1.7: Ions in Solution

Although salts are hard to melt and even harder to vaporize, many can be dissolved easily in a polar liquid such as water. The reason for this is simple. The water molecules help to dismantle the salt crystal, since the partial positive and negative charges on the polar water molecules (Figure 1-4) provide a substitute for the positive and negative charges that were present in the crystal lattice. Figure 1-7 illustrates what happens when a crystal such as sodium chloride is dissolved in water. Each positively charged Na⁺ is surrounded by water molecules with their negatively charged oxygens turned toward it, and each negatively charged Cl^- ion is surrounded by water molecules with their positively charged hydrogens closest. The ions from the salt crystal are said to be hydrated. If the stability that hydration gives the ions in solution is greater than the stability of the crystal lattice, then the salt will dissolve. Sodium chloride is a familiar example of a soluble salt. In contrast, if the hydration energy is too small, then the crystal will be the more stable form, and it will not dissolve in water. Silver chloride (AgCl) and barium sulfate (BaS0₄) are examples of insoluble salts. When a salt crystal dissolves, it does not simply come apart into ions; it is taken apart by the molecules of the liquid in which it is dissolved (the solvent). This is why salts will not dissolve in nonpolar liquids such as gasoline (octane, C₈H₁₈); there are no charges on the solvent molecules to make up for the loss of charge attractions within the crystalline salt.

Figure 1-7 Breakup of a salt crystal by water molecules. with hydration of ions. Each salt ion in solution is surrounded by polar water molecules with the opposite charge to that of the ion turned toward it. This electrostatic hydration energy compensates for the loss of attractions between ions in the salt crystal. From Dickerson and Geis. Chemistry, Matter, and the Universe.

Salt solutions conduct electricity, and this property was extremely important early in the development of theories of chemical bonding. Electrical conduction in metals takes place by means of moving electrons; the metal ions remain in place. Crystalline salts do not conduct electricity at all, but if the salt is melted, then positive ions can migrate one way through the liquid and negative ions can move the other way in the presence of an electric field. This mobility of ions is even greater if the salt is dissolved in water and the ions consequently are hydrated.

Some of the first concrete ideas about the nature of chemical bonding came from the electrolysis experiments of the English scientist Michael Faraday (1791-1867). (Electrolysis means "breaking apart with electricity.") If sodium chloride is melted (above 801°C) and if two electrodes (the cathode and the anode) are inserted into the melt as shown in Figure 1-8 and an electric current is passed through the molten salt, then chemical reactions take place at the electrodes: Sodium ions migrate to the cathode, where electrons enter the melt, and are reduced to sodium metal:

Figure 1-8 A commercial electrolysis cell for the production of metallic sodium and chlorine gas from molten NaCl. Liquid sodium floats to the top of the melt above the cathode and is drained off into a storage tank. Chlorine gas bubbles out of the melt above the anode. From Dickerson and Geis. Chemistry. Matter. and the Universe.

Figure 1-9 Schematic diagram of an electrolysis cell. For current to be carried , the fluid must contain mobile ions, either as a molten salt or as hydrated ions in solution , A substance capable of carrying current by migration of ions is called an electrolyte. If the electrolyte is a solution of CuCl₂ , which dissociates (breaks apart) to give Cu²⁺ and CI^- ions, then as current is passed through the cell. Cu²⁺ ions migrate to the cathode and are reduced to metallic copper, and CI^- ions migrate to the anode, where they are oxidized to Cl₂ gas. Platinum electrodes are used because they are chemically inert and will not react.

\[Na^+ + e^-(from\; cathode) → Na\]

Chloride ions migrate the other way, toward the anode, give up their electrons to the anode, and are oxidized to chlorine gas:

\[Cl^- → \frac{1}{2} Cl_{2} + e^-(to\; anode)\]

The overall reaction is the breakdown of sodium chloride into its elements:

\[Na^+ + CI^- → Na + \frac{1}{2}Cl_2\]

Sodium ions are reduced and chloride ions are oxidized. Electrolysis can also be carried out by passing electric current through solutions of salts (Figure 1-9). If a solution of sodium chloride in water is electrolyzed, chlorine gas is given off at the anode as in the case of molten sodium chloride, but the cathode product is hydrogen gas rather than metallic sodium:

\[Na^+ + Cl^- + H_2O \rightarrow Na + \frac{1}{2} Cl_2 + \frac{1}{2} H_2 + OH^- \label{1-1}\]

This is the same result that would be obtained if liquid sodium chloride was first electrolyzed to give metallic sodium:

\[Na^+ + Cl^- → Na + \frac{1}{2} Cl_2 \label{1-2}\]

and the sodium was then dumped into water:

\[Na + H_2O → Na^+ + \frac{1}{2}H_2 + OH^- \label{1-3}\]

Equation \(\ref{1-1}\) is just the sum of Equations \(\ref{1-2}\) and \(\ref{1-3}\), since the sodium metal that is produced in Equation \(\ref{1-2}\) is used up in Equation \(\ref{1-3}\). There is nothing mysterious about the different cathode products during electrolysis of sodium chloride in a melt or in solution, If water is present, some of the H₂O molecules will be dissociated into H⁺ and OH^- ions. Because H⁺ has a greater affinity for electrons than Na+ does, the H+ ions will take electrons away from metallic sodium, making the anode product H₂ rather than Na, and leaving Na⁺ ions in solution. In contrast, Cu²⁺ ions have a greater affinity for electrons than H⁺ ions do, so the anode product of electrolysis of CuCl₂ is metallic copper, whether the process is carried out in the melt or in solution (Figure 1-9). Typical products of electrolysis of solutions and melts are given in Table 1-8. Electrochemical reactions and cells are discussed in detail in Chapter 19. At the moment, we are focusing on what electrochemical reactions tell us about chemical bonding.

Table 1-8. Products of Electrolysis

Electrolytett	Cathode Producttt	Anode Product
Sulfuric acid (H2SO4) in H2Ott	H2tt	O2
Sodium sulfate (Na2SO4) in H2Ott	H2tt	O2
Sodium chloride (NaCl) in H2Ott	H2tt	Cl2
Potassium iodide (Kl) in H20tt	H2tt	I2
Copper sulfate (Cu2SO4) in H20tt	Cutt	O2
Silver nitrate (AgNO3) in H2Ott	Agtt	O2
Mercuric nitrate [Hg(NO3)2] in H20tt	Hgtt	O2
Lead nitrate [Pb(NO3)2] in H20tt	Pbtt	O2 and some PbO2
Molten lye (NaOH); not in H2Ott	Natt	O2

Faraday found that there was a quantitative relationship between the amount of electricity passed through an electrolytic cell and the amount of chemical change produced. He formulated Faraday's laws of electrolysis, which in terms of the modern theory of atoms and ions can be expressed as follows:

1. Passing the same quantity of electricity through a cell always leads to the same amount of chemical change for a given reaction. The weight of an element deposited or liberated at an electrode is proportional to the amount of electricity that is passed through.
2. It takes 96,485 coulombs of electricity to deposit or liberate 1 mole of a substance that gains or loses one electron during the cell reaction. If n electrons are involved in the reaction, then 96,485n coulombs of electricity are required to liberate a mole of product.

The quantity 96,485 coulombs of electricity has become known as 1 faraday in his honor, and has been given the symbol . Faraday's laws become self-evident when you realize that 1 is simply the charge on 1 mole of electrons, or 6.022 X 10²³ electrons. The scale-up factor of 6.022 X 10²³ from molecules to moles is paralleled by the same scale-up factor from 1 electron charge to 1 of charge. At the time, of course, Faraday knew neither the value of Avogadro's number nor the charge on an electron. His experiments did tell him, however, that charges on ions came in multiples of a fundamental unit, such that 96,485 coulombs corresponded to a mole of these units. The word electron first appeared in 1881, when the British physicist G. J. Stoney coined it to denote this fundamental unit of ionic charge. Its application to a real negatively charged particle came a decade later.

Example 16

Write equations for the reactions that occur when current is passed through molten NaCl. How many grams of sodium and chlorine are released when 1 of charge is passed through the cell?

Solution The cathode reaction is Na^+ + e^- → Na, and the anode reaction is Cl^-→ Cl2 + e^-. When 1 mole of electrons (1) passes through molten NaCI, each electron reduces one sodium ion, so 1 mole of sodium atoms is produced. Hence 22.990 g of Na are deposited at the cathode. At the anode, 1 mole of electrons is removed from 1 mole of chloride ions, leaving 1 mole of chlorine atoms, which combine pairwise to make mole of Cl2 molecules. Hence the weight of chlorine gas released is 35.453 g (the atomic weight of Cl, half the molecular weight of Cl2).

Example 17

How many grams of magnesium metal and chlorine gas are released when 1 of electricity is passed through an electrolytic cell containing molten magnesium chloride, MgC12?

Solution The cathode reaction is Mg^2+ + 2e^- → Mg, and the anode reaction is this: 2Cl^-→ Cl2 + 2e^-. Since two electrons are required to reduce each ion of Mg^2+, 1 mole of electrons will be sufficient to reduce half a mole of magnesium ions, depositing 12.153 g of magnesium. (The atomic weight of magnesium is 24.305 g mole^-1.) As in Example 16, 1 mole of Cl^- ions is oxidized, liberating half a mole or 35.453 g of Cl2 gas.

Example 18

The main commercial source of aluminum metal is the electrolysis of molten salts of Al^3+. How many faradays of charge, and how many coulombs, must be passed through the melt to deposit 1 kg of metal?

Solution One kilogram of aluminum is 1000 g/26.98 g mole^-1, or 37.06 moles. Since each atom of aluminum deposited requires three electrons, 37.06 moles will require 3 X 37.06, or 111.2, moles of electrons. Hence 111.2 or 10,730,000 coulombs will be needed.

Example 19

Electron flow at the rate of 1 coulomb per second (coulomb sec^-1) is a current of 1 ampere (A). Currents in industrial electrolytic production of aluminum are ordinarily in the range of 20,000 to 50,000 A. If a cell is operated at 40,000 A (40,000 coulombs sec^-1), how long will it take to produce the kilogram of aluminum metal mentioned in Example 18?

Solution The time required will be = 268 sec or 4.5 min

Figure 1-10 Illustrations of Faraday's laws of electrolysis (a) Two electrons are required to reduce each ion of Cu²⁺, or 2 moles of electrons (2) for each mole of copper. Each faraday is enough to oxidize 1 mole of Cl^- ions to mole of Cl₂ gas. (b) Only 1 of charge is required to reduce 1 mole of Ag⁺ ions to metallic silver. since the ionic charge on Ag⁺ is only +1. Chlorine gas is liberated at the same rate per faraday as before.

Faraday's laws are represented diagrammatically in Figure 1-10. We have been using these laws with a prior knowledge of the charges on different ions, and the knowledge that 96,485 coulombs is the total charge on 6.022 X 10²³ electrons. History actually operated in reverse: Faraday and others used electrolysis experiments to find out what the charges on ions were. The reasoning used is illustrated in Table 1-9. If twice as much electricity is required to liberate a mole of copper as a mole of silver (assuming that you know the atomic weights of the two metals and can calculate the weights of a mole of each), then the copper ion must have twice the charge of the silver ion. In Table 1-9, the number of faradays of charge required to liberate 1 mole of an element is the same as the number of charges, positive or negative, on the ion.

Table 1-9. Deduction of Ionic Charge by Electrolysis

Product of electrolysis	Electrode	Faradays per mole of atoms deposited	Ion in solution
Silver (Ag)	Cathode	1^a	Ag+
Chlorine (Cl2)	Anode	1	Cl^-
Copper (Cu)	Cathode	2	Cu^2+
Hydrogen (H2)	Cathode	1	H^+
Iodine (I2)	Anode	2	I^-
Oxygen (O2)^b	Anode	2	O^2-
Zinc (Zn)	Cathode	2	Zn^2+

^a For example, electrolysis of silver nitrate solution for 1 hour by using a current of 0.5 A deposits 2.015 g of silver; 2.015/107.9 = 0.0187 mole of silver.

(0.5 coulomb sec^-1) x 3600 sec = 0.0187

96.485 coulombs ^-1

^b Actually, oxygen (0₂) is produced by a complicated electrode reaction. The species O^2- can exist in molten oxides, but in water 0^2- becomes 2OH^- by reaction with a water molecule.

Contributors and Attributions

• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors.