Chapter 3 Principles of Electrochemistry

Question 1

Electrolysis method is often used to

Answer 1

• Extract metals from their metal ores when the metals cannot be extracted by heating their ores with carbon
• Purify metals
• Produce non-metals such as fluorine

Question 2

Electrolysis method is often used to

Answer 2

• Extract metals from their metal ores when the metals cannot be extracted by heating their ores with carbon
• Purify metals
• Produce non-metals such as fluorine

Question 3

Electrolysis is carried out in an electrolysis cell which consists of:

Answer 3

• • An electrolyte - this is the compound that is broken down during electrolysis and it is either a molten ionic compound or a concentrated aqueous solution of ions
    
    • Two electrodes - these are metal or graphite rods conduct electricity to the electrolyte and away from the electrolyte
      
      • The positive electrode is called the anode
      • The negative electrode is called the cathode
    • The power supply, which is direct current

Question 4

electrochemical cell is called an electrolytic cell

Answer 4

Question 5

Electrolysis of molten electrolytes (cations information)

Answer 5

• Cations (positively charged ions) move to the negatively charged cathode where they gain electrons
  
  • Reduction takes place at the cathode
  • If a metal is formed, a layer of metal is deposited on a cathode or it forms a molten layer in the cell
  • If hydrogen gas is formed, bubbles are seen
  • example:

Ag⁺ + e^- → Ag

2H⁺ + 2e^- → H₂

Question 6

Electrolysis of molten electrolytes (anion information)

Answer 6

• Anions (negatively charged ions) move to the positively charged anode where they lose electrons
  
  • Oxidation takes place at the anode
  • For example, bromine forms negatively charged ions which would be oxidised at the anode as follows:

2Br^- → Br₂ + 2e^-

Question 7

Electrolysis of aqueous solutions

Answer 7

• Aqueous solutions have more than one cation and anion in solution due to the presence of water
• Water contributes H⁺ and OH^- ions to the solution, which makes things more complicated
  
  • Water is a weak electrolyte and splits into H⁺ and OH^- ions as follows:

H₂O ⇌ H⁺ + OH^-

Question 8

• The actual ions that are discharged during electrolysis will depend on:

Answer 8

• The relative electrode potential of the ions
• The concentration of the ions

Question 9

Relative electrode potential of ions

Answer 9

• The relative electrode potential (E^ꝋ) of ions describes how easily an ion is discharged during electrolysis

Question 10

The positively charged cation with the most positive E^ꝋ will be

Answer 10

discharged at the cathode as this is the cation that is most easily reduced

• example:
• 2H⁺(aq) + 2e^- ⇌ H₂(g) E^ꝋ=0.00
• VNa⁺(aq) + e^- ⇌ Na(s) E^ꝋ=-2.71 V
• Since H⁺ ions have a higher E^ꝋ value, hydrogen gas (H₂) is formed at the cathode instead of sodium (Na)

Question 11

The negatively charged anion with the most negative E^ꝋ will be

Answer 11

discharged at the anode, as this is the anion that is most easily oxidised

• Example:
• 4OH^-(aq) → O₂(g) + 2H₂O(l) + 4e^- E^ꝋ = -0.40 V
• 2F^-(aq) → F₂(g) + 2e^- E^ꝋ =-2.87 V
• Since F^- ions have a lower E^ꝋ value than OH^- ions, fluorine (F₂) gas is formed at the anode

Question 12

Concentration of ions

Answer 12

• Ions that are present in higher concentrations are more likely to be discharged
• However, if a very dilute solution of NaF is electrolysed, there will be much more oxygen and much less fluorine gas formed at the anode
  
  • In reality, a mixture of both oxygen and fluorine gas is formed

Question 13

The amount of substance that is formed at an electrode during electrolysis is proportional to:

Answer 13

• • The amount of time where a constant current to passes
    
    • The amount of electricity, in coulombs, that passes through the electrolyte (strength of electric current)
    • The relationship between the current and time is:
• Q = I x t
• Q = charge (coulombs, C)
• I = current (amperes, A)
• t = time, (seconds, s)

Question 14

The amount or the quantity of electricity can also be expressed by

Answer 14

• the faraday (F) unit
  
  • One faraday is the amount of electric charge carried by 1 mole of electrons or 1 mole of singly charged ions
  • 1 faraday is 96 500 C mol^-1
• Thus, the relationship between the Faraday constant and the Avogadro constant (L) is:
• F = L x e
• F = Faraday’s constant (96 500 C mol^-1)
• L = Avogadro’s constant (6.022 x 10²³ mol^-1)
• e = charge on an electron

Question 15

Determining Avogadro’s Constant by Electrolysis

Answer 15

• The Avogadro’s constant (L) is the number of entities in one mole
  
  • L = 6.02 x 10²³ mol^-1
  • For example, four moles of water contains 2.41 x 10²⁴ (6.02 x 10²³ x 4) molecules of H₂O
• The value of L (6.02 x 10²³ mol^-1) can be experimentally determined by electrolysis using the following equation:

Question 16

Finding L experimentally Method

(L=Avogadro’s constant (6.022 x 10²³ mol^-1)

Answer 16

• The pure copper anode and pure copper cathode are weighed
• A variable resistor is kept at a constant current of about 0.17 A
• An electric current is then passed through for a certain time interval (e.g. 40 minutes)
• The anode and cathode are then removed, washed with distilled water, dried with propanone, and then reweighed

Question 17

Finding L experimentally Method

(L=Avogadro’s constant (6.022 x 10²³ mol^-1)

• Results

Answer 17

• The cathode has increased in mass as copper is deposited
• The anode has decreased in mass as the copper goes into solution as copper ions
• Often, it is the decreased mass of the anode which is used in the calculation, as the solid copper formed at the cathode does not always stick to the cathode properly
• Let’s say the amount of copper deposited in this experiment was 0.13 g

Question 18

Finding L experimentally Method

(L=Avogadro’s constant (6.022 x 10²³ mol^-1)

• Calculation

Answer 18

• The amount of charge passed can be calculated as follows:

Q = I x t= 0.17 x (60 x 40)= 408 C

• To deposit 0.13 g of copper (2.0 x 10^-3 mol), 408 C of electricity was needed
• The amount of electricity needed to deposit 1 mole of copper can therefore be calculated using simple proportion using the relative atomic mass of Cu

Question 19

Apparatus set-up for finding the value of L experimentally

Answer 19

Question 20

The electrode (reduction) potential (E) is a value which shows how

Answer 20

easily a substance is reduced

Question 21

Electric (reduction) potential are demonstrated using reversible half equations

Answer 21

• This is because there is a redox equilibrium between two related species that are in different oxidation states
• For example, if you dipped a zinc metal rod into a solution which contained zinc ions, there would be zinc atoms losing electrons to form zinc ions and at the same time, zinc ions gaining electrons to become zinc atoms
• This would cause a redox equilibrium

Question 22

When writing half equations for this topic, the electrons will always be written on the

Answer 22

• left-hand side (demonstrating reduction)
• The position of equilibrium is different for different species, which is why different species will have electrode (reduction) potentials
• The more positive (or less negative) an electrode potential, the more likely it is for that species to undergo reduction
  
  • The equilibrium position lies more to the right

Question 23

The more negative (or less positive) the electrode potential

Answer 23

• the less likely it is that reduction of that species will occur
• The equilibrium position lies more to the left
• For example, the negative electrode potential of sodium suggests that it is unlikely that the sodium (Na⁺) ions will be reduced to sodium (Na) atoms

Na⁺(aq) + e^- ⇌ Na(s) voltage = -2.71 V

Question 24

The position of equilibrium and therefore the electrode potential depends on factors such as:

Answer 24

• Temperature
• Pressure of gases
• Concentration of reagents

Question 25

• So, to be able to compare the electrode potentials of different species they need?

Answer 25

• they all have to be measured against a common reference or standard
• Standard conditions also have to be used when comparing electrode potentials

Question 26

Standard electrode potential conditions

Answer 26

• Ion concentration of 1.00 mol dm^-3
• A temperature of 298 K
• A pressure of 1 atm

Question 27

The electrode potentials are measured relative to something called a

Answer 27

• standard hydrogen electrode
• The standard hydrogen electrode is given a value of 0.00 V, and all other electrode potentials are compared to this standard
• This means that the electrode potentials are always referred to as a standard electrode potential (E^ꝋ)

Question 28

The standard electrode potential (E^ꝋ) is the

Answer 28

voltage produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions

Question 29

Once the E^ꝋof a half-cell is known, the

Answer 29

• voltage of an electrochemical cell made up of two half-cells can be calculated
  
  • These could be any half-cells and neither have to be a standard hydrogen electrode
• This is also known as the standard cell potential (E_cell^ꝋ)
  
  • The standard cell potential is the difference in E^ꝋ between two half-cells

Question 30

Standard Hydrogen Electrode

Answer 30

When a metal rod is placed in an aqueous solution, a redox equilibrium is established between the metal ions and atoms

Question 31

The position of the redox equilibrium is different for different metals

• Copper is more easily?

Answer 31

• reduced, thus the equilibrium lies further over to the right

Cu²⁺ (aq) + 2e^- ⇌ Cu (s)

Question 32

The position of the redox equilibrium is different for different metals

• Vanadium is more easily?

Answer 32

• oxidised, thus the equilibrium lies further over to the left

V²⁺ (aq) + 2e^- ⇌ V(s)

Question 33

The metal atoms and ions in solution cause an

Answer 33

• electric potential (voltage)
• This potential cannot be measured directly however the potential between the metal/metal ion system and another system can be measured

Question 34

electrode potential (E) and is measured in

Answer 34

• volts
  
  • The electrode potential is the voltage measured for a half-cell compared to another half-cell
  • Often, the half-cell used for comparison is the standard hydrogen electrode

Question 35

The standard hydrogen electrode is a

Answer 35

• half-cell used as reference electrodes and consists of:
  
  • Hydrogen gas in equilibrium with H⁺ ions of concentration 1.00 mol dm^-3 (at 1 atm)

2H⁺ (aq) + 2e^- ⇌ H₂ (g)

• • An inert platinum electrode that is in contact with the hydrogen gas and H⁺ ions

Question 36

When the standard hydrogen electrode is connected to another half-cell, the standard electrode potential of that half-cell can be read off a

Answer 36

voltmeter

Question 37

The standard electrode potential of a half-cell can be determined by connecting it to a standard hydrogen electrode

Answer 37

Question 38

what are three different types of half-cells that can be connected to a standard hydrogen electrode

Answer 38

• A metal / metal ion half-cell
• A non-metal / non-metal ion half-cell
• An ion / ion half-cell (the ions are in different oxidation states)

Question 39

Example of a metal / metal ion half-cell connected to a standard hydrogen electrode

Answer 39

Question 40

An example of a metal/metal ion half-cell is the Ag⁺/ Ag half-cell

Answer 40

• Ag is the metal
• Ag⁺ is the metal ion
• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:

Ag⁺ (aq) + e^- ⇌ Ag (s) E^ꝋ = + 0.80 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g) E^ꝋ = 0.00 V

• Since the Ag⁺/ Ag half-cell has a more positive E^ꝋ value, this is the positive pole and the H⁺/H₂ half-cell is the negative pole
• The standard cell potential (E_cell^ꝋ) is E_cell^ꝋ = (+ 0.80) - (0.00) = + 0.80 V
• The Ag⁺ ions are more likely to get reduced than the H⁺ ions as it has a greater E^ꝋ value
  
  • Reduction occurs at the positive pole
  • Oxidation occurs at the negative pole

Question 41

Non-metal/non-metal ion half-cell

Answer 41

• In a non-metal/non-metal ion half-cell platinum wire or foil is used as an electrode to make electrical contact with the solution
  
  • Like graphite, platinum is inert and does not take part in the reaction
  • The redox equilibrium is established on the platinum surface

Question 42

An example of a non-metal/non-metal ion is the Br₂/Br^- half-cell

Answer 42

• Br is the non-metal
• Br^- is the non-metal ion
• The half-cell is connected to a standard hydrogen electrode and the two half-equations are:

Br₂ (l) + 2e^- ⇌ 2Br^- (aq) E^ꝋ = +1.09 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g) E^ꝋ = 0.00 V

• The Br₂/Br^- half-cell is the positive pole and the H⁺/H₂ is the negative pole
• The E_cell^ꝋ is: E_cell^ꝋ = (+ 1.09) - (0.00) = + 1.09 V
• The Br₂ molecules are more likely to get reduced than H⁺ as they have a greater E^ꝋ value

Question 43

Example of a non-metal / non-metal ion half-cell connected to a standard hydrogen electrode

Answer 43

Question 44

Ion/Ion half-cell

Answer 44

• An example of such a half-cell is the MnO₄^-/Mn²⁺ half-cell
  
  • MnO₄^- is an ion containing Mn with oxidation state +7
  • The Mn²⁺ ion contains Mn with oxidation state +2
• This half-cell is connected to a standard hydrogen electrode and the two half-equations are:

MnO₄^- (aq) + 8H⁺ (aq) + 5e^- ⇌ Mn²⁺ (aq) + 4H₂O (l) E^ꝋ = +1.52 V

2H⁺ (aq) + 2e^- ⇌ H₂ (g) E^ꝋ = 0.00 V

• The H⁺ ions are also present in the half-cell as they are required to convert MnO₄^- into Mn²⁺ ions
• The MnO₄^-/Mn^{2+ -} half-cell is the positive pole and the H⁺/H₂ is the negative pole
• The E_cell^ꝋ = (+ 1.52) - (0.00) = + 1.52 V

Question 45

A platinum electrode is again

Answer 45

used to form a half-cell of ions that are in different oxidation states

Question 46

Ions in solution half cell

Answer 46

Question 47

The direction of electron flow can be determined by comparing the

Answer 47

• E^ꝋ values of two half-cells in an electrochemical cell

2Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq) E^ꝋ = +1.36 V

Cu²⁺ (aq) + 2e^- ⇌ Cu (s) E^ꝋ = +0.34 V

• The Cl₂ more readily accept electrons from the Cu²⁺/Cu half-cell
• This is the positive pole
• Cl₂ gets more readily reduced
• The Cu²⁺ more readily loses electrons to the Cl₂/Cl^- half-cell
  
  • This is the negative pole
  • Cu²⁺ gets more readily oxidised

Question 48

The electrons flow through the wires from the negative pole to the positive pole

Answer 48

Question 49

Feasibility

Answer 49

• The E^ꝋ values of a species indicate how easily they can get oxidised or reduced

Question 50

The more positive the E^ꝋ values, the easier it is to reduce the species on the

Answer 50

the left of the half-equation

• The reaction will tend to proceed in the forward direction

Question 51

The more negative the E^ꝋ values, the easier it is to reduce the species on the

Answer 51

on the right of the half-equation

• The reaction will tend to proceed in the backward direction
• A reaction is feasible (likely to occur) when the E_cell^ꝋ is positive

Question 52

• For example, two half-cells in the following electrochemical cell are:

Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq) E^ꝋ = +1.36 V

Cu²⁺ (aq) + 2e^- ⇌ Cu (s) E^ꝋ = +0.34 V

Answer 52

• Cl₂ molecules are reduced as they have a more positive E^ꝋ value
• The chemical reaction that occurs in this half cell is:

Cl₂ (g) + 2e^- → 2Cl^- (aq)

• Cu²⁺ ions are oxidised as they have a less positive E^ꝋ value
• The chemical reaction that occurs in this half cell is:

Cu (s) → Cu²⁺ (aq) + 2e^-

• The overall equation of the electrochemical cell is (after cancelling out the electrons):

Cu (s) + Cl₂ (g) → 2Cl^- (aq) + Cu²⁺ (aq)

OR

Cu (s) + Cl₂ (g) → CuCl₂ (s)

• The forward reaction is feasible (spontaneous) as it has a positive E^ꝋ value of +1.02 V ((+1.36) - (+0.34))
• The backward reaction is not feasible (not spontaneous) as it has a negative E^ꝋ value of -1.02 ((+0.34) - (+1.36))

Question 53

A reaction is feasible when the standard cell potential E^ꝋ is positive

Answer 53

Question 54

Constructing redox equations

\

Answer 54

• Step 1: Determine in which half-cell the oxidation and in which half-cell the reduction reaction takes place

Cl₂ (g) + 2e^- ⇌ 2Cl^- (aq) E^ꝋ = +1.36 V

Zn²⁺ (aq) + 2e^- ⇌ Zn (s) E^ꝋ = -0.76 V

• Reduction occurs in the Cl₂/Cl^- half-cell as it has the more positive E^ꝋ value
• Oxidation occurs in the Zn⁺/Zn half-cell as it has the least positive E^ꝋ value
• Step 2: Write down the half equations for each half-cell
  
  • Half-equation of the Cl₂/Cl^- half-cell

Cl₂ (g) + 2e^- → 2Cl^- (aq)

• Half-equation of the Zn⁺/Zn half-cell

Zn (s) → Zn²⁺ (aq) + 2e^-

• Step 3: Balance the number of electrons in both half-equations
  
  • The number of electrons is already balanced in both half-equations as they both contain two electrons
• Step 4 - Add up the two half-equations

Cl₂ (g) + 2e^- → 2Cl^- (aq)

Zn (s) → Zn²⁺ (aq) + 2e^-

______________________________________ +

Cl₂ (g) + Zn (s) + 2e → 2Cl^- (aq) + Zn²⁺ (aq) + 2e^-

• Step 5 - Cancel out the electrons (and H⁺ ions and H₂O molecules if any present) to find the overall redox reaction

Cl₂ (g) + Zn (s) → 2Cl^- (aq) + Zn²⁺ (aq)

OR

Cl₂ (g) + Zn (s) → ZnCl₂ (s)