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CHEM 112 Chapter Notes - Chapter 5: Reducing Agent, Disproportionation

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21 Aug 2016

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5-4: oxidation-reduction reactions oxidation: substance gains o atoms. O. s of some element increases as electrons are lost, electrons appear on right side of half equation reduction: substance loses o atoms. Os of some element decreases as electrons are lost, electrons appear on left side of half equation half reactions: 2 half reactions occurring at the same time, an oxidation and a reduction. **total number of electrons associated with oxidation must equal the total number associated with the reduction. Atoms are not created or lost, but rather rearranged: number of atoms of each type, number of electrons transferred, total charges on reactants and products. Write and balance separate half equations for oxidation and reduction. Adjust coefficients into two half equations so that the same number of electrons appear on both sides. Add together two half equations to obtain overall balanced equation (cancel out electrons) In a basic solution oh- must appear in final balanced equation (h+ for acidic)

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Related Questions

I don't understand what to do, please help.

REDOX Reaction of Glucose during Metabolism

C6H12O6 + 6O2 ââââ> 6CO2 + 6H2O

1. Which atoms are being oxidized? Which atoms are being reduced? What is the total number of electrons being transferred?

2. Assuming that this reaction occurs in acidic aqueous solution, what would we add to the reactants or products to balance the half reaction in terms of atoms? Write the half reaction, balanced with respect to atoms. You can use the FAD/FADH2 abbreviations.

3. Which side of the half-reaction do we put the electrons on? Is this an oxidation or a reduction? Write the equation for the balanced half reaction.

Having trouble with my Pre-Lab questions I don't understand how to get them.

Introduction An important type of reaction mechanism is the transfer of electrons from one species to another effecting a change in oxidation state. This gain or loss of electrons is referred to as reduction and oxidation, respectively. For example, consider the two balanced, chemical equations (1 & 2) that follow Cu (s) D Cu (aq) 2e (1) Cl2(aq) 2e D 2Cl (aq) (2) The first equation (1) shows elemental copper metal with a oxidation state of zero forming aqueous copper(Il) ions having an oxidation state of +2. This gain in oxidation number, 0 to +2, is called oxidation. The second equation (2) show elemental chlorine having an oxidation state of zero forming chloride ions in aqueous solution with an oxidation state of This decrease in oxidation number, 0 to -1, is called reduction. In any chemical reaction, the electrons gained or lost are conserved, i.e.. there is no net gain or loss of electrons. In fact, the two reactions illustrated above could occur simultaneously with the electrons lost by copper, gained by the chlorine in aqueous solution. Since the copper provides the electrons that subsequently reduce the chlorine copper is referred to as the reducing agent. Likewise, since the chlorine is accepting the electrons from copper, which subsequently oxidizes the copper, chlorine is referred to as he oxidizing agent. The two reactions above (1 & 2) are called half-reactions, as much as they each constitute one-half of the total reaction Equation 1 is the oxidation half reaction showing Cu(s) losing two electrons. In Equation 2, Cla(aq) gains two electrons and is the reduction half-reaction. If these two half-reaction are added together the result is the total oxidation-reduction reaction or "redox" reaction. Adding the two equations yields (3). Cu(s) Cl2(aq) D Cu (aq) 2CI (aq) (3) Whether or not the above reaction is spontaneous proceeds from reactants to products without input of energy into the system, may be determined by observing the reaction mixture if the reactants and products differ greatly in their appearance. Copper metal is a bright copper color and chlorine in aqueous solution will appear yellow. The Cu ion may turn the liquid part of the mixture blue, however, the Cris colorless and cannot be observed directly. Copper, like many of the transition metals, exhibits a variety of oxidation states depending upon the chemical environment in which it is found. In a strong reducing environment copper may exist as the free metal, Cu(s). An example of such a situation is illustrated in the following chemical equation (4). 4Cuo(s) CH4(g) D 4Cu(s) CO2(g) 2H2O(l) (4)