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General Chemistry Principles, Patterns, and Applications, 2011

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important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction<br />

reactions in aqueous solution that is generally applicable. One such method uses oxidation states, <strong>and</strong> a second is<br />

referred to as thehalf-reaction method. We show you how to balance redox equations using oxidation states in this<br />

section; the half-reaction method is described in Chapter 19 "Electrochemistry".<br />

Balancing Redox Equations Using Oxidation States<br />

To balance a redox equation using the oxidation state method, we conceptually separate the overall<br />

reaction into two parts: an oxidation—in which the atoms of one element lose electrons—<strong>and</strong> a<br />

reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of<br />

Cr 2+ (aq) with manganese dioxide (MnO2) in the presence of dilute acid. Equation 4.54 is the net ionic<br />

equation for this reaction before balancing; the oxidation state of each element in each species has been<br />

assigned using the procedure described in Section 3.5 "Classifying Chemical Reactions":<br />

Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from<br />

the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that<br />

are oxidized <strong>and</strong> reduced:<br />

Equation 4.55<br />

Cr 2+ + Mn 4+ → Cr 3+ + Mn 2+<br />

The oxidation can be written as<br />

Equation 4.56<br />

Cr 2+ → Cr 3+ + e −<br />

<strong>and</strong> the reduction as<br />

Equation 4.57<br />

Mn 4+ + 2e − → Mn 2+<br />

For the overall chemical equation to be balanced, the number of electrons lost by the reductant must<br />

equal the number gained by the oxidant. We must therefore multiply the oxidation <strong>and</strong> the reduction<br />

equations by appropriate coefficients to give us the same number of electrons in both. In this example, we<br />

must multiply the oxidation equation by 2 to give<br />

Equation 4.58<br />

2Cr 2+ → 2Cr 3+ + 2e −<br />

Note the Pattern<br />

In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons<br />

gained by the oxidant.<br />

Saylor URL: http://www.saylor.org/books<br />

Saylor.org<br />

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