Half-equation

When the compoimds involved are ionic, the equation for a redox reaction can be divided into two half equations, one representing the reduction of the oxidizing agent and the other the oxidation of the reducing agent. Thus equation (2a) or (2a ) can be divided into 

Note that this can only be done when the compounds are ionic it cannot be done in the case of an equation like (lb) or (2b). 

The division of equations for redox reactions into half equations is most useful for reactions in solution. This is because ions in solution are relatively independent of each other (e.g. the freezing points of dilute solutions correspond to the number of ions present). 

If the ions in solution are independent, this equation may be written 

International programs taught by professors and professionals from all over the world BBA in Global Business 

---

Thus an oxidising agent is identified as an electron acceptor and the oxidation of iron(II) by chlorine can be written as two "half equations, viz. 

This can be written as two simple ionic half equations... 

Identify the oxidising agent and the reducing agent in each reaction and write half-equations showing the donation or acceptance of electrons by each of these eight reagents. 

This equation can be split into two half-equations, one of oxidation and the other of reduction. Zinc atoms are oxidized to Zn2+ ions by losing electrons. The oxidation half-equation is... 

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

Before you can balance an overall redox equation, you have to be able to balance two halfequations, one for oxidation (electron loss) and one for reduction (electron gain). Sometimes that s easy. Given the oxidation half-equation... 

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.)... 

Sometimes, though, it is by no means obvious how a given half-equation is to be balanced. This commonly happens when elements other than those being oxidized or reduced take part in the reaction. Most often, these elements are oxygen (oxid. no. = — 2) and hydrogen (oxid. no. = +1). Consider, for example, the half-equation for the reduction of the permanganate ion,... 

To balance half-equations such as these, proceed as follows ... 

Example 4.9 shows how these rules are applied to balance half-equations. 

The process used to balance an overall redox equation is relatively straightforward, provided you know how to balance half-equations. Follow a systematic, four-step procedure ... 

Split the equation into two half-equations, one for reduction, the other for oxidation. 

Balance one of the half-equations with respect to both atoms and charge as described above (steps a through e). 

Combine the two half-equations in such a way as to eliminate electrons. 

Strategy Follow the four-step procedure described above. Actually, if you look carefully at the text preceding this example, you ll find that all the half-equations have already been balanced The color coding should help you find them. 

Multiplying the first half-equation by 3, the second by 2, and adding gives... 

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

Earlier, the balanced half-equations were found to be... 

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. 

Multiply the half-equations by coefficients (1,2,. . . ) so that there are equal numbers of electrons on both sides. 

Balance redox half-equations and overall equations. 

Classify each of the following half-equations as oxidation or reduction and balance. 

Balance the half-equations in Question 53. Balance (a) and (b) in basic medium, (c) and (d) in acidic medium. 

We add the half-equations to obtain the overall reaction and add the half-reaction voltages to obtain the overall voltage ... 

Strategy Split the redox equation into two half-equations (oxidation and reduction). This enables you to calculate both E° and n. Then use Equations 18.2 and 18.3 to find... 

The approach used in Example 18.5 to find the number of moles of electrons transferred, n, is generally useful. What you do is to break down the equation for the cell reaction into two half-equations, oxidation and reduction. The quantity n is the number of electrons appearing in either half-equation. 

There is a simple relationship between the amount of electricity passed through an electrolytic cell and the amounts of substances produced by oxidation or reduction at the electrodes. From the balanced half-equations... 

Relations of this type, obtained from balanced half-equations, can be used in many practical calculations involving electrolytic cells. You will also need to become familiar with certain electrical units, including those of... 

Consider a salt bridge cell in which the anode is a manganese rod immersed in an aqueous solution of manganese(II) sulfate. The cathode is a chromium strip immersed in an aqueous solution of chromium(in) sulfate. Sketch a diagram of die cell, indicating the flow of the current throughout. Write the half-equations for the electrode reactions, the overall equation, and die abbreviated notation for the celL... 

Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating ° for the cells. 

Follow the instructions of Question 29 for the following half-equations. 

Because E° is a positive quantity, +0.165 V, a redox reaction should occur. The balanced half-equations are... 

To obtain the final balanced equation, multiply the first half-equation by 3 and add to the second. The result is... 

As you might expect from the half-equation for its reduction, the oxidizing strength of the dichromate ion decreases as the concentration of H+ decreases (increasing pH). 

---