Mastering Electrolysis: From Equations to Real-World Applications
Dive into the world of electrolysis! Understand the conversion process in electrolytic cells, explore equations and formulas, and discover practical examples. Learn how energy transforms and powers this crucial chemical reaction.

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Now Playing:Electrolysis – Example 0a
Intros
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  1. What is electrolysis?
  2. What is electrolysis?
    Definition of electrolysis.
  3. What is electrolysis?
    Chemical cells for electrolysis.
Examples
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  1. Use the cell potential calculation to find conditions for electrolytic cells.
    For an electrolytic cell containing 1M NiI2 (aq) :
    • What reaction takes place at the anode?
    • What reaction takes place at the cathode?
    • What is the minimum voltage needed for this reaction to work?
    Introduction to electrochemistry
    Notes

    In this lesson, we will learn:

    • To recall the definition of electrolysis and some examples of its use.
    • To draw an electrolytic cell used in electrolysis.
    • How to find electrolysis operating conditions using cell potential calculations.
    • How to use Faraday’s law to relate charge flow to reaction stoichiometry.

    Notes:

    • Electrolysis is the breaking up (-lysis) of substances using electric current (electro-) – hence the name.
      It can be done with dissolved or molten ionic solutions such as sodium chloride (NaCl) or aluminum oxide (Al2O3) to obtain the pure elements from their ions. This is very useful because some valuable substances like aluminum metal are not found naturally in their elemental form because they are too reactive. They must be produced from their ions.
      In terms of energy, electrolysis reactions are the opposite of a redox cell reaction:
      • In a redox or galvanic cell, the reaction is spontaneous and chemical potential energy is used to produce electrical energy. In terms of thermodynamics, Gibbs free energy of the system drives the forward reaction to make products.
      • In an electrolysis reaction, the process is not spontaneous, so electrical energy is needed for the reaction to make products. In terms of thermodynamics, the forward reaction results in a loss of Gibbs free energy of the surroundings, so it is unfavourable.

    • Electrolysis is often used on binary salts – containing only two different ions – in order to obtain the constituent ions as their pure elements. This is how NaCl is electrolyzed to obtain chlorine gas (Cl2) and sodium which reacts to form NaOH, another important chemical.
      Because only two oppositely charged ions are involved with binary salts, both ions will take part in the opposite processes of reduction and oxidation to return to their elemental form. When binary salts are electrolyzed, there are a few differences to the cell compared to redox (galvanic) cells:
      • Because the compounds being electrolyzed are water-soluble, inert electrodes are used, normally carbon or platinum.
      • No salt bridge is necessary. This is because electrolyte can mix as there’s usually only one electrolyte; the ionic substance that is being electrolyzed! It is in aqueous or molten form so a current can flow easily.
      This is only true of binary salts being electrolyzed. A regular electrolytic cell involving active metal species will be set up like a redox cell with a power source.

      See the diagram below:
      Electrolysis

    • As said above, electrolysis reactions are not spontaneous!
      This means that instead of the cells producing electricity, electrolysis cells need voltage to be applied. You can calculate the voltage just like you calculate redox cell potential in Calculating cell potential. Your calculated E0cell should be less than zero for an electrolytic cell.
      For example, with sodium chloride, NaCl, being electrolyzed to produce sodium and chlorine:

      Na+ + e-\enspace \enspace Na \quad \quad \quad \quad E0red = -2.71 V
      Cl2 + 2e-\enspace \enspace 2Cl- \quad \quad \quad E0red = +1.36 V

      NaCl electrolysis reduces Na+ to Na metal and oxidizes Cl- ions to Cl2 so the Cl half-equation and E0red value needs reversing. Applying this to the cell potential equation:

      E0cell = E0red (reduction) + E0ox (oxidation)

      We can work out:

      E0cell = -2.71 + (-1.36) = -2.71 -1.36 = -4.07 V

      This means that at least 4.07 V must be applied to the cell for the reaction to start occurring spontaneously. Higher voltage is applied usually because there is resistance in the cell and it is not at standard conditions.

    • Remember that electrolysis in aqueous solution means in water – the water solvent can be reduced or oxidized itself. Taking the example of NaCl again, in aqueous solution there are four possible reactions going on in the cell, each with different reduction potentials.1 Assuming neutral pH:
      • The oxidation of chlorine: \, 2Cl-\enspace \enspace Cl2 + 2e- where E0ox = -1.36 V
      • The oxidation of water: \, 2H2O (l) \enspace \enspace O2 + 4e- + 4 H+(aq) where E0ox = -1.23 V
      • The reduction of water: \, 2H2O (l) + 2e-\enspace \enspace H2 (g) + 2OH- (aq) where E0red = -0.83 V
      • The reduction of sodium: \, Na+(aq) + e-\enspace \enspace Na where E0red = -2.71 V

      If we applied 4.07 V to the system, what would react? The reaction that is preferable is the one with the least required voltage! Like a spontaneous redox reaction, the half-equations with the greatest E0red and E0ox are preferred.
      ‘Preferred’ means this reaction is much more likely to happen. The other reaction(s) can occur but are less likely to.
      One reduction and one oxidation ‘half reaction’ must happen. Think about the voltage differences (potential difference!) between the half-equations:
      • The reduction potential of water is less negative than sodium metal, so only water gets reduced to hydrogen gas at the cathode (where reduction occurs).
      • The oxidation potential for water is less negative than chlorine but they are very similar.
      Remember, again that this is not run at ideal (standard) conditions though! E0red, or Reduction potential, assumes 1M concentration, but increasing the concentration of NaCl lowers the real voltage needed to oxidize Cl- and make Cl2. Extra voltage, called overvoltage or overpotential is still applied to the reaction to make it react faster (like the rate at ideal conditions).
      The real process of NaCl electrolysis is this:

      2 NaCl (aq) + 2H2O(l)\enspace \enspace 2Na+ (aq) + 2OH- (aq) + H2 (g) + Cl2 (g)

      Chlorine gas (Cl2) is obtained, as is hydrogen gas and NaOH which can both be isolated and sold.

    • You need to look at the conditions that occur in your electrolytic cell. When you have a substance to be electrolyzed, check the ions that will be produced:
      • If H+ is produced, your conditions will be acidic.
      • If OH- is made, it will be basic.
      • If neither, assume it is neutral.
      This is important when looking for half-equations because any equations for the ions in solution must have the correct conditions! For example, for the oxidation half-reaction:

      2OH- + Br-\enspace \enspace H2O + BrO- + 2e- E0 = +0.76 V

      This half-reaction could only occur in basic conditions.

    • As electrical energy is either used up, as in electrolytic cells, or generated in voltaic cells, we can measure the flow of electrons (the current) using an ammeter.
      Faraday’s law relates this current to the stoichiometry and rate of a reaction:

    • q=Itq = I * t

      Where:
      • qq = charge (in coulombs)
      • II = current (in amps)
      • tt = time (seconds)

      This equation is important because when rearranged for I=q/tI = q/t, we measure the current as a rate of charge over time. Quantity of charge is proportional to quantity of mass change at an electrode.
      • In other words, the higher the current, the more charge per unit time passes through the cell and the faster a reaction at an electrode will occur (e.g. more Cu (s) being deposited, gaining mass, or a Ni (s) electrode being oxidised, losing mass).

      This is where the value of the Faraday constant comes from – 96485 C mol-1. It means that one mole of electrons carries 96485 Coulombs of charge. For example, if a copper electrode was being oxidised in the half-reaction:

      Cu (s) \, \, Cu2+ (aq) + 2e-

      We know that one mole of copper liberates two moles of electrons, which has 2 × 96485 C of charge. Since moles are related to the mass of a sample substance, we now have a relationship between:
      • The mass change of a copper electrode during an experiment;
      • The number of moles of copper this mass equates to;
      • The number of moles of electrons liberated;
      • The amount of charge this amount of electrons carries
      • The time taken or the current reading of the cell.

      Faraday’s laws of electrochemistry can be summarized by the following:

      m=m = q.M.F.z.\large \frac{q. \, M.} {F. z.}

      Where:
      • mm = mass change of substance at electrode (in g)
      • qq = electric charge (Coulombs)
      • MM = molar mass of substance at electrode
      • FF = the Faraday constant, 96485 C mol-1
      • zz = the number of electrons transferred per ion (e.g. 2 for Ca2+ ions).

      As the equation above shows, increasing q will increase the rate of deposition (Mn+ (aq) ions becoming M (s) at the electrode) or liberation, the reverse process.

      Substituting q for ItI * t in the equation above gives:

      m=m = I.t.M.F.z.\large \frac{I. \, t. \, M.} {F. z.}

      This relates all the factors mentioned above.

    • WORKED EXAMPLE: Using Faraday’s laws.
      Calculate how long a 7A current must be applied to deposit 1.5 g of Zn metal at an electrode.

      We are solving for time, so rearranging this equation for t gives us this:

    • t=t = m.F.z.I.M.\large \frac{m. \, F. \, z.} {I. M.}

      We have the values m = 1.5; F = 96485; z = 2 (for 2e- per Zn2+ ion); I = 7; M = 65.38.
      With so many variables, a good way to ‘proof’ your answer is to solve for the units. We are solving for t, so our units should end in s (seconds).

      t=t = (1.5g).(96485Cmol1).(2)(7Cs1).(65.38gmol1)\large \frac{(1.5 \, g ). \, (96485 \, C \, mol^{-1}). \, (2)} {(7 \, C \, s^{-1}). \, (65.38 \, g \, mol^{-1} ) } = 632.5 s = 10.54 min

      The calculation shows that it takes ten and a half minutes for this much of a mass change to take place. Alternatively, you can use the conversion factor method to go from mass to time:

      1.5 gZn=g \, Zn = 1molZn65.38g2mole1molZn96485C1mole1A.s1C1s7A\large \frac{1 \, mol \, Zn } {65.38 \, g } \, * \, \frac{2 \, mol \, e^{-}} {1 \, mol \, Zn } \, * \, \frac{96485 \, C } {1 \, mol \, e^{-} } \, * \, \frac{1 \, A. \, s} {1 \, C} \, * \, \frac{1 \, s} {7 \, A} = 632.5 s = 10.54min

      Whichever method you use, we received the same answer because the units cancel!


      1 Sourcefor data: ATKINS, P. W., & DE PAULA, J. (2006).Atkins' Physical chemistry. Oxford, Oxford University Press
    Concept

    Introduction to Electrolysis

    Electrolysis is a fundamental process in chemistry and industry, playing a crucial role in various applications. This electrochemical technique uses electrical energy to drive non-spontaneous chemical reactions, enabling the production of important materials and the purification of metals. At the heart of electrolysis is the electrolytic cell, a device consisting of electrodes (anode and cathode), an electrolyte solution, and a power source. The cell facilitates the movement of ions and the transfer of electrons, resulting in chemical changes at the electrodes. Electrolysis finds widespread use in manufacturing, metal extraction, and even in everyday items like batteries. To help visualize this process, we've included an introduction video that demonstrates electrolysis in action. This video will provide a clear, step-by-step explanation of how an electrolytic cell operates, making it easier to understand this fascinating aspect of electrochemistry. Whether you're a student or simply curious about science, this overview of electrolysis will illuminate its significance in our world.

    FAQs

    Here are some frequently asked questions about electrolysis:

    1. What is the process of conversion in electrolysis?

    In electrolysis, electrical energy is converted into chemical energy. The external power source provides electrical energy, which drives non-spontaneous chemical reactions at the electrodes, resulting in the formation of new chemical substances.

    2. What is the chemical equation for electrolysis?

    The chemical equation for electrolysis depends on the specific substances involved. For example, in the electrolysis of molten sodium chloride (NaCl), the equation is: 2NaCl (l) 2Na (l) + Cl2 (g). This shows the decomposition of NaCl into sodium metal and chlorine gas.

    3. How is electrolysis used in everyday life?

    Electrolysis has numerous everyday applications, including: - Electroplating jewelry and automotive parts - Purifying metals like copper and aluminum - Producing chlorine for water treatment - Extracting reactive metals like sodium and magnesium - Charging rechargeable batteries

    4. What is the formula for calculating electrolysis?

    The main formula used in electrolysis calculations is derived from Faraday's laws: m = (Q × M) / (n × F), where m is the mass of substance produced, Q is the total electric charge passed, M is the molar mass of the substance, n is the number of electrons transferred per ion, and F is Faraday's constant (96,485 C/mol).

    5. What is the necessary voltage to power the electrolysis of molten sodium chloride?

    The theoretical minimum voltage required for the electrolysis of molten sodium chloride is about 4.07 V. This is calculated from the difference in standard reduction potentials of sodium and chlorine. However, in practice, a higher voltage (typically 6-7 V) is used to overcome various resistances in the system and ensure efficient electrolysis.

    Prerequisites

    Understanding the fundamental concepts that lay the groundwork for more advanced topics is crucial in any field of study, especially in chemistry. When it comes to the topic of electrolysis, having a solid grasp of prerequisite subjects is essential for comprehending its intricacies and applications. One of the most important prerequisite topics for understanding electrolysis is the introduction to chemical reactions.

    Electrolysis is a complex process that involves the use of electrical energy to drive non-spontaneous chemical reactions. To fully appreciate the principles behind electrolysis, students must first have a strong foundation in basic chemical reactions. This prerequisite knowledge enables them to understand how electrical energy can be harnessed to facilitate chemical changes that wouldn't occur naturally.

    The study of non-spontaneous chemical reactions is particularly relevant to electrolysis. In nature, most chemical reactions occur spontaneously, moving towards a state of lower energy. However, electrolysis deals with reactions that require an input of energy to proceed. By understanding the concepts of spontaneous and non-spontaneous reactions, students can better grasp why electrolysis is necessary and how it functions.

    Moreover, the ability to write and balance chemical equations is a crucial skill when studying electrolysis. This fundamental aspect of chemistry, covered in the introduction to chemical reactions, allows students to represent the processes occurring during electrolysis accurately. It helps in visualizing the movement of ions, the transfer of electrons, and the formation of products at the electrodes.

    Understanding redox reactions, which are also part of the introduction to chemical reactions, is another vital prerequisite for electrolysis. Electrolysis involves the oxidation and reduction of species at the electrodes, and a solid understanding of these concepts is essential for comprehending the electron transfer processes that occur during electrolysis.

    By mastering these prerequisite topics, students build a strong foundation that enables them to tackle the complexities of electrolysis with confidence. They can more easily understand the role of the electrolyte, the function of electrodes, and the principles governing the decomposition of compounds through electrical means. This knowledge not only aids in academic understanding but also prepares students for practical applications of electrolysis in various fields, including metallurgy, electroplating, and the production of chemicals.

    In conclusion, the importance of understanding prerequisite topics like the introduction to chemical reactions cannot be overstated when studying electrolysis. It provides the necessary context and foundational knowledge that allows students to fully appreciate and engage with this fascinating aspect of electrochemistry. By building on these fundamental concepts, students can develop a deeper, more comprehensive understanding of electrolysis and its wide-ranging applications in science and industry.