Redox reactions involve electron transfer between species. Oxidation means losing electrons, while reduction means gaining them. These processes are key to understanding how chemicals interact and change during electrochemical reactions.
Half-cell reactions break down redox reactions into separate oxidation and reduction steps. Balancing these reactions is crucial for analyzing electrochemical systems and predicting the overall reaction outcomes in various applications.
Redox Reactions
Oxidation and reduction definitions
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Oxidation
Involves the loss of electrons from a species
Results in an increase in the oxidation state of the species (F e 2 + → F e 3 + Fe^{2+} \rightarrow Fe^{3+} F e 2 + → F e 3 + )
The species undergoing oxidation acts as the reducing agent by donating electrons
Reduction
Involves the gain of electrons by a species
Results in a decrease in the oxidation state of the species (C l 2 → 2 C l − Cl_2 \rightarrow 2Cl^- C l 2 → 2 C l − )
The species undergoing reduction acts as the oxidizing agent by accepting electrons
Oxidizing and reducing agents
Oxidizing agent
Accepts electrons from another species causing its oxidation
Is reduced in the process as it gains electrons (M n O 4 − → M n 2 + MnO_4^- \rightarrow Mn^{2+} M n O 4 − → M n 2 + )
Examples include O 2 O_2 O 2 , C l 2 Cl_2 C l 2 , and M n O 4 − MnO_4^- M n O 4 −
Reducing agent
Donates electrons to another species causing its reduction
Is oxidized in the process as it loses electrons (2 I − → I 2 2I^- \rightarrow I_2 2 I − → I 2 )
Examples include H 2 H_2 H 2 , N a Na N a , and F e 2 + Fe^{2+} F e 2 +
Half-Cell Reactions
Balancing half-cell reactions
Oxidation half-reaction
Represents the loss of electrons by a species
The reactant loses electrons to form the oxidized product
Electrons appear as products on the right side (M g → M g 2 + + 2 e − Mg \rightarrow Mg^{2+} + 2e^- M g → M g 2 + + 2 e − )
Reduction half-reaction
Represents the gain of electrons by a species
The reactant gains electrons to form the reduced product
Electrons appear as reactants on the left side (A g + + e − → A g Ag^+ + e^- \rightarrow Ag A g + + e − → A g )
Steps for balancing half-cell reactions in acidic solution:
Balance all atoms except H and O (C r 2 O 7 2 − → 2 C r 3 + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} C r 2 O 7 2 − → 2 C r 3 + )
Balance O atoms by adding H 2 O H_2O H 2 O (C r 2 O 7 2 − + 14 H + → 2 C r 3 + + 7 H 2 O Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O C r 2 O 7 2 − + 14 H + → 2 C r 3 + + 7 H 2 O )
Balance H atoms by adding H + H^+ H + (C r 2 O 7 2 − + 14 H + + 6 e − → 2 C r 3 + + 7 H 2 O Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O C r 2 O 7 2 − + 14 H + + 6 e − → 2 C r 3 + + 7 H 2 O )
Balance charge by adding electrons (M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O )
Combining half-cell reactions
Identify the oxidation and reduction half-reactions
Multiply each half-reaction by a factor to equalize the electrons transferred
Add the half-reactions together and cancel out the electrons
Simplify the overall reaction by canceling common terms on both sides
Check that the final equation is balanced in terms of atoms and charge
Example:
Oxidation: 2 A l → 2 A l 3 + + 6 e − 2Al \rightarrow 2Al^{3+} + 6e^- 2 A l → 2 A l 3 + + 6 e −
Reduction: 3 C u 2 + + 6 e − → 3 C u 3Cu^{2+} + 6e^- \rightarrow 3Cu 3 C u 2 + + 6 e − → 3 C u
Overall: 2 A l + 3 C u 2 + → 2 A l 3 + + 3 C u 2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu 2 A l + 3 C u 2 + → 2 A l 3 + + 3 C u