Chapter 21

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

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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 t**he *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** ## Footnote \

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)**

Question 55

Balancing Redox Reactions in Basic Solution

Answer 55