Chapter 8
Electrochemistry: Using Electrons
In This Chapter
Finding out about redox reactions
Balancing redox equations
Taking a look at electrochemical cells
Combustion is a redox reaction. So are respiration, photosynthesis, and many other biochemical processes people depend on for life. In this chapter, I explain redox reactions, go through the balancing of this type of equation, and then show you some applications of redox reactions in an area of chemistry called electrochemistry. Electrochemistry is an area of chemistry in which we use chemical reactions to produce electrons (electricity) or use electrons (electricity) to cause a desired chemical reaction to take place.
Transferring Electrons with Redox Reactions
Redox reactions — reactions in which there’s a simultaneous transfer of electrons from one chemical species (chemical entity such as an atom or molecule) to another — are really composed of two different reactions:
Oxidation: A loss of electrons
Reduction: A gain of electrons
These reactions are coupled, because the electrons that are lost in the oxidation reaction are the same electrons that are gained in the reduction reaction. In fact, these two reactions (reduction and oxidation) are commonly called half-reactions, because you need these two halves to make a whole reaction, and the overall reaction is called a redox (reduction/oxidation) reaction. In Chapter 7, I describe a redox reaction that occurs between zinc metal and the cupric (copper II, Cu2+) ion. The zinc metal loses electrons and the copper II ion gains them.
Oxidation
You can use three definitions for oxidation:
The loss of electrons
The gain of oxygen
The loss of hydrogen
Because I typically deal with electrochemical cells, I normally use the definition that describes the loss of the electrons. The other definitions are useful in processes such as combustion and photosynthesis.
Loss of electrons
One way to define oxidation is with the reaction in which a chemical substance loses electrons in going from reactant to product. For example, when sodium metal reacts with chlorine gas to form sodium chloride (NaCl), the sodium metal loses an electron, which chlorine then gains. The following equation shows sodium losing the electron:
Na(s) → Na+ + e–
When it loses the electron, chemists say that the sodium metal has been oxidized to the sodium cation. (A cation is an ion with a positive charge due to the loss of electrons — see Chapter 5.)
Reactions of this type are quite common in electrochemical reactions, reactions that produce or use electricity.
Gain of oxygen
In certain oxidation reactions, it’s obvious that oxygen has been gained in going from reactant to product. Reactions where the gain of oxygen is more obvious than the gain of electrons include combustion reactions (burning) and the rusting of iron. Here are two examples:
C(s) + O2(g) → CO2(g) (burning of coal)
2 Fe(s) + 3 O2(g) → 2 Fe2O3(s) (rusting of iron)
In these cases, chemists say that the carbon and the iron metal have been oxidized to carbon dioxide and rust, respectively.
Loss of hydrogen
In other reactions, you can best see oxidation as the loss of hydrogen. Methyl alcohol (wood alcohol) can be oxidized to formaldehyde:
CH3OH(l) → CH2O(l) + H2(g)
In going from methanol to formaldehyde, the compound goes from having four hydrogen atoms to having two hydrogen atoms.
Reduction
You can use three definitions to describe reduction:
The gain of electrons
The loss of oxygen
The gain of hydrogen
Gain of electrons
Chemists often see reduction as the gain of electrons. In the process of electroplating silver onto a teapot, for example, the silver cation is reduced to silver metal by the gain of an electron. The following equation shows the silver cation’s gaining the electron:
Ag+ + e– → Ag
When it gains the electron, chemists say that the silver cation has been reduced to silver metal.
Loss of oxygen
In some reactions, seeing reduction as the loss of oxygen in going from reactant to product is easy. For example, a reaction with carbon monoxide in a blast furnace reduces iron ore (primarily rust, Fe2O3) is to iron metal:
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
The iron has lost oxygen, so chemists say that the iron ion has been reduced to iron metal.
Gain of hydrogen
In certain cases, you can describe a reduction as the gain of hydrogen atoms in going from reactant to product. For example, carbon monoxide and hydrogen gas can be reduced to methyl alcohol:
CO(g) + 2 H2(g) → CH3OH(l)
In this reduction process, the CO has gained the hydrogen atoms.
One’s loss is the other’s gain
Neither oxidation nor reduction can take place without the other. When those electrons are lost, something has to gain them. Consider, for example, the net-ionic equation (the equation showing just the chemical substances that are changed during a reaction — see Chapter 7) for a reaction with zinc metal and an aqueous copper(II) sulfate solution:
Zn(s) + Cu2+ → Zn2+ + Cu
This overall reaction is really composed of two half-reactions:
Zn(s) → Zn2+ + 2e– (oxidation half-reaction — the loss of electrons)
Cu2+ + 2e– → Cu(s) (reduction half-reaction — the gain of electrons)
To help yourself remember which reaction is oxidation and which is reduction in terms of electrons, memorize the phrase “LEO goes GER” (Lose Electrons Oxidation; Gain Electrons Reduction).
Zinc loses two electrons; the copper(II) cation gains those same two electrons. Zn is being oxidized. But without Cu2+ present, nothing will happen. That copper cation is the oxidizing agent. It’s a necessary agent for the oxidation process to proceed. The oxidizing agent accepts the electrons from the chemical species that’s being oxidized.
Cu2+ is reduced as it gains electrons. The species that furnishes the electrons is the reducing agent. In this case, the reducing agent is zinc metal.
The oxidizing agent is the species that’s being reduced, and the reducing agent is the species that’s being oxidized. Both the oxidizing and reducing agents are on the left (reactant) side of the redox equation.
Oxidation numbers
Oxidation numbers are bookkeeping numbers. They allow chemists to do things such as balance redox equations. Oxidation numbers are positive or negative numbers, but don’t confuse them with charges on ions or valences. Chemists assign oxidation numbers to elements using these rules:
For free elements: The oxidation number of an element in its free (uncombined) state is zero (for example, Al(s) or Zn(s)). This is also true for elements found in nature as diatomic (two-atom) elements (H2, O2, N2, F2, Cl2, Br2, or I2) and for sulfur, found as S8.
For single-atom ions: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion (for example, Na+ = +1, S2– = –2).
For compounds: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. This rule often allows chemists to calculate the oxidation number of an atom that may have multiple oxidation states, if the other atoms in the ion have known oxidation numbers. (See Chapter 6 for examples of atoms with multiple oxidation states.)
For alkali metals and alkaline earth metals in compounds: The oxidation number of an alkali metal (IA family) in a compound is +1; the oxidation number of an alkaline earth metal (IIA family) in a compound is +2.
For oxygen in compounds: The oxidation number of oxygen in a compound is usually –2. If, however, the oxygen is in a class of compounds called peroxides (for example, hydrogen peroxide, or H2O2), then the oxygen has an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1.
For hydrogen in compounds: The oxidation state of hydrogen in a compound is usually +1. If the hydrogen is part of a binary metal hydride (compound of hydrogen and some metal), then the oxidation state of hydrogen is –1.
For halogens: The oxidation number of fluorine is always –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they’re in combination with an oxygen or fluorine. (For example, in ClO–, the oxidation number of oxygen is –2 and the oxidation number of chlorine is +1; remember that the sum of all the oxidation numbers in ClO– has to equal –1.)
These rules give you another way to define oxidation and reduction — in terms of oxidation numbers. For example, consider this reaction, which shows oxidation by the loss of electrons:
Zn(s) → Zn2+ + 2e–
Notice that the zinc metal (the reactant) has an oxidation number of zero (the first rule), and the zinc cation (the product) has an oxidation number of +2 (the second rule). In general, you can say that a substance is oxidized when there’s an increase in its oxidation number.
Reduction works the same way. Consider this reaction:
Cu2+ + 2e– → Cu(s)
The copper is going from an oxidation number of +2 to zero. A substance is reduced if there’s a decrease in its oxidation number.
Balancing Redox Equations
Redox equations are often so complex that the inspection method (the fiddling-with-coefficients method) of balancing chemical equations doesn’t work well with them. (See Chapter 7 for a discussion of this balancing method.) So chemists developed other methods of balancing redox equations, such as the ion electron (half-reaction) method.
Here’s an overview of how it works: You convert the unbalanced redox equation to the ionic equation and then break it down into two half-reactions — oxidation and reduction. Balance each of these half-reactions separately and then combine them to give the balanced ionic equation. Finally, put the spectator ions into the balanced ionic equation, converting the reaction back to the molecular form. (For a discussion of molecular, ionic, and net-ionic equations, see Chapter 7.)
To successfully balance redox equations with the ion electron method, you need to follow the steps precisely and in order. Here’s what to do:
1. Convert the unbalanced redox reaction to the ionic form.
Suppose you want to balance this redox equation:
Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO(g) + H2O(l)
In this reaction, you show the nitric acid in the ionic form, because it’s a strong acid (for a discussion of strong acids, see Chapter 11). Copper(II) nitrate is soluble (indicated by (aq)), so show it in its ionic form (see Chapter 7). Because NO(g) and water are molecular compounds, they remain in the molecular form:
Cu(s) + H+ + NO3– → Cu2+ + 2 NO3– + NO(g) + H2O(l)
2. If necessary, assign oxidation numbers; then write two half-reactions (oxidation and reduction) that show the chemical species that have had their oxidation numbers changed.
In some cases, telling what’s been oxidized and reduced is easy; in other cases, it isn’t. Start by going through the example reaction and assigning oxidation numbers (see the earlier section “Oxidation numbers” for details):
Cu(s) + H+ + NO3– → Cu2+ + 2 NO3– + NO(g) + H2O(l)
0 +1 +5 (–2)3 +2 +5 (–2)3 +2 –2 (+1)2 –2
To write your half-reactions, look closely for places where the oxidation numbers have changed, and write down those chemical species. Copper changes its oxidation number (from 0 to 2) and so does nitrogen (from –5 to +2), so your unbalanced half-reactions are
Cu(s) → Cu2+
NO3– → NO
3. Balance all atoms, with the exception of oxygen and hydrogen.
Starting with the two unbalanced half-reactions above, you can balance atoms other than oxygen and hydrogen by inspection — fiddling with the coefficients. (You can’t change subscripts; you can only add coefficients.) In this case, both the copper and nitrogen atoms already balance, with one each on both sides:
Cu(s) → Cu2+
NO3– → NO
4. Balance the oxygen atoms.
How you balance these atoms depends on whether you’re dealing with acid or basic solutions:
• In acidic solutions, take the number of oxygen atoms needed and add that same number of water molecules to the side that needs oxygen.
• In basic solutions, add two OH– to the side that needs oxygen for every oxygen atom that is needed. Then, to the other side of the equation, add half as many water molecules as OH– anions used.
An acidic solution shows some acid or H+; a basic solution has an OH– present. The example equation is in acidic conditions (nitric acid, HNO3, which is H+ + NO3– in ionic form). You don’t have to do anything on the half-reaction involving the copper, because no oxygen atoms are present. But you do need to balance the oxygen atoms in the second half-reaction:
Cu(s) → Cu2+
NO3– → NO + 2 H2O
5. Balance the hydrogen atoms.
Again, how you balance these atoms depends on whether you’re dealing with acid or basic solutions:
• In acidic solutions, take the number of hydrogen atoms needed and add that same number of H+ to the side that needs hydrogen.
• In basic solutions, add one water molecule to the side that needs hydrogen for every hydrogen atom that’s needed. Then, to the other side of the equation, add as many OH– anions as water molecules used.
The example equation is in acidic conditions. You need to balance the hydrogen atoms in the second half-reaction:
Cu(s) → Cu2+
4 H+ + NO3– → NO + 2 H2O
6. Balance the ionic charge on each half-reaction by adding electrons.
Cu(s) → Cu2+ + 2 e– (oxidation)
3 e– + 4 H+ + NO3– → NO + 2 H2O (reduction)
The electrons should end up on opposite sides of the equation in the two half-reactions. Remember that you’re using ionic charge, not oxidation numbers.
7. Balance electron loss with electron gain between the two half-reactions.
The electrons that are lost in the oxidation half-reaction are the same electrons that are gained in the reduction half-reaction, so the number of electrons lost and gained must be the same. But Step 6 shows a loss of two electrons and a gain of three. So you must adjust the numbers using appropriate multipliers for both half-reactions. In this case, you have to find the lowest common multiple of 2 and 3. It’s 6, so multiply the first half-reaction by 3 and the second half-reaction by 2.
3 × [Cu(s) → Cu2+ + 2 e–] = 3 Cu(s) → 3 Cu2+ + 6 e–
2 × [3 e– + 4 H+ + NO3– → NO + 2 H2O] = 6 e– + 8 H+ + 2 NO3– → 2 NO + 4 H2O
8. Add the two half-reactions together and cancel anything common to both sides.
The electrons should always cancel (the number of electrons should be the same on both sides).
3 Cu + 6 e– + 8 H+ + 2 NO3– → 3 Cu2+ + 6 e– + 2 NO + 4 H2O
9. Convert the equation back to the molecular form by adding the spectator ions.
If it’s necessary to add spectator ions (ions not involved in the reaction, but there to ensure electrical neutrality – Chapter 7) to one side of the equation in order to convert it back to the molecular equation, add the same number to the other side of the equation. For example, there are eight H+ on the left side of the equation. In the original equation, the H+ was in the molecular form of HNO3. You need to add the NO3– spectator ions back to it. You already have 2 on the left, so simply add 6 more. You then add 6 NO3– to the right-hand side to keep things balanced. Those are the spectator ions that you need for the Cu2+ cation to convert it back to the molecular form that you want.
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
10. Check to make sure that all the atoms are balanced, all the charges are balanced (if working with an ionic equation at the beginning), and all the coefficients are in the lowest whole-number ratio.
That’s how it’s done. Reactions that take place in bases are just as easy, as long as you follow the rules.
Exploring Electrochemical Cells
Redox reactions sometimes involve direct electron transfer, in which one substance immediately picks up the electrons another has lost. For instance, if you put a piece of zinc metal into a copper(II) sulfate solution, zinc gives up two electrons (becomes oxidized) to the Cu2+ ion that accepts the electrons (reducing it to copper metal). The copper metal begins spontaneously plating out on the surface of the zinc. The equation for the reaction is
Zn(s) + Cu2+ → Zn2+ + Cu
But if you separate those two half-reactions so that when the zinc is oxidized, the electrons it releases are forced to travel through a wire to get to the Cu2+, you get something useful: a galvanic or voltaic cell, a redox reaction that produces electricity. In this section, I show you how that Zn/Cu2+ reaction may be separated out so that you have an indirect electron transfer and can produce some useable electricity. I also show you how electrolytic cells do the reverse, using electricity to cause a redox reaction. Finally, you see how rechargeable batteries both generate electricity and cause chemical reactions.
Galvanic cells: Getting electricity from chemical reactions
Galvanic cells use redox reactions to produce electricity. These cells are commonly called batteries, but sometimes this name is somewhat incorrect, because a battery is composed of two or more cells connected together. You put a battery in your car, but you put a cell into your flashlight.
A Daniell cell is a type of galvanic cell that uses the Zn/Cu2+ reaction to produce electricity. In the Daniell cell, a piece of zinc metal is placed in a solution of zinc sulfate in one container, and a piece of copper metal is placed in a solution of copper(II) sulfate in another container. These strips of metal are called the cell’s electrodes. They act as a terminal, or a holding place, for electrons.
A wire connects the electrodes, but nothing happens until you put a salt bridge between the two containers. The salt bridge, normally a U-shaped hollow tube filled with a concentrated salt solution, provides a way for ions to move from one container to the other to keep the solutions electrically neutral. It’s like running only one wire up to a ceiling light; the light won’t work unless you put in a second wire to complete the circuit.
With the salt bridge in place, electrons can start to flow through the same basic redox reaction as the one I show you at the beginning of this section. Zinc is being oxidized, releasing electrons that flow through the wire to the copper electrode, where they’re available for the Cu2+ ions to use in forming copper metal. Copper ions from the copper(II) sulfate solution are being plated out on the copper electrode, while the zinc electrode is being consumed. The cations in the salt bridge migrate to the container with the copper electrode to replace the copper ions being consumed, while the anions in the salt bridge migrate toward the zinc side, where they keep the solution containing the newly formed Zn2+ cations electrically neutral.
The zinc electrode is called the anode, the electrode at which oxidation takes place, and it’s labeled with a – sign. The copper electrode is called the cathode, the electrode at which reduction takes place, and it’s labeled with a + sign.
This Daniell cell produces a little over 1 volt. You can get just a little more voltage if you make the solutions that the electrodes are in very concentrated. But what can you do if you want, for example, 2 volts? You have a couple of choices: You can hook two of these cells up together and produce 2 volts, or you can choose two different metals that are farther apart than zinc and copper on the activity series chart (see Chapter 7). The farther apart the metals are on the activity series, the more voltage the cell produces.
Electrolytic cells: Getting chemical reactions from electricity
An electrolytic cell uses electricity to produce a desired redox reaction. For instance, water can be decomposed by the use of electricity in an electrolytic cell. The overall cell reaction is
2 H2O(l) → 2 H2(g) + O2(g)
In a similar fashion, you can produce sodium metal and chlorine gas by the electrolysis of molten sodium chloride.
Producing chemical changes by passing an electric current through an electrolytic cell is called electrolysis. This reaction may be the recharging of a battery (as you see the next section) or one of many other applications. For instance, ever wonder how the aluminum in that aluminum can is mined? Aluminum ore is primarily aluminum oxide (Al2O3). People produce aluminum metal by reducing the aluminum oxide in a high-temperature electrolytic cell using approximately 250,000 amps. That’s a lot of electricity. Taking old aluminum cans, melting them down, and reforming them into new cans is far cheaper than extracting the metal from the ore. That’s why the aluminum industry is strongly behind the recycling of aluminum. It’s just good business.
Electrolytic cells are also used in a process called electroplating. In electroplating, a more-expensive metal is plated (deposited in a thin layer) onto the surface of a cheaper metal by electrolysis. Back before plastic auto bumpers became popular, chromium metal was electroplated onto steel bumpers. Those five-dollar gold chains you can buy are really made of some cheap metal with an electroplated surface of gold.
Having it both ways with rechargeable batteries
Redox reactions can be reversed to regenerate the original reactants, allowing people to make rechargeable batteries. Nickel-cadmium (Ni-Cad) and lithium batteries fall into this category, but the most familiar type of rechargeable battery is probably the automobile battery.
The ordinary automobile battery, or lead storage battery, consists of six cells connected in series. The anode of each cell (where oxidation takes place) is lead, and the cathode (where reduction takes place) is lead dioxide (PbO2). The electrodes are immersed in a sulfuric acid (H2SO4) solution. When you start your car, the following cell reactions take place:
Anode: Pb(s) + H2SO4(aq) → PbSO4(s) + 2 H+ + 2 e–
Cathode: 2 e– + 2 H+ + PbO2(s) + H2SO4(aq) → PbSO4(s) + 2 H2O(l)
Overall reaction: Pb(s) + PbO2(s) + 2 H2SO4(aq) → 2 PbSO4 + 2 H2O(l)
When this happens, both electrodes become coated with solid lead(II) sulfate, and the sulfuric acid is used up.
After you start the automobile, the alternator or generator takes over the job of producing electricity (for spark plugs, lights, and so on) and also recharges the battery. During charging, the automobile battery acts like an electrolytic cell. The alternator reverses both the flow of electrons into the battery and the original redox reactions, and it regenerates the lead and lead dioxide:
2 PbSO4(s) + 2 H2O(l) → Pb(s) + PbO2(s) + 2 H2SO4(aq)
The lead storage battery can be discharged and charged many times. But the shock of running over bumps in the road or into the curb flakes off a little of the lead(II) sulfate and eventually causes the battery to fail.