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QUANTITATIVE ELECTROLYSIS CALCULATIONS

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully13. Electrolysis product calculations (negative cathode and positive anode products)

Quantitative chemistry calculations Help for problem solving in doing electrolysis calculations, using experiment data, making predictions. Practice revision questions on quantitative calculations of electrolysis electrode products (given electrode equations), mass of products at cathode or anode electrodes, moles and volumes of gases formed at the cathode and anode electrodes. This page describes and explains, with fully worked out examples, methods of calculation involving moles, masses or volumes of gases formed in an electrolysis process. You need to understand electrode equations, interconvert mass and moles and use the molar volume where gases are formed at the electrodes in electrolytic processes. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses

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Explaining electrolysis and descriptions of experimental methods

study examples carefully13. Electrolysis product calculations (negative cathode and positive anode products)

Relative atomic masses needed: Na = 23, Cl = 35.5, H = 1, Cu = 63.5, Al = 27, O = 16 and the molar volume of any gas is 24 dm3 at room temperature and pressure. The common electrode equations you may come across are listed below.

electrode involved: (-) negative cathode or (+) positive anode for the Electrode Equation below

moles of electrons involved (mass of product formed)

example of industrial process where this electrode reaction happens

sodium (-) Na+(l) + e- ==> Na(l)

1 (23g) = 1.0 mol Na metal per mol e-s

electrolysis of molten chloride salts to make chlorine and the metal

chlorine (+) 2Cl-(l/aq) - 2e- ==> Cl2(g)

2 (71g) = 0.5 mol Cl2 gas (12 dm3) released per mol e-s

electrolysis of molten chloride salts or their aqueous solution to make chlorine

hydrogen (-) 2H+(aq) + 2e- ==> H2(g)

2 (2g) = 0.5 mol H2 gas (12 dm3) released per mol e-s

electrolysis of many salt solutions to make hydrogen

copper (-) Cu2+(aq) + 2e- ==> Cu(s)

2 (63.5g) = 0.5 mol Cu deposited per mol e-s

deposition of copper in its electrolytic purification or electroplating

copper (+) Cu(s) - 2e- ==> Cu2+(aq)

2 (63.5g) = 0.5 mol Cu dissolves per mol e-s

dissolving of copper in its electrolytic purification or electroplating

aluminium (-) Al3+(l) + 3e- ==> Al(l)

3 (27g) = 0.33 mol Al metal per mol e-s

extraction of aluminium in the electrolysis of its molten oxide ore

oxygen (+) 2O2-(l) - 4e- ==> O2(g)

4 (32g) = 0.25 mol O2 (12 dm3) gas released per mol e-s electrolysis of molten oxides

oxygen (+) 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g)

4 (32g) = 0.25 mol O2 gas (6 dm3) released per mol e-s electrolysis of many salt solutions such as sulphates, sulphuric acid etc. gives oxygen (but chloride salts ==> chlorine)

top sub-indexExplaining electrolysis and descriptions of experimental methods


Part one: The (+) anode and (-) cathode electrode product ratio

  • What controls the rate of electrolysis? What controls how much product is formed?

  • The amount of material in moles formed at the electrode in electrolysis depends on three factors.

    • The quantity of any product produced from electrolysis, and its rate of production, depends on the number of electrons that are transferred in the external circuit and how fast the electrons flow in the circuit.

    • The charge on the ion - the bigger the charge on the ions, the more electrons must be transferred to give one mole of the product compare the effect of one mole of electrons in the table above and see examples 13.1.1 to 13.1.5 below in Part one)

    • The current flow, current is the rate of flow of charge, the higher the current flow (in amps), the more electrons are transferred per unit time e.g. seconds (the current flowing in amperes, A, see examples in Part two). Therefore, the rate of product formation is proportional to current.

    • The time duration of the electrolysis, the longer the electrolysis runs for, the more product is formed (time in seconds, minutes or hours, see examples in Part two).  Therefore, the amount of product formed is proportional to time.

  • If you know how much of a substance is made at one electrode, you can theoretically calculate the amount of substance formed at the other electrode.

  • The basis of these calculations is the ratio of the electrons involved in both electrode reactions (hence the introductory table of electrode equations above).

  • The electrode equations in the table above are referred to in the examples below.

  • In studying the examples below you must refer to the electrode equations in the table above,

    • and remember 1 mole = formula mass in grams and 1 mol of gas = 24 dm3 at room temperature/pressure..

  • Electrolysis calculation Example 13.1.1

    • The electrolysis of brine, aqueous sodium chloride solution, NaCl(aq) produces hydrogen gas, H2(g) at the -ve electrode and chlorine gas, Cl2(g) at the positive electrode. Atomic masses: H = 1, Cl = 35.5

    • 2H+(aq) + 2e- ==> H2(g) and 2Cl-(l/aq) - 2e- ==> Cl2(g)

    • 2 electrons are involved in both the formation of a hydrogen molecule [Mr(H2) = 2] or a chlorine molecule [Mr(Cl2) = 71].

    • The ratio of the products for H2(g) : Cl2(g)  is 1 mol : 1 mol or 24dm3 : 24 dm3 or 2g : 71g

      • If during the electrolysis of sodium chloride solution, 25 cm3 of hydrogen were produced, what volume of chlorine is theoretically formed?

      • Since the mole ratio is 1 : 1 for H2 : Cl2 for every 25 cm3 of hydrogen formed, 25 cm3 of chlorine will be formed.

  • Electrolysis calculation Example 13.1.2

    • The electrolysis of molten aluminium oxide Al2O3 is a more complicated affair. 

    • Its best to think of the ratio effect of a current of 12 moles of electrons passing through the electrolyte.

      • 4Al3+(l) + 12e- ==> 4Al(l) and 6O2-(l) - 12e- ==> 3O2(g)

      • It takes 3 moles of electrons to form 1 mole of Al from 1 mole of Al3+ ions.

      • and 4 moles of electrons to form 1 mole of O2 molecules from 2 moles of O2- ions.

      • Atomic masses: Al = 27, O = 16

    • The ratio of the products from 12 moles of electrons is therefore

      •   Al(l) : O2(g) is 4 mol : 3 mol or 108g : 96g or 72dm3.

    • If 0.1 mol of molten aluminium oxide is completely electrolysed (i) what mass of aluminium is formed and (ii) what volume of oxygen is formed (at RTP)

      • atomic mass Al = 27, molar volume of any gas at RTP = 24 dm3

      • (i) From 1 mole of Al2O3 you get 2 moles of Al

        • therefore from 0.1 mol of Al2O3 you get 0.2 mol of Al

        • mass Al = moles Al x atomic mass

        • mass Al = 0.2 x 27 = 0.54 g Al

      • (ii) From 1 mole of Al2O3 you get the equivalent of 3 moles of O atoms,

        • BUT you must treat this as 1.5 mol O2 molecules

        • therefore from 1 mol Al2O3 you get 1.5 mol of O2 molecules

        • so from 0.1 mol Al2O3 you get 0.15 mol of O2 gas

        • volume of oxygen gas = moles of oxygen gas x 24

        • volume of oxygen = 0.15 x 24 = 3.6 dm3 (3600 cm3)

  • Electrolysis calculation Example 13.1.3

    • In the electrolysis of dilute sulphuric acid, 36 cm3 of hydrogen, H2 was formed at the negative electrode (cathode).

    • What volume of oxygen, O2 would be formed at the positive electrode (anode)?

    • 2H+(aq) + 2e- ==> H2(g) and 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g)

      • It takes an electron transfer of 2 electrons to form each hydrogen molecule from 2 hydrogen, H+ ions and the transfer of 4 electrons to make 1 molecule of oxygen from 4 hydroxide, OH- ions.

      • Therefore, from the same amount of electrons (current), the ratio of hydrogen : oxygen formed is 2 : 1

      • so the volume of oxygen formed is 18 cm3. (36 : 18 have the ratio 2 : 1)

  • Electrolysis calculation Example 13.1.4

    • In the electrolysis of copper sulphate solution using carbon electrodes, what mass and volume of oxygen would be formed at the positive electrode if 254g of copper was deposited on the negative electrode? Atomic masses: Cu = 63.5, O = 16.

    • Cu2+(aq) + 2e- ==> Cu(s) and 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g)

      • It takes a transfer of 2 moles of electrons to form 1 mole of solid copper (63.5g) from 1 mole of copper(II) ions, Cu2+

      • and a transfer of 4 moles of electrons to form 1 mole of oxygen from 4 moles of hydroxide, OH- ions.

      • Therefore the expected mole ratio of Cu(s) : O2(g) from the electrolysis is 2 : 1

      • The moles of Cu deposited = 254/63.5 = 4 moles

      • so moles oxygen formed = 2 moles, since Mr(O2) = 2 x 16 = 32

    • mass of oxygen formed = 2 x 32 = 64g, volume of oxygen = 2 x 24 = 48 dm3 

  • Electrolysis calculation Example 13.1.5

    • In the industrial manufacture of aluminium by electrolysis of the molten oxide (plus cryolite) 250kg of aluminium are formed.

    • What volume of oxygen would be theoretically formed at room temperature and pressure?

    • [ Ar(Al) = 27 and 1 mole of gas at RTP = 24 dm3 (litres) ]

      • Aluminium oxide is Al2O3, so on splitting in electrolysis the atomic ratio for Al : O is 2 : 3,

      • and a mole ratio of Al : O2 or 4 : 3

      • 4Al3+(l) + 12e- ==> 4Al(l) and 6O2-(l) - 12e- ==> 3O2(g)

      • Note: It takes 12 electrons added to four Al3+ ions to make four atoms of Al, and 12 electrons removed from six oxide ions, O2-, to form six oxygen atoms, which combine to form three O2 molecules (see next line).

      • BUT, oxygen exists as O2 molecules, so the mole ratio of Al atoms : O2 molecules is 4 : 3

      • 250kg Al = 250000g, Al = 250000/27 moles = 9259.26 moles Al metal.

      • Therefore scaling for moles O2 = 9259.6 x 3/ 24 = 6944.44 moles O2 molecules.

      • Since volume of 1 mole of gas at RTP = 24 dm3 (litres)

      • Volume of oxygen formed = 6944.44 x 24 = 166667 dm3 

 

Self-assessment Quizzes [pec, based on part 1 only]

type in answer Honly or multiple choice click me for QUIZ!Honly


Explaining electrolysis and descriptions of experimental methods

top sub-index


 

Part Two: The relationship between current and the quantity of electrode product

More mathematics to do with electrolysis

  • As described and explained at the start the amount of product formed is proportional to time and current.
  • This is expressed by the following formula ...
    • Quantity of charge transferred (Coulombs, C) = current flow (amperes, A) x time (seconds, s)
    • Q = I x t
    • The amount of product formed is proportional to Q, the charge transferred to oxidise or reduce ions in the electrolyte.
    • So you can logically deduce the following sort of proportionality from the Q = It equation ..
      • Doubling the current with double the rate of production etc., halving the current halves the rate ... etc.
      • Doubling the time will double the amount of electrode products formed etc., halving the time will halve the amount of product formed ... etc.
      • Substituting numbers into the equation enables you to predict how charge flows, then by a subsequent calculation, you can predict by calculation how much product will be formed e.g.
        • e.g. if you run a current of 1.50 A for 2 minutes, how charge has flowed?
        • Q = I x t = 1.50 x (2 x 60) = 180 C
        • The 180 Coulombs can be converted into moles of electrons, and then, adjusting for the charge on the ion, converted into moles of ion discharged at the electrode, and finally moles of product.
        • The following examples show you how to do all these calculations in a logical manner.
        • You may also need to rearrange the equation to determine current needed or time taken.
        • Q = I x t,   I = Q / t  and  t = Q / I   (do we need a triangle guys?)
        • How long would it take to pass 5000 Coulombs, with a current of 4A?
        • t = Q / I = 5000 / 4 = 1250 s (or 20.83 minutes, 20 mins & 50 seconds)
        • What current do you need to pass 96500 Coulombs in 2 hours?
        • I = Q / t  = 96500 / (20 x 60 x 60) = 1.34A
  • 1 Faraday (F) = 96 500 Coulombs (C) = 1 mole of electrons.

    • This can be expressed as the Faraday Constant = 96500 Cmol-1

    • A current of 1A = 1C/s

    • 1 mole of any gas = 24 dm3 or 24000cm3 at 25oC/1 atmosphere.

  • Quantity of electricity in coulombs = current in amps x time in seconds

    • Q (C) = I (A) x t (s)

  • Example 13.2.1: A current was passed through an electrolysis circuit of silver nitrate solution and O.54g of silver was formed.

    • Ar(Ag)= 108 and the electrode equation is Ag+ (aq) + e- ==> Ag(s) 

    • Ar(Ag)= 64 and the electrode equation is Cu2+ (aq) + 2e- ==> Cu(s) 

    • If in the same circuit a copper(II) sulphate and copper electrodes cell was connected, how much copper is deposited at the negative (-) cathode?

    • 0.54g Ag = 0.54 / 108 = 0.005 mol Ag

    • now 1 mole electrons deposits 1 mol of silver, but only 0.5 mol of copper for the same electrons.

    • so mol copper deposited = 0.005 / 2 = 0.0025 mol Cu, mass Cu = 0.0025 x 64 = 0.16g Cu

    • -

  • Example 13.2.2: How much copper is deposited if a current of 0.2 Amps is passed for 2 hours through a copper(II) sulphate solution ?

    • Electrode equation: (-) cathode Cu2+(aq) + 2e- ==> Cu(s) and Ar(Cu) = 64

    • the quantity of electricity passed in Coulombs

      • = current in A x time in secs (Q = I x t)

      • = 0.2 x 2 x 60 x 60 = 1440 Coulombs, and 1 mole electrons = 96500 Coulombs

      • therefore moles of electrons passed through circuit = 1440 / 96500 = 0.01492

      • it takes two moles of electrons to form one mole of copper

      • therefore moles copper = 0.01492 / 2 = 0.00746

      • mass of copper = moles of copper x atomic mass of copper

      • = 0.00746 x 64 = 0.4775g of copper deposited.

      • -

  • Example 13.2.3: In the electrolysis of molten sodium chloride 60 cm3 of chlorine was produced.

    • Electrode equations:

      • (-) cathode Na+ + e- ==> Na or 2Na+ + 2e- ==> 2Na for electron 'molar' comparison

      • (+) anode 2Cl- -2e- ==> Cl2 

    • Calculate ...

    • (a) how many moles of were chlorine produced?

      • mol = vol / molar vol = 60 / 24000 = 0.0025 mol Cl2 

      • -

    • (b) what mass of sodium would be formed?

      • from the electrode equations 2 mol sodium will be made for every mole of chlorine

      • so 0.0025 x 2 = 0.005 mol sodium will be formed. Ar(Na) = 23

      • mass = mol x atomic or formula mass = 0.005 x 23 = 0.115g Na

      • -

    • (c) for how long would a current of 3 A in the electrolysis circuit have to flow to produce the 60cm3 of chlorine?

      • To produce 0.0025 mol of Cl2 you need 0.005 mol of electrons

      • 0.005 mol electrons = 0.005 x 96500 coulombs = 482.5 C

      • Q = I x t, so 482.5 = 2 x t, therefore t = 482.5 / 3 = 161 s (to nearest second)

      • -

  • Example 13.2.4: In an electrolysis of sodium chloride solution experiment a current of 2 A was passed for 2 minutes.

    • Electrode equations:

      • (-) cathode 2H+ + 2e- ==> H2 and (+) anode 2Cl- -2e- ==> Cl2 

    • (a) Calculate the volume of chlorine gas produced.

      • Q = I x t, so Q = 2 x 2 x 60 = 240 C

      • 240 C = 240 / 96500 = 0.002487 mol electrons

      • this will produce 0.002487 / 2 = 0.001244 mol Cl2 (two electrons/molecule)

      • vol = mol x molar volume = 0.001244 x 24000 = 29.8 cm3 of Cl2 

      • -

    • (b) What volume of hydrogen would be formed?

      • 29.8 cm3 of H2 because two electrons transferred per molecule, same as chlorine.

      • -

    • (c) In practice the measured volume of chlorine can be less than the theoretical value. Why?

      • chlorine is moderately soluble in water and also reacts with the sodium hydroxide formed.

      • -

  • Example 13.2.5: In a copper(II) sulphate electrolysis experiment ...

    • Electrode equation: (-) cathode Cu2+(aq) + 2e- ==> Cu(s) and Ar(Cu) = 64

    • (a) how much copper is deposited on the cathode by a 0.2A current flowing for 10 minutes?

      • Q = I x t, Q = 0.2 x 10 x 60 = 120 C, mole electrons = 120 / 96500 = 0.001244 mol e- 

      • 2 mole electrons deposits 1 mol of Cu, so mol Cu deposited = 0.001244 / 2 = 0.000622

      • mass = mol x atomic or formula mass = 0.000622 x 64 = 0.0398g Cu

      • -

    • (b) how long must a 0.1 A current be passed to deposit 1g of copper on the cathode?

      • 1g Cu = 1 / 64 = 0.015625 mol, needs 0.015625 x 2 mol electrons = 0.03125 mol e- 

      • 0.03125 mol e- = 0.03125 x 96500 = 3016 C

      • Q = I x t, 3016 = 0.1 x t, t = 3016 / 0.1 = 30160s, 30160 / 3600 = 8.38 hours.

      • -

  • Example 13.2.6: What volume of oxygen is formed by passing a current of 5A through acidified water for 25 minutes at a temperature of 25oC and 101kPa (1 atmosphere pressure)

    • Electrode equations:

      • (-) cathode 2H+ + 2 e- ==> H2 or 4H+ + 4e- ==> 2H2 for molar electron comparison

      • (+) anode 4OH- -2e- ==> 2H2O(l) + O2 

    • Quantity of electricity in Coulombs = current in A x time in seconds

    • Q = I x t = 5 x 25 x 60 = 7500 C, now 1 mole of electrons = 96500 C

    • so moles of electrons = 7500 / 96500 = 0.07772 moles

    • it takes 4 moles of electrons to form 1 mole of oxygen gas

    • therefore moles of oxygen formed = 0.07772 / 4 = 0.01943

    • 1 mole of gas = 24000 cm3, therefore volume of gas = 0.01943 x 24000 = 466.3 cm3 of O2

    • -

  • Example 13.2.7: How long will it take to produce 2 dm3 of chlorine gas by passing a 6A current through concentrated sodium chloride solution at 25C and 101kPa (1 atmosphere pressure)

    • (+) anode 2Cl- -2e- ==> Cl2 

    • therefore chlorine to be produced = 2/24 = 0.08333 moles of chlorine

    • 2 moles of electrons must be removed from 2 moles of chloride ions to produce 1 mole of chlorine gas,

    • therefore moles of electrons required = 0.08333 x 2 = 0.1666

    • 1 mole of electrons = 96500 Coulombs, therefore quantity of electricity required

    • = 0.1666 x 96500 = 16077 Coulombs

    • quantity of electricity in Coulombs = current in A x time in seconds

    • 16077 = 6 x time in seconds, so time in seconds = 16077 / 6 = 2679.5 seconds

    • or 2679.5 / 60 = 44.66 minutes to produce 2 dm3 of chlorine gas.

    • -

  • --


Explaining electrolysis and descriptions of experimental methods

OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

  9. Moles and the molar volume of a gas, Avogadro's Law

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

  13. Electrolysis products calculations (negative cathode and positive anode products) (this page)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials

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