Reduction half-equations

At the same time, H+ ions are reduced to H2 molecules by gaining electrons the reduction half-equation is... 

In another case, this time a reduction half-equation,... 

Throughout this discussion we will show balanced oxidation half-equations in yellow, balanced reduction half-equations in green.)... 

To eliminate electrons, multiply the oxidation half-equation by 5 and add to the reduction half-equation. 

Multiply the reduction half-equation by 3, the oxidation half-equation by 2, then add. This will produce 6e on both sides, so they will cancel. 

Multiplying the reduction half-equation by 2 and adding to the oxidation halfequation, you should obtain, after simplification,... 

A We write down the oxidation half-equation with the method of Chapter 5, and obtain the reduction half-equation from Table 21-1, along with the reduction half-cell potential. 

Here, the oxidation half-equation must be multiplied by 3 (so that 24 electrons are produced), and the reduction half-equation by 8 (so that the same 24 electrons are consumed) ... 

The following balanced equations represent reactions that occur in aqueous acid. Break them down into balanced oxidation and reduction half-equations. 

Reduction half-equation E / in aqueous solution EV in liquid ammonia... 

To analyze and tabulate the results of such an experiment, the overall oxidation-reduction equation is often separated into what are known as half-equations or half-reactions, one representing the reduction part and the other the oxidation part. For the Zn—Cu " reaction, Equation (12.8a) shows the oxidation half-equation and Equation (12.8b) the reduction half-equation. Notice that the zinc metal releases or loses two electrons (and so is oxidized) and that these electrons are in turn transferred to the Cu " ion, which is reduced. Note also that these two half-equations can be added together again to yield Equation (12.7), the overaU equation for the reaction. 

Electrons are added to the left for a reduction half-equation and to the right for an oxidation half-equation. 

PRACTICE EXAMPLE A Represent the reaction of aluminum with hydrochloric acid to produce AlCl3(aq) and H2(g) by oxidation and reduction half-equations and an overall equation. 

Step 3. Balance each half-equation for electric charge. Add the number of electrons necessary to get the same electric charge on both sides of each half-equation. By doing this, you will see that the half-equation in which electrons appear on the right side is the oxidation half-equation. The other half-equation, with electrons on the left side, is the reduction half-equation. 

Step 4. Combine the half-equations to obtain an overall redox equation. Multiply the reduction half-equation by two and the oxidation half-equation by three to obtain the common multiple 6 e in each halfequation. Make the appropriate cancellations of and H2O. 

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