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Heat capacity and specific heat capacity

The heat capacity C of a substance of mass m is the net amount of heat qnet needed to change its temperature T by one Kelvin.

As mentioned in the previous article, the net amount of heat absorbed or released by a system at constant pressure is equal to the change in enthalpy of the system, qnet = ΔH, which can also be written as qsys = ΔHsys. This net amount of heat absorbed or released, respectively raises or lowers the temperature of a substance in the system Tsys. The heat capacity of a substance of mass m in a system at constant pressure, by definition, is therefore:

C=\frac{\Delta H_{sys}}{\Delta T_{sys}}

where C has units of J K-1.

The specific heat capacity c of a substance in a system is the heat capacity of the substance divided by the mass of the substance. In other words, it is the net amount of heat absorbed or released by the substance in the system that changes one kilogramme of the substance by a temperature of one Kelvin. At constant pressure, we have

c=\frac{\Delta H_{sys}}{m\Delta T_{sys}}\; \; \; \Rightarrow \; \; \; \Delta H_{sys}=cm\Delta T_{sys}\; \; \; \; \; \; \; \; 3

where c has units of J K-1 kg-1.

It is important to remember that ΔHsys is the transfer of heat at constant pressure between a system and its surroundings. If ΔHsys > 0, heat is transferred from the surroundings to the system, in which case, the temperature of the system increases, while the temperature of the surroundings decreases (assuming that the surroundings is a limited volume of solid, liquid or gas). Conversely, if ΔHsys < 0, heat is transferred from the system to its surroundings, in which case, the temperature of the system decreases, while the temperature of the surroundings increases.

An example of the application of eq3 is the measure of the change in enthalpy of a known mass of liquid that is heated from T1 to T2 (see diagram above). In this experiment, we consider the system as the liquid in the container and the surroundings as the Bunsen burner. Using eq3,

\Delta H_{sys}=cm\left ( T_2-T_1 \right )

Since T2 > T1 and both c and m are positive, ΔHsys > 0 and the liquid in the system undergoes an endothermic process.

The above experiment measures the change in temperature of the system to arrive at the value of ΔHsys. We can also determine ΔHsys indirectly by measuring the change in temperature of the surroundings. Consider the following experiment to measure ΔHsys of an exothermic reaction:

If the system and its surroundings are initially at the same temperature, then any thermal energy produced by the reaction increases the temperature of the system, causing the flow of heat q from the system to its surroundings. Since we are measuring the temperature change of the surroundings, eq3 becomes,

q_{sur}=\Delta H_{sur}=cm\Delta T_{sur}\; \; \; \; \; \; \; \; 4

where c and m are the specific heat capacity of material in the surroundings and the mass of material in the surroundings respectively.

By the theory of conservation of energy, any heat gained in the surroundings is heat lost by the system, i.e.

q_{sys}+q_{sur}=0\; \; \Rightarrow \; \; q_{sur}=-q_{sys}

Substituting qsur = –qsys in eq4, we have qsys = – cmΔTsur or

\Delta H_{sys}=-cm\Delta T_{sur}\; \; \; \; \; \; \; \; 5

Since most experiments are designed to indirectly measure the change in temperature of the surroundings to obtain ΔHsys, eq5 is usually used to analyse the enthalpy change of a chemical reaction in a system instead of eq3. For an exothermic reaction, the temperature of the surroundings increases (ΔTsur > 0) and according to eq5, ΔHsys < 0. For an endothermic reaction, ΔTsur < 0 and therefore ΔHsys > 0.

In the next article, we shall look at some actual experiment setups for measuring enthalpy changes of reactions.

 

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Using bond enthalpies to calculate the change in enthalpy of reactions

In the article on ‘Measuring the change in enthalpy of a reaction (Part 2)’, the change in enthalpy of a reaction is determined via calorimetry. Enthalpy change of reactions can also be estimated using bond enthalpy data.

Rewriting eq2 in terms of enthalpy,

\Delta H=\Delta H_{bf}+\Delta H_{bb}\; \; \; \; \; \; \; \; 6

where ΔH is the change in enthalpy of the reaction, ΔHbf is the bond enthalpy of formation (ΔHbf < 0) and ΔHbb is the bond enthalpy of breaking (ΔHbb > 0).

Since both ΔHbb and ΔHbf are taken from the same bond enthalpy data set, which has all numbers quoted as positive values, eq6 becomes

\Delta H=\left|\Delta H_{bb} \right |-\left | \Delta H_{bf} \right |\; \; \; \; \; \; \; \; 7

or to be specific, if there are multiple bonds broken and multiple bonds formed:

\Delta H=\sum _i \left |\Delta H_{bb} \right |_i-\sum _j\left | \Delta H_{bf} \right |_j\; \; \; \; \; \; \; \; 8

In other words,

The change in enthalpy of a reaction is the difference between the sum of the magnitude of bond enthalpies of all bonds broken and the sum of the magnitude of bond enthalpies of all bonds formed.

 

Question

With reference to the bond enthalpy data in the previous article,

a) what is the change in enthalpy for the reaction:

C(g)+4H(g)\rightarrow CH_4(g)

b) what is the change in enthalpy for the hydrogenation of propene?

CH_3CH=CH_2(g)+H_2(g)\rightarrow C_3H_8(g)

Answer

a) Using eq8, ΔH = 0 – (416 x 4) = -1664 kJmol-1.

b) 1 mole of π (C – C) bond and one mole of H – H bond are broken and two moles of C – H bonds are formed. Using the bond enthalpy data, the bond enthalpy of one mole of π (C – C) bond can be estimated from the difference between one mole of C = C bond and one mole of σ (C – C) bond. Thus, using eq8,

\Delta H=(612-348+435)-(416\times 2)=-133\: kJmol^{-1}

 

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Comparison between electrochemical cell and electrolytic cell

The comparison between electrochemical cells and electrolytic cells highlights their distinct features and functions.

Electrochemical cell

Electrolytic cell

Converts chemical energy to electrical energy

Converts electrical energy to chemical energy

Flow of electrons determined by reactivity of chemical species

Flow of electrons determined by external power supply

Anode is negative, cathode is positive by convention

Anode is positive, cathode is negative by convention

May or may not require a salt bridge

May or may not require separate cells

Oxidation occurs at anode while reduction occurs at cathode

 

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Applications of electrolysis: manufacture of chemicals

Electrolysis is widely used in the manufacture of chemicals, facilitating the production of essential substances like chlorine, sodium hydroxide, and hydrogen through the controlled decomposition of ionic compounds in solution.

1) The chloralkali process

The chloralkali process is an electrolytic process for the manufacture of H2, Cl2 and NaOH.

Concentrated NaCl (saturated brine) is pumped into the left compartment, and water is pumped into the right compartment. Chloride ions are oxidised at the inert titanium anode to form chlorine gas:

2Cl– (aq) → Cl2 (g) + 2e–

while water is reduced at the cathode to give hydrogen gas:

2H2O (l) + 2e– → H2 (g) + 2OH– (aq)

A cation exchange membrane, which allows sodium ions but not hydroxide and chloride ions to flow from the left compartment to the right compartment, is used. Hence, NaOH is only formed in the right compartment with the overall reaction being:

2NaCl (aq) + 2H2O (l) → Cl2 (g) + H2 (g) + 2NaOH (aq)

Chlorine gas, which is formed in the left compartment, is piped out, where it reacts with water to give hydrochloric and hypochlorous acids. Finally, NaCl can be replaced with KCl or CaCl2 for the manufacture of KOH or Ca(OH)2, respectively.

 

Question

A diaphragm, instead of a cation-exchange membrane, is sometimes used in the electrolytic cell (diaphragm cell) of the chloralkali process. How does that affect the processes within the cell?

Answer

The diaphragm is porous and usually made of asbestos or a polymer fibre. Unlike the cation-exchange membrane, which only allows sodium ions to pass through, it permits the passage of both sodium and chloride ions. The other processes remain the same. However, the use of a diaphragm results in a lower purity of NaOH due to the presence of NaCl in the cathodic compartment.

 

2) The Castner-Kellner process

Another way to manufacture H2, Cl2 and NaOH is via the Castner-Kellner process. The main difference between this process and the chloralkali process is the use of a liquid mercury cathode, which lowers the activation energy for the reduction of Na+ to Na, allowing the otherwise thermodynamically infeasible reaction to proceed.

Upper compartment:

Concentrated NaCl is fed into the upper compartment where Cl– is oxidised at the inert titanium anode to liberate Cl2.

2Cl– (aq) → Cl2 (g) + 2e–

At the same time, Na+ is reduced at the Hg cathode to Na, which then dissolves in the cathode to form an amalgam, Na/Hg, which in turn circulates to the lower compartment.

Na+ (aq) + e– → Na (s)

Na (s) + Hg (l) → Na/Hg (l)

Lower compartment:

Water that is fed into the lower compartment reacts with the Na/Hg amalgam to give NaOH, H2 and Hg, which is pumped back up to the upper compartment for reuse. Note that Na is oxidised to Na+ while hydrogen in water is reduced to H2.

2Na/Hg (l) + 2H2O (l) → 2NaOH (aq) + H2 (g) + 2Hg (l)

The overall redox reaction is:

2NaCl (aq) + 2H2O (l) → 2NaOH (aq) + H2 (g) + 2Cl2 (g)

 

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Applications of electrolysis: anodising

Anodising is an electrolytic process to create an inert oxide layer on a metal, which is the anode and hence the name anodising. The diagram below shows the passivation of aluminium by the formation of a layer of aluminium oxide, which protects the aluminium from corrosion. Anodised aluminium is used in many products, such as aircraft parts, consumer electronic components, and carabiners.

At the anode:

2H2O (l) → O2 (g) + 4H+ (aq)+ 4e–

The oxygen produced reacts with the anode to form the protective layer:

3O2 (g) + 4Al (s) → 2Al2O3 (s)

 

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Applications of electrolysis: refining of metals and electroplating

Electrolysis is widely used in the refining of metals and electroplating. In metal refining, it helps purify ores by separating impurities from the desired metal, such as copper. In electroplating, electrolysis is used to coat a surface with a thin layer of metal, improving its appearance, corrosion resistance, and wear resistance. Both processes are essential in various industries, from electronics to manufacturing.

Electrolytic cell 4 in a previous article is an example of an electrolytic cell for refining metals. Copper ore, which contains small amounts of other metals like Pt and Au, is the anode, while a piece of pure Cu is the cathode.

At the anode, Cu, which competes with H2O and SO42-, is preferentially oxidised to aqueous Cu2+ because it lies above H2O and SO42- in the electrochemical series and is more concentrated relative to the other two species (i.e. high number of Cu atoms in the electrode per unit volume as compared to the other two species in the electrolyte per unit volume).

Cu (s) →Cu2+ (aq) + 2e–

At the cathode, Cu2+, which competes with H2O, is preferentially reduced to Cu, as it is lower in the electrochemical series than H2O. Additionally, it is more concentrated relative to H2O.

Cu2+ (aq) + 2e– → Cu (s)

The overall redox reaction results in the coating of a layer of pure copper on the surface of the pure copper cathode:

Cu (s) → Cu (s)

If we replace the anode with a pure metal, such as Ni, and the electrolyte with an aqueous salt of the metal, such as NiSO4, a layer of Ni will coat the cathode (see diagram below). This is the principle of electroplating.

Examples of electroplating are:

Plating Metal, M

Plated object

Ni, Cr

Automotive parts, electrical appliances

Sn

Ornaments, jewelry

Ag, Au, Cu

Cutlery, ornaments, jewelry, electronic components

 

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Applications of electrolysis: extraction of metals

Electrolysis is a vital process in the extraction of metals from their ores, enabling the efficient separation of valuable elements through the application of electric current, which transforms metal ions into their solid form.

A theoretical way to obtain pure aluminium is to electrolyse molten Al2O3 (alumina). However, the melting point of Al2O3 is 2072 oC, making this method impractical. In the 1800s, French scientist Paul Héroult and American scientist Charles Hall developed a better method known as the Hall-Heroult process. In this process, Al2O3 is first dissolved in a solvent mixture of Na3AlF6 (cryolite) and AlF3 at about 1000 oC before being electrolysed to give pure aluminium.

Dissolution reaction:

Al2O3 (s) + 4[AlF6]3- (l) → 3[Al2OF6]2- (l) + 6F– (l)

Solid alumina is dissolved in cryolite and AlF3 at about 1000 oC to give an oxyflouridealuminate complex.

 

At the anode:

2[Al2OF6]2- (l) + 12F– (l) → 4[AlF6]3- (l) + O2 (g) + 4e–

O2 (g) + C (s) → CO2 (g)

The oxyflouridealuminate complex reacts with excess fluoride ions to give the hexafluoroaluminate complex and oxygen, which then oxidises the carbon anodes to yield carbon dioxide (note that at such high temperatures, the carbon anodes are no longer inert). The overall anode reaction is:

2[Al2OF6]2- (l) + 12F– (l) + C (s) → 4[AlF6]3- (l) + CO2 (g) + 4e–

 

At the cathode:

[AlF6]3- (l) + 3e– → Al (l) + 6F– (l)

The hexafluoroaluminate complex is reduced to give pure molten aluminium.

 

The overall redox reaction is

2Al2O3 (s) + 3C (s) → 4Al (l) + 3CO2 (g)

As the carbon anodes are oxidised to carbon dioxide, they have to be replaced over time.

 

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Analysing reactions in an electrolytic cell

To analyse reactions in the PbBr2 electrolytic cell (see previous article), we follow these steps:

Step 1:  Determine the direction of flow of electrons

The flow of electrons in an electrolytic cell is determined by the way the battery (direct current source) is connected to the circuit. The battery is represented by two parallel lines with the longer line symbolising the positive terminal and the shorter one, the negative terminal.

For this example, the positive terminal of the battery is connected to the left electrode, while the negative terminal to the right electrode. Electrons flow from the negative terminal to the right electrode making it negatively charged. The negatively charged electrode attracts cations in the electrolyte towards its surface, where they undergo reduction. Hence, the electrode that is connected to the negative terminal of the battery is the cathode. Using the same logic, oxidation takes place at the left electrode, which is the anode.

Step 2:  Find out the reduction reaction at the cathode by

    • Identifying the reducible species (usually positive ions), which in this case is Pb2+. There is only one reducible species, as the electrolyte is molten PbBr2 and the electrodes are inert.
    • Distinguishing the species that is most readily reduced if there is more than one competing species by:
      • comparing the species on the electrochemical series: N.A.
      • comparing the concentrations of the species if their positions in electrochemical series are relatively near: N.A.
    • Writing the ionic half-reaction equation for the reduction:

Pb2+ (l) + 2e– → Pb (l)

Step 3:  Find out the oxidation reaction at the anode by

    • Identifying the oxidisable species (usually negative ions): Br–
    • Distinguishing the species that is most readily oxidised if there is more than one competing species by:
      • comparing the species on the electrochemical series: N.A.
      • comparing the concentrations of the species if their positions in electrochemical series are relatively near: N.A.
    • Writing the ionic half-reaction equation for the reduction:

2Br– (l) → Br2 (g) + 2e– 

Combining the two half reaction equations, the overall redox reaction in the electrolytic cell is:

Pb2+ (l) + 2Br– (l) → Pb (l) + Br2 (g) 

Note that the electrolytic circuit is complete by the flow of electrons from the anode to the positive terminal of the battery via the wire and a net flux of negative ions towards the anode and positive ions towards the cathode in the electrolyte.

 

Question

What are the products at the anode and cathode for the following electrolytic cells?

Cell Electrode pair

Electrolyte

1

 C/C

Concentrated NaCl

2

 Pt/Pt

Acidified H2O

3

 C/C

Dilute CuSO4

4

 Cu/Cu

Dilute CuSO4
Answer

 

Cell

 Species at anode

Products at anode

1

 Cl–, H2O

Cl2, since [Cl–] >> [H2O]

2

 H2O

O2

3

 H2O, SO42–

O2, since H2O is much higher in the electrochemical series than SO42-

4

 Cu, H2O, SO42–

Cu2+, since Cu is higher in the electrochemical series than H2O & SO42-, and [Cu] >> [SO42-] & [H2O]

H2O is always assumed to be the species instead of OH– in non-hydroxide aqueous solutions, as the concentration of OH– from H2O is very small. Similarly, H2O is assumed to be the species instead of H+ in non-acidic aqueous solutions, as the concentration of H+ from H2O is very small.

Cell

 Species at cathode

Products at cathode

1

 Na+, H2O

H2, since H2O is much lower in the electrochemical series than Na+

2

 H+, H2O

 Both species are reduced to H2, with H+ undergoing reduction first, as it is much lower in the electrochemical series than H2O. H2O will be reduced when [H+] falls appreciably.

3

 Cu2+, H2O

Cu, since Cu2+ is much lower in the electrochemical series than H2O

4

 Cu2+, H2O

Cu, since Cu2+ is lower in the electrochemical series than H2O

 

 

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Electric double layer and electrode potential

An electric double layer consists of two layers of opposite charges that accumulate at the phase boundary between an electrode (a solid electrical conductor, e.g. a piece of zinc) and an electrolyte (a molten or an aqueous electrical conductor, e.g. aqueous zinc sulphate).

Zinc, according to the metal reactivity series, is a relatively reactive metal. When a zinc rod (an electrode) is placed in aqueous zinc sulphate, zinc atoms are easily oxidised to zinc ions, which enter the solution leaving free electrons on the surface of the electrode. At the same time, some zinc ions in the solution combine with free electrons in the electrode and are reduced to zinc atoms. Initially, more zinc atoms are oxidised to zinc ions than zinc ions are reduced to zinc atoms. As the concentration of zinc ions in the solution increases, an equilibrium is attained such that the rate of oxidation equals to the rate of reduction:

Zn^{2+}(aq)+2e^-\rightleftharpoons Zn(s)

The equilibrium position for zinc, when compared to less reactive metals, lies to the left of the above equation, resulting in an excess of electrons on the surface of the electrode, which attract zinc ions in the solution that are close to the surface to form two parallel array of charges called an electric double layer. Just like a ball at a certain height has the capacity to do work, opposite charges that are separated at a distance have the potential to do work too. We call such a potential, which arises due to the formation of the electric double layer, an electrode potential.

As the formation of the ions and electrons in the electric double layer depends on the reactivity of the metal, the electrode potential of a Zn/ZnSO4 system is greater than that of a less reactive metal, such as copper, which is characterised by an equilibrium position in aqueous copper sulphate that lies to the right of the following equation:

Cu^{2+}(aq)+2e^-\rightleftharpoons Cu(s)

If we connect the zinc and copper electrodes with a wire, the electrons are no longer contained in the respective electrodes. They flow from a region of higher electrode potential (higher energy) to a region of lower electrode potential (lower energy), i.e. from the zinc electrode to the copper electrode (see diagram below). This difference in electrode potentials is called the potential difference between the two systems. The further apart the two metals are in the metal reactivity series, the greater the potential difference. We call the linked electrodes and their respective salt solutions, an electrochemical cell, and the individual metal/metal salt solution system, a half-cell.

The overall electrochemical reaction is:

Zn(s)+Cu^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Cu(s)

 

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Electrolysis and the electrolytic cell

Electrolysis is the use of electrical energy to carry out a redox reaction in an electrolytic cell, which is composed of a direct current source, e.g. a battery, connected to an electrochemical cell. The term ‘electrolysis’ was coined by Michael Faraday, who developed two laws of electrolysis in 1833.

The simplest electrolytic cell consists of two electrodes dipped in an electrolyte that may be a molten or aqueous compound. The electrodes are connected to each other via a direct current source, which supplies electrical energy for the redox reaction.

The choice of electrolyte is dependent on the objective of the electrolytic process. It could be a compound that assists the extraction of a useful metal, or simply one to maintain the flow of charges in the electrochemical cell. Similarly, the choice of electrodes is dictated by the aim of the process. Atoms of active electrodes may participate directly in the redox reaction in some processes, while inert electrodes like platinum or graphite are used when we want them to function purely as conductors. For example, the diagram above shows the electrolysis of PbBr2, with molten lead bromide being the electrolyte and graphite rods as the inert electrodes.

 

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