As you work through this section, if you find that you need a bit more background material to help you understand the topics at hand, you can consult “Chemistry: The Molecular Science” (5th ed. Moore and Stanitski) Chapter 3-5, 17-1 and 17-2, and/or Chapter 16.9 and Chapter 17.1-17.3 in the Additional Reading Materials section.

D38.1 Electrochemistry

Electrochemistry deals with chemical reactions that produce electricity and chemical reactions caused by passage of electrical current through matter. The reactions involve transfer of electron density. Such reactions are called oxidation-reduction (or redox) reactions. Many metals may be purified or electroplated using electrochemical methods. Batteries use chemical reactions that produce electricity spontaneously and electricity can do useful work.

The SI unit of electric charge is the coulomb, symbol C. The elementary unit of charge is the charge of an electron, 1.602 × 10⁻¹⁹ C. Movement of electrons or ions carries electric charge from one place to another. Electric current is the quantity of charge transferred per unit time, with SI unit ampere, symbol A, the transfer of one coulomb per second (1 A = 1 C/s). Typically, electric current flows in a closed path, an electric circuit; it is necessary to maintain a closed circuit for current to flow. A flow of charge is caused by an electrical potential difference between two points in the circuit. Electrical potential, SI unit volt, symbol V, is the ability of an electric field to do work on a charge. When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule (J) of energy.

D38.2 Redox Reactions and Oxidation Number

Some redox reactions involve transfer of electrons between reactant atoms to yield ionic products. For example, formation of a crystal lattice of sodium chloride (a lattice of Na⁺ ions and Cl⁻ ions) requires transfer of electrons from sodium atoms to chlorine atoms:

It is useful to represent the electron transfer in two steps, one in which electrons are lost and one in which electrons are gained. These steps are called half-reactions. The half reactions for reaction of sodium with chlorine are

The half-reactions allow us to see that the number of electrons lost equals the number of electrons gained—we cannot create or destroy electrons in a chemical reaction. The half-reactions also show which species gains electrons (Cl atoms in Cl₂) and which species loses electrons (Na atoms). The species that loses electrons is said to be oxidized and the loss of electrons is called oxidation; the species that gains electrons is said to be reduced and the gain of electrons is called reduction.

The species that causes reduction to occur is called the reducing agent (or reductant). In this reaction sodium is the reducing agent because it causes Cl₂ to gain electrons. The species that causes oxidation to occur is called the oxidizing agent (or oxidant). In this reaction chlorine functions as an oxidant, causing sodium to lose electrons.

Some redox processes, however, do not involve obvious transfer of electrons. For example, when the product is a covalent compound, transfer of electrons to form ions is not involved:

To deal with such cases we define oxidation number (or oxidation state) of an element in a compound as the charge its atoms would possess if the compound was ionic, that is, if all electrons in each polar bond were assigned to the more electronegative atom. These stepwise rules can be used to assign oxidation numbers to each element in a compound.

1. 1. 1. The oxidation number of an atom in an elemental substance is zero.
      2. The oxidation number of a monoatomic ion is equal to the ion’s charge.
      3. The sum of oxidation numbers over all atoms in a complete formula for a compound is zero.
      4. The sum of the oxidation numbers over all atoms in a polyatomic ion equals the ion’s charge. If a compound includes more than one polyatomic ion, the oxidation number of an atom in one ion can differ from the oxidation number in the other polyatomic ion.
      5. Atoms of some elements have the same oxidation number in almost all compounds.
         1. Fluorine always has oxidation number −1 in its compounds
         2. Atoms of alkali metals (Li, Na, K, Rb, Cs) have oxidation number +1 in nearly all compounds
         3. Atoms of alkaline earth metals (Be, Mg, Ca, Sr, Ba) have oxidation number +2 in nearly all compounds
         4. Hydrogen has oxidation number +1 when combined with nonmetals, −1 when combined with metals
      6. Apply these two rules only if rules 1-5 have not determined all oxidation numbers.
         1. Oxygen has oxidation number −2 unless rules 1-5 have already given O a different oxidation number
         2. In binary compounds of nonmetals, the more electronegative element is given a negative oxidation number equal to the charge on its monoatomic ion; for example in PCl₃, the more electronegative Cl is assigned oxidation number −1  and P is assigned +3 (by rule 3).

Using oxidation numbers we can recognize oxidation-reduction (redox) reactions by looking for one or more elements whose oxidation numbers change. When its oxidation number increases, an element has been oxidized; when its oxidation number decreases, the element has been reduced.

In the 2 Na(s) + Cl₂(g) ⟶ 2 NaCl(s) reaction, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in NaCl). In the H₂(g) + Cl₂(g) ⟶ 2 HCl(g) reaction, hydrogen is oxidized (its oxidation number increases from 0 in H₂ to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in HCl).

Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (fuel) and oxidant (often, O₂) react vigorously and produce significant quantities of heat, often in the form of a flame. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:

D38.3 Balancing Redox Reactions

Chemical equations for oxidation-reduction reactions are often difficult to balance. Splitting the process into half-reactions makes it easier because you can balance the oxidation “half” separately from the reduction “half”, make certain that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction, and then sum the half-reaction to find the overall equation. Redox reactions frequently occur in solutions, which can be acidic, basic, or neutral. Because they are available in solution, both H₃O⁺ ions and H₂O(l) molecules may be reactants or products in acidic solution. Both OH⁻ ions and H₂O(l) molecules may be reactants or products in basic solution.

Acidic Solution

We will balance an example redox reaction in acidic solution and one in basic solution to highlight how the nature of the solution can play a role. (A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.) Consider the unbalanced reaction:

Here, Fe²⁺ has lost an electron to become Fe³⁺; therefore, the iron underwent oxidation. The manganese underwent reduction; it gained five electrons to change from an oxidation state of +7 to an oxidation state of +2.

It is a useful simplification to represent H₃O⁺ ions as H⁺(aq). In the example reaction, the reduction half-reaction involves oxygen, and we can use H⁺(aq) and H₂O(l) to balance the elements O and H. Because there are four O atoms on the left side and none on the right, add 4 H₂O(l) to the right. Then, to balance H atoms, add 8 H⁺(aq) on the left:

Once the atoms have been balanced we need to make certain that the total electric charge is the same on each side of each half-reaction. The iron atoms in the oxidation half-reaction are already balanced, but there are two positive charges on the left and three on the right, so charge is unbalanced. We use electrons to balance the charge. Adding a single electron on the right side gives a balanced oxidation half-reaction:

You should check the half-reaction for the number of each atom type and the total charge on each side of the equation. The charges include the actual charges of the ions times the number of ions and the charge on an electron times the number of electrons.

If the atoms and charges balance, the half-reaction is balanced.

In oxidation half-reactions, electrons appear on the right side of the equation. Because iron was oxidized, iron is the reducing agent.

For the reduction half-reaction, we have balanced the atoms but we need to check for charge balance. The total charge on the reactant-side is (1 × (−1)) + (8 × (+1)) = +7; the total charge on the product-side is (1 × (+2)) + (4 × (0)) = +2. Therefore, it is necessary to add five negative charges, that is, five electrons, to the left side to achieve charge balance:

You should always do a final check that the half-reaction is balanced for each atom type and the charge.

Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when MnO₄⁻ was reduced to Mn²⁺. In all reduction half-reactions, electrons appear on the left side. The species that was reduced, MnO₄⁻, is the oxidizing agent.

We now have two balanced half-reactions:

The key to combining the half-reactions is the electrons: the number of electrons transferred from the oxidation half-reaction must equal the number of electrons transferred to the reduction half-reaction. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. Multiplying each half-reaction by the appropriate factor and summing the resulting half-reactions gives:

This is the overall balanced equation in acidic solution. Electrons do not appear in the equation because all electrons lost in the oxidation are gained in the reduction. On each side of the overall equation there should be the same number of atoms of each kind and the same total electric charge. Be sure to carefully check each side to verify everything has been balanced correctly.

Basic Solution

In a basic solution, there is an excess of OH⁻ ions instead of H₃O⁺ ions. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. For example, balance this reaction in basic solution:

[latex]\text{MnO}_4^{-}(\text{aq})\;+\;\text{Cr(OH)}_3(\text{s})\;{\longrightarrow}\;\text{MnO}_2(\text{s})\;+\;\text{CrO}_4^{2-}(\text{aq})[/latex]

Start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction.

[latex]\begin{array}{rrcl} \text{oxidation (unbalanced):} & \text{Cr(OH)}_3(\text{s}) &\longrightarrow& \text{CrO}_4^{2-}(\text{aq}) \\[0.5em] \text{reduction (unbalanced):} & \text{MnO}_4^{-}(\text{aq}) &\longrightarrow& \text{MnO}_2(\text{s}) \end{array}[/latex]

Starting with the oxidation half-reaction, note that chromium atoms are already balanced:

[latex]\text{oxidation (unbalanced):}\;\text{Cr(OH)}_3(\text{s})\;{\longrightarrow}\;\text{CrO}_4^{2-}(\text{aq})[/latex]

If this were an acidic solution, we could use H⁺(aq) and H₂O(l) to balance H and O. Adding one water molecule to the left side provides the necessary oxygen; to balance H we need five H⁺(aq) on the right side:

[latex]\text{oxidation (unbalanced):}\;\text{Cr(OH)}_3(\text{s})\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{CrO}_4^{2-}(\text{aq})\;+\;5\;\text{H}^{+}(\text{aq})[/latex]

The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 × (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

[latex]\text{oxidation (balanced, acidic): Cr(OH)}_3(\text{s})\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{CrO}_4^{2-}(\text{aq})\;+\;5\text{H}^{+}(\text{aq})\;+\;3\text{e}^{-}[/latex]

Checking the half-reaction:

[latex]\text{Cr: Does}\;(1\;\times\;1) = (1\;\times\;1)\text{? Yes.} \\[0.5em] \text{H: Does}\;(1\;\times\;3\;+\;1\;\times\;2) = (5\;\times\;1)\text{? Yes.} \\[0.5em] \text{O: Does}\;(1\;\times\;3\;+\;1\;\times\;1) = (4\;\times\;1)\text{? Yes.} \\[0.5em] \text{Charge: Does}\;[0 = [1\;\times\;(-2)\;+\;5\;\times\;(+1)\;+\;3\;\times\;(-1)]\text{? Yes.}[/latex]

Now work on the reduction. On the left side there are four O atoms in the MnO₄⁻ ; on the right side there are two O atoms in MnO₂. To balance O we need two water molecules on the right. Then to balance H we need to add four H⁺(aq) on the left.

[latex]\text{reduction (unbalanced):}\;\text{MnO}_4^{-}(\text{aq})\;+\;4\;\text{H}^{+}(\text{aq})\;{\longrightarrow}\;\text{MnO}_2(\text{s})\;+\;2\;\text{H}_2\text{O}(l)[/latex]

Then add three electrons to the left side to balance the charge:

[latex]\text{reduction (balanced, acidic):}\;\text{MnO}_4^{-}(\text{aq})\;+\;4\;\text{H}^{+}(aq)\;+\;3\;\text{e}^{-}\;{\longrightarrow}\;\text{MnO}_2(\text{s})\;+\;2\;\text{H}_2\text{O}(l)[/latex]

Make sure to check the half-reaction:

[latex]\text{Mn: Does}\;(1\;\times\;1) = (1\;\times\;1)\text{? Yes.} \\[0.5em] \text{H: Does}\;(4\;\times\;1) = (2\;\times\;2)\text{? Yes.} \\[0.5em] \text{O: Does}\;(1\;\times\;4) = (1\;\times\;2\;+\;2\;\times\;1)\text{? Yes.} \\[0.5em] \text{Charge: Does}\;[1\;\times\;(-1)\;+\;4\;\times\;(+1)\;+\;3\;\times\;(-1)] = [0]\text{? Yes.}[/latex]

Collecting what we have so far:

[latex]\begin{array}{lrcl} \text{oxidation:} & \text{Cr(OH)}_3(\text{s})\;+\;\text{H}_2\text{O}(l) &\longrightarrow& \text{CrO}_4^{2-}(aq)\;+\;5\;\text{H}^{+}(aq)\;+\;3\;\text{e}^{-} \\[0.5em] \text{reduction:} & \text{MnO}_4^{-}(aq)\;+\;4\;\text{H}^{+}(aq)\;+\;3\;\text{e}^{-} &\longrightarrow& \text{MnO}_2(\text{s})\;+\;2\;\text{H}_2\text{O}(l) \end{array}[/latex]

In this case, both half reactions involve the same number of electrons; therefore, simply add the two half-reactions together and simplify.

[latex]\text{MnO}_4^{-}(\text{aq})\;+\;4\;\text{H}^{+}(\text{aq})\;+\;\text{Cr(OH)}_3(\text{s})\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\\[0.5em] \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{CrO}_4^{2-}(\text{aq})\;+\;\text{MnO}_2(\text{s})\;+\;2\;\text{H}_2\text{O}(l)\;+\;5\;\text{H}^{+}(\text{aq})[/latex]
[latex]\text{MnO}_4^{-}(\text{aq})\;+\;\text{Cr(OH)}_3(\text{s})\;{\longrightarrow}\;\text{CrO}_4^{2-}(\text{aq})\;+\;\text{MnO}_2(\text{s})\;+\;\text{H}_2\text{O}(l)\;+\;\text{H}^{+}(\text{aq})[/latex]

Checking each side of the equation:

[latex]\text{Mn: Does}\;(1\;\times\;1) = (1\;\times\;1)\text{? Yes.} \\[0.5em] \text{Cr: Does}\;(1\;\times\;1) = (1\;\times\;1)\text{? Yes.} \\[0.5em] \text{H: Does}\;(1\;\times\;3) = (2\;\times\;1\;+\;1\;\times\;1)\text{? Yes.} \\[0.5em] \text{O: Does}\;(1\;\times\;4\;+\;1\;\times\;3) = (1\;\times\;4\;+\;1\;\times\;2\;+\;1\;\times\;1)\text{? Yes.} \\[0.5em] \text{Charge: Does}\;[1\;\times\;(-1)] = [1\;\times\;(-2)\;+\;1\;\times\;(+1)]\text{? Yes.}[/latex]

This is the balanced equation in acidic solution. To “convert” to basic solution, add enough OH⁻(aq) to each side of the equation to react with all the H⁺(aq) on one side. This converts H⁺(aq) to H₂O(l) on one side and adds OH⁻(aq) on the other side. Then simplify:

[latex]\text{OH}^{-}(\text{aq})\;+\;\text{MnO}_4^{-}(\text{aq})\;+\;\text{Cr(OH)}_3(\text{s})\;{\longrightarrow}\;\text{CrO}_4^{2-}(\text{aq})\;+\;\text{MnO}_2(\text{s})\;+\;\text{H}_2\text{O}(l)\;+\;(\text{H}^{+}\;+\;\text{OH}^{-})(\text{aq})[/latex]
[latex]\text{OH}^{-}(\text{aq})\;+\;\text{MnO}_4^{-}(\text{aq})\;+\;\text{Cr(OH)}_3(\text{s})\;{\longrightarrow}\;\text{CrO}_4^{2-}(\text{aq})\;+\;\text{MnO}_2(\text{s})\;+\;2\text{H}_2\text{O}(l)[/latex]

Checking each side of the equation:

[latex]\text{Mn: Does}\;(1\;\times\;1) = (1\;\times\;1)\text{? Yes.} \\[0.5em] \text{Cr: Does}\;(1\;\times\;1) = (1\;\times\;1)\text{? Yes.} \\[0.5em] \text{H: Does}\;(1\;\times\;1\;+\;1\;\times\;3) = (2\;\times\;2)\text{? Yes.} \\[0.5em] \text{O: Does}\;(1\;\times\;1\;+\;1\;\times\;4\;+\;1\;\times\;3) = (1\;\times\;4\;+\;1\;\times\;2\;+\;2\;\times\;1)\text{? Yes.} \\[0.5em] \text{Charge: Does}\;[1\;\times\;(-1)\;+\;1\;\times\;(-1)] = [1\;\times\;(-2)]\text{? Yes.}[/latex]

This is the balanced equation in basic solution.

D38.4 Introduction to Voltaic Cells

A voltaic cell (or galvanic cell) is an electrochemical cell in which a product-favored redox reaction produces an electric current. Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate (Figure 1). As soon as the copper metal contacts the solution, silver metal begins to form on the copper surface and copper ions pass into the solution. The blue-green color of the solution on the far right of Figure 1 indicates the presence of copper ions. Thus the net ionic equation for the reaction is

Cu(s)  +  2 Ag⁺(aq)  →  Cu²⁺(aq)  +  2 Ag(s)

The redox reaction may be split into its two half-reactions that sum to the overall reaction:

The half-reactions make clear that two electrons are transferred from a copper atom to form a copper(II) ion. Each of the two electrons is transferred to one silver(I) ion to form a silver atom. Because a flow of electrons constitutes an electric current, this electron transfer could generate an electric current if we could devise a way to carry out one half-reaction in one place and connect it with a wire to the other half-reaction in a different place. This is how a voltaic cell is designed. Figure 2 shows one way to do this.

The beaker on the left contains a 1-M solution of copper(II) nitrate [Cu(NO₃)₂] with a strip of copper metal partially submerged in the solution. The copper is an electrode, a means for conducting electrons into or out of the solution. At the surface of the copper strip copper atoms lose electrons to form copper(II) ions that go into solution. The electrons are conducted out of the solution by the copper strip and move through the wire. Notice that copper, copper(II) ions, and electrons are in the first half-reaction above: the oxidation half-reaction. The flow of electrons (electric current) passes through the light bulb and moves are toward the silver strip in the beaker on the right. In the right-hand beaker silver(I) ions near the surface of the silver strip gain electrons and form silver atoms, which attach to other silver atoms in the silver strip. Notice that silver(I) ions, electrons, and silver metal are in the second half-reaction above: the reduction half-reaction. Thus, with the two half-reactions in separate beakers, an electric current can be generated. The container in which each half-reactions occurs is called a half-cell.

If this were all that happened, the electric current could not flow for long. In the left-hand beaker one copper(II) ion is added to the solution for every two electrons conducted into the external wire. This means that the solution in the beaker has more positive ions than negative ions and an electric charge would build up on the solution unless either some positive ions move out of the beaker or some negative ions move into the beaker. A similar issue arises in the right-hand beaker, where silver(I) ions are being removed from the solution. Balancing total ionic charges requires either that negative ions move out of the right-hand solution or positive ions move in. This balancing of ion charges in the beaker is maintained by the salt bridge, a solution of a salt that does not mix with either half-cell solution but allows ions to pass into or out of the half-cells. By allowing ions to conduct charge into or out of the half-cells, the salt bridge completes the electrical circuit involving the two half-cells. Without it no current can flow and the bulb cannot light. Notice that if the salt bridge is considered part of an electrical circuit, negative ions in the salt bridge more in the same direction as electrons around the circuit and positive ions move in the opposite direction.

The half-cell in which oxidation occurs is called the anode. The half-cell in which reduction occurs is called the cathode. It is easy to remember that the anode involves oxidation because both words begin with vowels. It is easy to remember that the cathode involves reduction because both words begin with consonants. These definitions, anode/oxidation and cathode/reduction, are true for any electrochemical cell, not just a voltaic cell.