into half-reactions to show oxidation and reduction separately. The half-reaction showing oxidation (electron loss) of a species is the oxidation half-reaction. The half-reaction showing reduction (electron gain) of a species is the reduction half-reaction. In the reaction of nitrogen gas and hydrogen gas, the oxidation half-reaction shows the electron loss of hydrogen.
H22H++2e{\rm{H}}_2\rightarrow2{\rm{H}}^++2{\rm{e}}^-
The reduction half-reaction shows the electron gain of nitrogen.
N2+6e2N3{\rm{N}}_2+6{\rm{e}}^-\rightarrow2{\rm{N}}^{3-}
Redox reactions often occur in aqueous solutions. For example, in the presence of an acid, chromium ions (Cr2+) react with manganese dioxide (MnO2). The overall reaction is:
2Cr2+(aq)+MnO2(s)+4H+(aq)2Cr3+(aq)+Mn2+(aq)+2H2O(l){2\rm{Cr}}^{2+}(aq)+{\rm{MnO}_2}(s)+{4\rm{H}^+}(aq)\rightarrow{2\rm{Cr}^{3+}}(aq)+{\rm{Mn}}^{2+}(aq)+{2\rm{H}}_{2}{\rm{O}}(l)
In this reaction, chromium loses an electron and is oxidized. Manganese gains two electrons and is reduced. The oxidation half-reaction is:
Cr2+Cr3++e{\rm{Cr}}^{2+}\rightarrow{\rm{Cr}}^{3+}+{\rm{e}}^-
And the reduction half-reaction is:
Mn4++2eMn2+{\rm{Mn}}^{4+}+2{\rm{e}}^-\rightarrow{\rm{Mn}}^{2+}
When determining whether a reaction is a redox reaction, consider the oxidation states, the degree of electron loss or gain of the species involved. If the oxidation states of all species are the same on both sides of the reaction, the reaction is not a redox reaction. Consider the reaction HNO3+KOHKNO3+H2O{\rm{HNO}}_3+{\rm{KOH}}\rightarrow{\rm{KNO}}_3+{\rm{H}}_2{\rm{O}}. The oxidation state of each element in the reactants is the same as the oxidation state of that element in the products. H has an oxidation number of +1 on both sides, O has an oxidation number of -2 on both sides, and so on. This is not a redox reaction. Only if species gain or lose electrons (that is, if their overall charge changes) is the reaction a redox reaction.

Balancing Oxidation-Reduction Reactions

Balancing oxidation-reduction reactions involves electron transfer and takes eight steps: dividing the reaction into half-reactions, balancing elements other than oxygen and hydrogen, balancing oxygen, balancing hydrogen, balancing the charges, multiplying each half-reaction with coefficients to balance the number of electrons transferred, adding the two half-equations, and canceling common terms on both sides of the reaction.

Balancing a redox reaction is similar to balancing other chemical reactions, except that the electron transfer must be considered. There are eight steps to follow when balancing redox reactions. These eight steps have variations depending on whether the reaction occurs in a neutral, acidic, or basic environment. Acidic conditions have excess hydrogen ions (H+). Basic conditions have excess hydroxide ions (OH). Neutral conditions do not have excess hydrogen (H+) or hydroxide ions (OH).

1. Divide the reaction into its two half-reactions.

2. Balance all elements in the half-reaction other than oxygen and hydrogen.

3. Balance oxygen by adding the appropriate number of water (H2O) molecules to the other side of the reaction. (If the condition is neutral, skip this step.)

4. Balance hydrogen (including those added in step 3) by adding hydrogen ions (H+), also known as protons, to the opposite side of the reaction. (If the condition is neutral, skip this step.)

5. Balance the charges on both sides of the half-reactions by adding electrons (e) to the side with the greater positive charge.

6. Multiply each half-reaction by the smallest whole number required to equalize the electrons gained by reduction and lost by oxidation.

7. Add the half-reactions together to form a complete reaction and cancel any common terms. (If the condition is basic, go on to step 8. Otherwise stop at this step.)

8. Add hydroxide ions (OH) to turn remaining hydrogen ions (H+) to water molecules, and cancel any common terms on either side of the reaction.

Step-By-Step Example
Balancing an Oxidation-Reduction Reaction Equation in Neutral Conditions
Consider the following equation that describes the oxidation of iron by copper ions:
Cu+(aq)+Fe(s)Fe3+(aq)+Cu(s){\rm{Cu}}^+(aq)+{\rm{Fe}}(s)\rightarrow{\rm{Fe}}^{3+}(aq)+{\rm{Cu}}(s)
In this reaction, the oxidation states of the copper (Cu) and iron (Fe) both change. The copper ion gains electrons and is reduced. The iron atom loses electrons and is oxidized. Follow the steps for balancing this redox reaction in neutral conditions.
Step 1
Divide the reaction into half-reactions.
Cu+(aq)Cu(s)Fe(s)Fe3+(aq)\begin{gathered}{\rm{Cu}}^+(aq)\rightarrow{\rm{Cu}}(s)\\{\rm{Fe}}(s)\rightarrow{\rm{Fe}}^{3+}(aq)\end{gathered}
Step 2
Balance all elements other than oxygen and hydrogen. For these half-reactions, this step is already complete because the coefficient of each species is 1.
Step 3
Balance oxygen by adding the appropriate number of water (H2O) molecules to the other side of the reaction. The reaction is neutral, so this step is skipped.
Step 4
Balance hydrogen by adding ions (H+) to the opposite side of the reaction. The reaction is neutral, so this step is skipped.
Step 5
Balance the charges by adding electrons (e) to the side with greater positive charge.
Cu+(aq)+eCu(s)Fe(s)Fe3+(aq)+3e\begin{gathered}{\rm{Cu}}^+(aq)+{\rm{e}}^-\rightarrow{\rm{Cu}}(s)\\\\{\rm{Fe}}(s)\rightarrow{\rm{Fe}}^{3+}(aq)+3{\rm{e}}^-\end{gathered}
Step 6
Multiply each half-reaction by the smallest whole number required to equalize the electrons gained by reduction and lost by oxidation. The reduction reaction is multiplied by a coefficient of 3 to balance out the three electrons lost by iron in the oxidation reaction.
3Cu+(aq)+3e3Cu(s)Fe(s)Fe3+(aq)+3e\begin{gathered}3{\rm{Cu}}^+(aq)+3{\rm{e}}^-\rightarrow3{\rm{Cu}}(s)\\\\{\rm{Fe}}(s)\rightarrow{\rm{Fe}}^{3+}(aq)+3{\rm{e}}^-\end{gathered}
Step 7
Add the half-reactions together to form a complete reaction, and cancel common terms.
3Cu+(aq)+Fe(s)+3eFe3+(aq)+3Cu(s)+3e3Cu+(aq)+Fe(s)Fe3+(aq)+3Cu(s)\begin{aligned}3{\rm{Cu}}^+(aq)+{\rm{Fe}}(s)+\xcancel{3{\rm{e}}^-}&\rightarrow{\rm{Fe}}^{3+}(aq)+3{\rm{Cu}}(s)+\xcancel{3{\rm{e}}^-}\\3{\rm{Cu}}^+(aq)+{\rm{Fe}}(s)&\rightarrow{\rm{Fe}}^{3+}(aq)+3{\rm{Cu}}(s)\end{aligned}
Solution
The reaction is not basic, so there is no need to continue to Step 8 (adding OH ions to turn remaining H+ ions to H2O molecules). Balancing the reaction is now complete.
Step-By-Step Example
Balancing an Oxidation-Reduction Reaction Equation in Acidic Conditions
Consider the following redox reaction between oxalic acid (H2C2O4) and permanganate(VII) (MnO4):
H2C2O4(aq)+MnO4(s)CO2(g)+Mn2+(aq){\rm{H}}_2{\rm{C}}_2{\rm O}_4(aq)+{\rm{MnO}_4}^-(s)\rightarrow{\rm{CO}}_2(g)+{\rm{Mn}}^{2+}(aq)
In this reaction, the oxidation states of the carbon and manganese both change. The manganese gains electrons and is reduced. The carbon loses electrons and is oxidized. Follow the steps for balancing this redox reaction in acidic conditions.
Step 1
Divide the reaction into half-reactions.
H2C2O4(aq)CO2(g)MnO4(s)Mn2+(aq)\begin{gathered}{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)\rightarrow{\rm{CO}}_2(g)\\{\rm{MnO}_4}^-(s)\rightarrow{\rm{Mn}}^{2+}(aq)\end{gathered}
Step 2
Balance all elements other than oxygen and hydrogen.
H2C2O4(aq)2CO2(g)MnO4(s)Mn2+(aq)\begin{gathered}{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)\rightarrow2{\rm{CO}}_2(g)\\{\rm{MnO}_4}^-(s)\rightarrow{\rm{Mn}}^{2+}(aq)\end{gathered}
Step 3
Balance oxygen by adding the appropriate number of water (H2O) molecules to the other side of the reaction. This is necessary only in the manganese half-reaction because the oxygen in the carbon dioxide accounts for the oxygen in the carbon half-reaction.
H2C2O4(aq)2CO2(g)MnO4(s)Mn2+(aq)+4H2O(l)\begin{gathered}{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)\rightarrow2{\rm{CO}}_2(g)\\{\rm{MnO}_4}^-(s)\rightarrow{\rm{Mn}}^{2+}(aq)+4{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 4
Balance hydrogen (including those added in step 3) by adding hydrogen ions (H+) to the opposite side of the reaction.
H2C2O4(aq)2CO2(g)+2H+(aq)8H+(aq)+MnO4(s)Mn2+(aq)+4H2O(l)\begin{gathered}{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)\rightarrow2{\rm{CO}}_2(g)+2{\rm{H}}^+(aq)\\8{\rm{H}}^+(aq)+{\rm{MnO}_4}^-(s)\rightarrow{\rm{Mn}}^{2+}(aq)+4{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 5
Balance the charges by adding electrons (e) to the side with greater positive charge.
H2C2O4(aq)2CO2(g)+2H+(aq)+2e8H+(aq)+MnO4(s)+5eMn2+(aq)+4H2O(l)\begin{gathered}{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)\rightarrow2{\rm{CO}}_2(g)+2{\rm{H}}^+(aq)+2{\rm{e}}^-\\8{\rm{H}}^+(aq)+{\rm{MnO_4}^-}(s)+5{\rm{e}}^-\rightarrow{\rm{Mn}}^{2+}(aq)+4{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 6
Multiply each half-reaction by the smallest whole number required to equalize the electrons gained by reduction and lost by oxidation. The carbon half-reaction was multiplied by a coefficient of 5, and the manganese half-reaction was multiplied by a coefficient of 2 to equalize the number electrons in each reaction at 10.
5H2C2O4(aq)10CO2(g)+10H+(aq)+10e16H+(aq)+2MnO4(s)+10e2Mn2+(aq)+8H2O(l)\begin{gathered}5{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)\rightarrow10{\rm{CO}}_2(g)+10{\rm{H}}^+(aq)+10{\rm{e}}^-\\16{\rm{H}}^+(aq)+2{\rm{MnO_4}^-}(s)+10{\rm{e}}^-\rightarrow2{\rm{Mn}^{2+}}(aq)+8{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 7
Add the half-reactions together to form a complete reaction, and cancel any common terms. The 10 hydrogen ions on the products side cancel out 10 of the 16 hydrogen ions on the reactants side.
5H2C2O4(aq)+16H+(aq)+2MnO4(s)+10e10CO2(g)+10H+(aq)+10e+2Mn2+(aq)+8H2O(l)5H2C2O4(aq)+6H+(aq)+2MnO4(s)10CO2(g)+2Mn2+(aq)+8H2O(l)\begin{aligned}5{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)+\xcancel{1}6{\rm{H}}^+(aq)+2{\rm{MnO_4}^-}(s)+\xcancel{10{\rm{e}}^-}&\rightarrow10{\rm{CO}}_2(g)+\xcancel{10{\rm{H}}^+(aq)}+\xcancel{10{\rm{e}}^-}+2{\rm{Mn^{2+}}}(aq)+8{\rm{H}}_2{\rm{O}}(l)\\5{\rm{H}}_2{\rm{C}}_2{\rm{O}}_4(aq)+6{\rm{H}}^+(aq)+2{\rm{MnO_4}^-}(s)&\rightarrow10{\rm{CO}}_2(g)+2{\rm{Mn^{2+}}}(aq)+8{\rm{H}}_2{\rm{O}}(l)\end{aligned}
Solution
This reaction is not basic, so there is no need to continue to Step 8 (adding OH ions to turn remaining H+ ions to H2O molecules). Balancing the reaction is now complete.
Step-By-Step Example
Balancing an Oxidation-Reduction Reaction Equation in Basic Conditions
Consider the following redox reaction of aluminum (Al) and nitrite (NO2):
Al(s)+NO2(aq)AlO2(aq)+NH3(aq){\rm{Al}}(s)+{\rm{NO}_2}^-(aq)\rightarrow{\rm{AlO}_2}^-(aq)+{\rm{NH}}_3(aq)
In this reaction, the oxidation states of aluminum and nitrogen both change. The nitrogen gains electrons and is reduced. The aluminum loses electrons and is oxidized. Follow the steps for balancing this redox reaction in basic conditions.
Step 1
Divide the reaction into half-reactions.
Al(s)AlO2(aq)NO2(aq)NH3(aq)\begin{gathered}{\rm{Al}}(s)\rightarrow{\rm{AlO}_2}^-(aq)\\{\rm{NO}_2}^-(aq)\rightarrow{\rm{NH}}_3(aq)\end{gathered}
Step 2
Balance all elements other than O and H. For these half-reactions, this is already complete because all coefficients of the species are 1.
Step 3
Balance oxygen by adding the appropriate number of water (H2O) molecules to the other side of the reaction.
Al(s)+2H2O(l)AlO2(aq)NO2(aq)NH3(aq)+2H2O(l)\begin{gathered}{\rm{Al}}(s)+2{\rm{H}}_2{\rm{O}}(l)\rightarrow{\rm{AlO}_2}^-(aq)\\{\rm{NO}_2}^-(aq)\rightarrow{\rm{NH}}_3(aq)+2{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 4
Balance hydrogen (including those added in step 3) by adding hydrogen ions (H+) to the opposite side of the reaction.
Al(s)+2H2O(l)AlO2(aq)+4H+(aq)NO2(aq)+7H+(aq)NH3(aq)+2H2O(l)\begin{gathered}{\rm{Al}}(s)+2{\rm{H}}_2{\rm{O}}(l)\rightarrow{\rm{AlO}_2}^-(aq)+4{\rm{H}}^+(aq)\\{\rm{NO}_2}^-(aq)+7{\rm{H}}^+(aq)\rightarrow{\rm{NH}_3}(aq)+2{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 5
Balance the charges by adding electrons (e) to the side with greater positive charge.
Al(s)+2H2O(l)AlO2(aq)+4H+(aq)+3eNO2(aq)+7H+(aq)+6eNH3(aq)+2H2O(l)\begin{gathered}{\rm{Al}}(s)+2{\rm{H}}_2{\rm{O}}(l)\rightarrow{\rm{AlO}_2}^-(aq)+4{\rm{H}}^+(aq)+3{\rm{e}}^-\\{\rm{NO}_2}^-(aq)+7{\rm{H}}^+(aq)+6{\rm{e}}^-\rightarrow{\rm{NH}}_3(aq)+2{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 6
Multiply each half-reaction by the smallest whole number required to equalize the electrons gained by reduction and lost by oxidation.
2Al(s)+4H2O(l)2AlO2(aq)+8H+(aq)+6eNO2(aq)+7H+(aq)+6eNH3(aq)+2H2O(l)\begin{gathered}2{\rm{Al}}(s)+4{\rm{H}}_2{\rm{O}}(l)\rightarrow2{\rm{AlO}_2}^-(aq)+8{\rm{H}}^+(aq)+6{\rm{e}}^-\\{\rm{NO}_2}^-(aq)+7{\rm{H}}^+(aq)+6{\rm{e}}^-\rightarrow{\rm{NH}}_3(aq)+2{\rm{H}}_2{\rm{O}}(l)\end{gathered}
Step 7
Add the half reactions together to form a complete reaction, and cancel any common terms.
2Al(s)+42H2O(l)+NO2(aq)+7H+(aq)+6e2AlO2(aq)+8H+(aq)+6e+NH3(aq)+2H2O(l)2Al(s)+2H2O(l)+NO2(aq)2AlO2(aq)+H+(aq)+NH3(aq)\begin{aligned}2{\rm{Al}}(s)+{\stackrel{2}{\xcancel{4}}}{\rm{H_2O}}(l)+{\rm{NO_2}^{-}}(aq)+\xcancel{7{\rm{H^+}}(aq)}+\xcancel{6{\rm{e^-}}}&\rightarrow2{\rm{AlO_2}^{-}}(aq)+\xcancel{8}{\rm{H^+}}(aq)+\xcancel{6{\rm{e^-}}}+{\rm{NH_3}}(aq)+\xcancel{2{\rm{H_2O}}(l)}\\2{\rm{Al}}(s)+2{\rm{H_2}}{\rm{O}}(l)+{\rm{NO_2}^{-}}(aq)&\rightarrow2{{\rm{AlO}_2}^{-}}(aq)+{\rm{H^+}}(aq)+{\rm{NH_3}}(aq)\end{aligned}
Solution
Because the reaction conditions are basic, add hydroxide ions (OH) to turn remaining H+ ions to water molecules, and cancel any common terms on either side of the reaction.
2Al(s)+2H2O(l)+NO2(aq)+OH(aq)2AlO2(aq)+H+(aq)+OH(aq)+NH3(aq)2Al(s)+2H2O(l)+NO2(aq)+OH(aq)2AlO2(aq)+H2O(l)+NH3(aq)2Al(s)+H2O(l)+NO2(aq)+OH(aq)2AlO2(aq)+NH3(aq)\begin{aligned}{\rm{2Al}}(s)+2{\rm{H_2O}}(l)+{\rm{NO_2}^{-}}(aq)+{\rm{OH^-}}(aq)&\rightarrow{2\rm{AlO_2}^{-}}(aq)+{\rm{H^+}}(aq)+{\rm{OH^-}}(aq)+{\rm{NH_3}}(aq)\\ {\rm{2Al}}(s)+\xcancel{2}{\rm{H_2O}}(l)+{\rm{NO_2}^{-}}(aq)+{\rm{OH^-}}(aq)&\rightarrow{2\rm{AlO_2}^{-}}(aq)+\xcancel{{\rm{H_2O}}(l)}+{\rm{NH_3}}(aq)\\ {2\rm{Al}}(s)+{\rm{H_2O}}(l)+{\rm{NO_2}^{-}}(aq)+{\rm{OH^-}}(aq)&\rightarrow{2\rm{AlO_2}^{-}}(aq)+{\rm{NH_3}}(aq)\end{aligned}