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Redox Reactions Study Guide

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Introduction

Most aqueous reaction equations can be balanced by trial and error. Oxidation-reduction reactions require a more systematic approach to balancing equations using either an acidic or basic solution.

Balancing Redox Reactions

The oxidation and reduction parts of the reaction can be broken down into their half reactions. The half-reaction contains only the compounds that contain the species being reduced or the species being oxidized. As an example, copper metal reacts with a silver ion to yield silver metal and copper ions:

Cu(s) + 2Ag+ (aq) → 2Ag(s) + Cu2+ (aq) The oxidation half-reaction involves the oxidation of copper: Cu(s) → Cu2+ (aq) + 2e The reduction half-reaction involves the reduction of the silver ions: Ag+ (aq) + e → Ag(s)

Balancing oxidation-reduction reactions depends on whether the solution is acidic or basic. The method for balancing redox reactions in an acidic solution is as follows:

 1. Write the reduction and oxidation half-reactions. For each half-reaction: a. Balance all the elements except oxygen and hydrogen. b. Balance the oxygen using H2O. c. Balance the hydrogens using H+. d. Balance the charge using electrons (e–). 2. If the number of electrons needed previously in 1d for each half-reaction is different, equalize the electrons in each half-reaction by multiplying the appropriate reaction(s) by an integer. 3. Add the half-reactions and check that the elements and charges are balanced.

Example:

Balance the following reaction in an acidic solution:

H2SO4 + C → CO2 + SO2

Solution:

Sulfur is being reduced (S6+ to S4+) and carbon is being oxidized (C0 to C4+).

 1. Write the reduction and oxidation half-reactions: Oxidation: C → CO2 Reduction: H2SO4 → SO2 For each half-reaction: a. Balance all the elements except oxygen and hydrogen: Oxidation: C → CO2 Reduction: H2SO4 → SO2 b. Balance the oxygen using H2O: Oxidation: C + 2H2O → CO2 Reduction: H2SO4 → SO2 + 2H2O c. Balance the hydrogens using H+: Oxidation: C + 2H2O → CO2 + 4H+ Reduction: 2H+ + H2SO4 → SO2 + 2H2O d. Balance the charge using electrons (e–): Oxidation: C + 2H2O → CO2 + 4H+ + 4e– Reduction: 2e– + 2H+ + H2SO4 → SO2 + 2H2O 2. If the number of electrons needed in 1 d for each half-reaction is different, equalize the electrons in each half-reaction by multiplying the appropriate reaction(s) by an integer. In this case, the reduction half-reaction can be multiplied by 2 to yield both half-reactions with four electrons. Notice that the electrons must be on opposite sides to cancel. Oxidation: C + 2H2O → CO2 + 4H+ + 4e– Reduction: 4e– + 4H+ + 2H2SO4 → 2SO2 + 4H2O 3. Add the half-reactions and check that the elements and charges are balanced. C + 2H2O → CO2 + 4H+ + 4e–

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