M18Q8: Electrolysis

Learning Objectives

  • Use reduction potentials to predict products of the electrolysis of molten salts and aqueous salts.
    | Molten Salt | Aqueous Salt |
  • Calculate the quantity of product formed at an electrode during an electrolysis reaction or the amount of time required to generate a given amount of a species.
    | Quantitative Electrolysis |

| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |


In galvanic cells, chemical energy is converted into electrical energy. The opposite is true for electrolytic cells. In electrolytic cells, electrical energy causes nonspontaneous redox reactions to occur in a process known as electrolysis.

The same principles are involved in electrolytic cells as in galvanic cells, with two differences:

  • Electrons flow from the anion to the cation.
  • Oxidation occurs at the anode.
  • Reduction occurs at the cathode.
  • The E°cell is negative and indicates a nonspontaneous process. Applying electrical energy to the system allows the nonspontaneous reaction to occur (Note, this is different from galvanic or voltaic cells.)
  • The anode is the POSITIVE electrode and the cathode is the NEGATIVE electrode (Note, this is different from galvanic or voltaic cells.)

When determining the anode and cathode in an electrolytic cell:

  • Identify all species in your system (usually ions and water (if an aqueous system))
  • Determine which can be oxidized and which can be reduced to predict what happens at cathode and anode. Cations tend to not lose additional electrons and are reduced. Likewise, anions tend to not gain additional electrons and are oxidized. Water can be both oxidized and reduced, depending on the conditions.

Another way to analyze this is that the anode with a positive potential will attract negative ions, which will lose electrons as they are oxidized. The cathode with a negative potential will attract cations, which will gain electrons as they are reduced. A generic electrolytic cell can be seen in Figure 1.

Figure 1. This is a general depiction of an electrolytic cell, where electrical energy is applied to force a nonspontaneous redox reaction to occur.

In this section, we will look at three electrolytic cells and the quantitative aspects of electrolysis.

The Electrolysis of Molten Sodium Chloride

Electrolysis of molten ionic compounds, or salts, can be used to recreate the original elements in their non-ion form. A solid ionic compound has cations and anions locked in a crystal lattice structure that will not conduct electricity, however, melting the salt will allow the cations and anions of the salt to move freely and conduct electricity. This means it is possible to apply an electrical current to the molten salt and drive the nonspontaneous redox reactions to form the original elements in their standard state.

In molten sodium chloride, liquid sodium and chloride ions are present. Sodium ions already have a positive charge and are reduced. Likewise, chloride ions already have a negative charge and are oxidized. In the electrolytic cell, the ions are free to migrate to the electrodes. A simplified diagram of the cell commercially used to produce sodium metal and chlorine gas is shown in Figure 2. Sodium is a strong reducing agent and chlorine is used to purify water, and is used in antiseptics and in paper production. The reactions are:

Anode: 2 Cl(ℓ) Cl2(g) + 2 e anode = +1.3 V
Cathode: Na+(ℓ) + e Na(ℓ) cathode = -2.7 V
Overall: 2 Na+(ℓ) + 2 Cl(ℓ) 2 Na(ℓ) + Cl2(g) cell = E°cathode – E°anode = -4.0 V

The power supply (battery) must supply a minimum of 4 V.

This diagram shows a tank containing a light blue liquid, labeled “Molten N a C l.” A vertical dark grey divider with small, evenly distributed dark dots, labeled “Porous screen” is located at the center of the tank dividing it into two halves. Dark grey bars are positioned at the center of each of the halves of the tank. The bar on the left, which is labeled “Anode” has green bubbles originating from it. The bar on the right which is labeled “Cathode” has light grey bubbles originating from it. An arrow points left from the center of the tank toward the anode, which is labeled “C l superscript negative.” An arrow points right from the center of the tank toward the cathode, which is labeled “N a superscript plus.” A line extends from the tops of the anode and cathode to a rectangle centrally placed above the tank which is labeled “Voltage source.” An arrow extends upward above the anode to the left of the line which is labeled “e superscript negative.” A plus symbol is located to the left of the voltage source and a negative sign it located to its right. An arrow points downward along the line segment leading to the cathode. This arrow is labeled “e superscript negative.” The left side of below the diagram is the label “2 C l superscript negative right pointing arrow C l subscript 2 ( g ) plus 2 e superscript negative.” At the right, below the diagram is the label “2 N a superscript plus plus 2 e superscript negative right pointing arrow 2 N a ( l ).”
Figure 2. Passing an electric current through molten sodium chloride decomposes the material into sodium metal and chlorine gas. Care must be taken to keep the products separated to prevent the spontaneous formation of sodium chloride.

The Electrolysis of Water

It is possible to split water into hydrogen and oxygen gas by electrolysis. Acids are typically added to increase the concentration of hydrogen ion in solution (Figure 3). The reactions are:

Anode: 2 H2O(ℓ) O2(g) + 4 H+(aq) + 4 e anode = +1.23 V
Cathode: 2 H+(aq) + 2 e H2(g) cathode = 0 V
Overall: 2 H2O(ℓ) O2(g) + 2 H2(g) cell = E°cathode – E°anode = -1.23 V

The overall voltage is -1.23 V, which indicate a non-spontaneous reaction. A minimum voltage of 1.23 V must be applied to the reaction system in order to cause the electrolysis of water to happen.

This figure shows an apparatus used for electrolysis. A central chamber with an open top has a vertical column extending below that is nearly full of a clear, colorless liquid, which is labeled “H subscript 2 O plus H subscript 2 S O subscript 4.” A horizontal tube in the apparatus connects the central region to vertical columns to the left and right, each of which has a valve or stopcock at the top and a stoppered bottom. On the left, the stopper at the bottom has a small brown square connected just above it in the liquid. The square is labeled “Anode plus.” A black wire extends from the stopper at the left to a rectangle which is labeled “Voltage source” on to the stopper at the right. The left side of the rectangle is labeled with a plus symbol and the right side is labeled with a negative sign. The stopper on the right also has a brown square connected to it which is in the liquid in the apparatus. This square is labeled “Cathode negative.” The level of the solution on the left arm or tube of the apparatus is significantly higher than the level of the right arm. Bubbles are present near the surface of the liquid on each side of the apparatus, with the bubbles labeled as “O subscript 2 ( g )” on the left and “H subscript 2 ( g )” on the right.
Figure 3. Water decomposes into oxygen and hydrogen gas during electrolysis. Sulfuric acid was added to increase the concentration of hydrogen ions and the total number of ions in solution, but does not take part in the reaction. The volume of hydrogen gas collected is twice the volume of oxygen gas collected, due to the stoichiometry of the reaction.

The Electrolysis of Aqueous Sodium Iodide

The electrolysis of aqueous solutions are an interesting example of electrolysis. When the ionic compound dissolves into cations and anions, a solution capable of conducting electricity is produced. The cations, anions and water (solvent) will create multiple possible oxidation and reductions reaction. Ultimately, the oxidation and reduction reactions that occur will be the ones that create the smallest Ecell value .

  • At the anode, either the anion of the salt or the water will be oxidized, whichever is the stronger reducing agent.
  • At the cathode either the cation or the water will be reduced, whichever is the strongest oxidizing agent.

Consider the electrolysis of 1 M aqueous sodium iodide, the sodium iodide dissolves in the water so sodium ions, iodide ions, and water are all present in the solution.

Looking at the anode first, either the iodide ion or the water will be oxidized. The possible reactions are (note: E°red values given from the standard reduction potential table):

(i) 2 I(aq) I2(g) + 2 e anode = +0.54 V
(ii) 2 H2O(ℓ) O2(g) + 4 H+(aq) + 4 e anode = +1.23 V

For electrolysis to occur, voltage must be applied to overcome the negative Ecell. When multiple reactions are possible, the reactions that result in the least negative Ecell will be the ones that occur. This means that the oxidation reaction with the least positive E° will happen. The values above suggest that iodide ions will be oxidized at the anode as using reaction (i) for the oxidation would give a less-negative cell potential.

Now consider the cathode. Two possible reductions could occur:

(iii) 2 H2O(ℓ) + 2 e H2(g)  +  2 OH(aq) cathode = -0.83 V
(iv) Na+(aq) + e Na(s) cathode = -2.71 V

The reduction reaction with the least negative (or most positive) will occur.  This means when voltage is applied to the reaction system, reaction (iii) will occur.

The overall reaction is then:

Anode: 2 I(aq) I2(g) + 2 e anode = +0.54 V
Cathode: 2 H2O(ℓ) + 2 e H2(g) + 2 OH(aq) cathode = -0.83 V
Overall: 2 H2O(ℓ) + 2 I(aq) H2(g) + 2 OH(aq) + I2(g) cell = E°cathode – E°anode = -1.37 V

The overall potential for this process is -1.37 V, which indicates a non-spontaneous process. This means at least +1.37 V of electricity must be applied to the system to cause the reaction to move forward.

Chemistry in Real Life: Electroplating

An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. We can get an idea of how this works by investigating how silver-plated tableware is produced (Figure 4).

This figure contains a diagram of an electrochemical cell. One beakers is shown that is just over half full. The beaker contains a clear, colorless solution that is labeled “A g N O subscript 3 ( a q ).” A silver strip is mostly submerged in the liquid on the left. This strip is labeled “Silver (anode).” The top of the strip is labeled with a red plus symbol. An arrow points right from the surface of the metal strip into the solution to the label “A g superscript plus” to the right. A spoon is similarly suspended in the solution and is labeled “Spoon (cathode).” It is labeled with a black negative sign on the tip of the spoon’s handle above the surface of the liquid. An arrow extends from the label “A g superscript plus” to the spoon on the right. A wire extends from the top of the spoon and the strip to a rectangle labeled “Voltage source.” An arrow points upward from silver strip which is labeled “e superscript negative.” Similarly, an arrow points down at the right to the surface of the spoon which is also labeled “e superscript negative.” A plus sign is shown just outside the voltage source to the left and a negative is shown to its right.
Figure 4. The spoon, which is made of an inexpensive metal, is connected to the negative terminal of the voltage source and acts as the cathode. The anode is a silver electrode. Both electrodes are immersed in a silver nitrate solution. When a steady current is passed through the solution, the net result is that silver metal is removed from the anode and deposited on the cathode.

In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. As the potential is increased, current flows. Silver metal is lost at the anode as it goes into solution:

Ag(s)  →  Ag+(aq) + e

The mass of the cathode increases as silver ions from the solution are deposited onto the spoon:

Ag+(aq) + e  →  Ag(s)

The net result is the transfer of silver metal from the anode to the cathode. The quality of the object is usually determined by the thickness of the deposited silver and the rate of deposition.

Quantitative Aspects of Electrolysis

The current applied over a given amount of time can be used to quantitatively find the amount of material electroplated. The amount of material electroplated also depends on the number of moles of electrons transferred during the reaction, using the equation:

Amps × time = \dfrac{(\text{mass plated})(\text{mol e}^{-}\ \text{transferred})(\text{F})}{\text{Molar Mass}}

Where Amps = Coulombs/sec, time = seconds, mass plated = grams, F = 96485 C/mol e, and Molar Mass = g/mol of molecule electroplated (usually a metal).

Example 1

Converting Current to Mass of Solid Electroplated
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. What mass of silver was deposited at the cathode from a silver nitrate solution?

We can use the equation from above in order to calculate the mass of silver electroplated. First, to find the number of electrons transferred in the process, we need to look at AgNO3(aq) being converted in Ag(s). The balanced half-reaction for this process would be: Ag+(aq) + e  →  Ag(s). One mole of electrons is transferred in the process.

We also need to convert the time into seconds. One hour is 3600 seconds.

Lastly, the molar mass of silver is 107.86 g/mol.

Amps × time = \dfrac{(\text{mass plated})(\text{mol e}^{-}\ \text{transferred})(\text{F})}{\text{Molar Mass}}

(10.23 Amps) (3600 sec) = \dfrac{(\text{mass plated})(1\ \text{mol e}^{-})(96485\ \frac{\text{C}}{\text{mol e}^{-}})}{107.86\ \frac{\text{g}}{\text{mol}}}

Solving for the mass of silver plated, we find that 41.19 g of Ag is electroplated.

Check Your Learning
Aluminum metal can be made from aluminum ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 2.50 × 103 A passed through the solution for 15.0 minutes?


Al3+(aq) + 3 e  →  Al(s); 210. g Al

Example 2

Time Required for Deposition
In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m2 from a solution containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current is 33.46 A? The density of chromium (metal) is 7.19 g/cm3.

This problem brings in a number of topics covered earlier. First, we must find the amount of chromium deposited using the density and the volume of Cr required:

Volume chromium needed = length × width × height = surface area × height

Volume chromium needed = (3.3 m2) × (0.010 mm × \frac{1\ \text{m}}{1000\ \text{mm}}) = 3.3 × 10-5 m3

Mass chromium deposited = density × volume

Mass chromium deposited = (7.19 \frac{\text{g}}{\text{cm}^{3}}) × (3.3 × 10-5 m3) × (\frac{100\ \text{cm}}{1\ \text{m}})3 = 237.3 g chromium

Additionally, since the solution contains chromium(III) ions (Cr3+), 3 moles of electrons are required per mole of Cr. Now, we can calculate the amount of time required:

Amps × time = \dfrac{(\text{mass plated})(\text{mol e}^{-}\ \text{transferred})(\text{F})}{\text{Molar Mass}}

(33.46 Amps) (time) = \dfrac{(237.3\ \text{grams})(3\ \text{mol e}^{-})(96485\ \frac{\text{C}}{\text{mol e}^-})}{51.996\ \frac{\text{g}}{\text{mol}}}

time = 39000 seconds, or 11 hours

Key Concepts and Summary

Electrolytic cells are often thought of as the opposite of galvanic cells, where nonspontaneous reactions occur through an application of electrical energy. Many of the same principles that apply to a galvanic cell also apply to an electrolytic cell, except for the “charges” given to the anode and cathode.

When considering the electrolysis of a molten salt, the anion is always oxidized at the anode and the cation is always reduced at the cathode. In order to make a reaction occur, a voltage greater than the electrochemical cell potential (Eºcell) must be applied. During electrolysis of an aqueous salt, there are two possible anodes (the anion and water) and two possible cathodes (the cation and water). In order to know whether the anion or water will be oxidized, we look at the standard reduction potential table and choose the species that is the stronger reducing agent, or has the least positive standard reduction potential. In order to know whether the cation or water will be reduced, we choose the species that is the stronger oxidizing agent, or the species that has the least negative standard reduction potential. By doing this, we minimize the voltage that must be applied in order to allow the nonspontaneous reaction to occur.

We can also calculate many aspects of electrolysis quantitatively, including the current that must be applied (in Amps) the time required, and the mass of metal plated.

Key Equations

  • Amps × time = \dfrac{(\text{mass plated})(\text{mol e}^{-}\ \text{transferred})(\text{F})}{\text{Molar Mass}}


a process where electrical energy causes nonspontaneous redox reactions to occur
electrolytic cell
an electrochemical cell which involves a nonspontaneous reaction, as opposed to the spontaneous reactions in a galvanic cell
a process that results in a thin coating of one metal on top of a conducting surface, often another metal

Chemistry End of Section Exercises

  1. Consider the electroylsis of molten CaI2, what reactions will occur at the cathode and the anode? How many grams of calcium metal will be produced by supplying the system with 0.75 amps for 50 minutes?
  2. An attempt is made to produce Na metal by electrolyzing an aqueous Na2SO4 solution with a pH close to 7. Will it be successful? Explain.
  3. Considering only the cost of electricity, would it be most expensive to produce 100.0 grams of solid aluminum (from Al(NO3)3), gold (from AuNO3), or zinc (from Zn(NO3)2?
  4. Hydrogen for fuel cells can be produced via electrolysis.
    1. What is the potential required for this reaction?
    2. With a cost of 15 cents per kilowatt-hour, how much will it cost to produce 1.00 kilograms of hydrogen gas?
  5. Olympic gold medals are actually gold plated in an electrolytic process. How long would it take to plate the 6.0 g of gold onto a medal from a solution of AuCl3(aq) with a current of 2.5 A, assuming 100% efficiency for the plating process?
  6. How many coulombs of charge are required to reduce 0.20 mol of Cr3+ to Cr?
  7. How many grams of nickel would be electroplated by passing a constant current of 7.2 A through a solution of NiSO4 for 90.0 min?
  8. What is produced at the anode and at the cathode in the electrolysis of CuCl2(aq)?
    Cu2+(aq) + 2 e  →  Cu(s) E° = 0.34  V
    2 H2O(ℓ) + 2 e  →  H2(g) + 2 OH(aq) E° = -0.83 V
    O2(g) + 4 H+(aq) + 4 e  →  2 H2O(ℓ) E° = +1.23 V
    Cl2(g)  +  2 e  →  2 Cl(aq) E° = +1.36 V
    1. Anode: O2(g)        Cathode: H2(g)
    2. Anode: Cl2(g)       Cathode: Cu2+(aq)
    3. Anode: Cl(aq)      Cathode: Cu(s)
    4. Anode: Cl(aq)      Cathode: Cu2+(aq)
    5. Anode: O2(g)       Cathode: Cu(s)
  9. Predict the products that would be observed at the anode and cathode in the electrolysis of NaBr(aq).
    Na+(aq) + e  →  Na(s) E° = -2.71 V
    O2(g) + 4 H+(aq) + 4 e  →  2 H2O(ℓ) E° = +1.23 V
    Br2(ℓ) + 2 e  →  2 Br(aq) E° = +1.09 V
    2 H2O(ℓ) + 2 e  →  H2(g) + 2 OH(aq) E° = -0.83 V
    1. Anode: O2(g)     Cathode: H2(g)
    2. Anode: Na+(aq)     Cathode: Br(aq)
    3. Anode: O2(g)        Cathode: H2(g)
    4. Anode: Br2(l)       Cathode: H2(g)
    5. Anode: Br(aq)     Cathode: Na+(aq)

Answers to Chemistry End of Section Exercises

  1. anode:  2 I(ℓ)  →  I2(s) + 2 e
    cathode:  Ca2+(ℓ) + 2 e  →  Ca(s)
    0.47 g
  2. No. Reduction of sodium at the cathode (Na+(aq) + e  →  Na(s); E° = -2.71 V) competes with the reduction of H+ (2 H+(aq) + 2 e  →  H2(g); E° = 0.00 V) and the reduction of H2O (2 H2O(ℓ) + 2 e  →  H2(g) + OH(aq); E° = -0.83 V). The sodium reaction has the most negative E° and so will not occur.
  3. Aluminum has the most negative E° so it will require the most electricity to reduce.
  4. (a) E° = -1.23 V; The minimum potential required is 1.23 V.
    (b) $4.92
  5. 59 min
  6. 5.8 × 104 C
  7. 12 g
  8. E
  9. D
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