Chapter 4 Electrochemistry Calculations & Applications

Question 1

Faraday’s constant can be used to calculate

Answer 1

• The mass of a substance deposited at an electrode
• The volume of gas liberated at an electrode

Question 2

Calculating the mass of a substance deposited at an electrode

Answer 2

• Write the half-equation at the electrode
• Determine the number of coulombs needed to form one mole of substance at the specific electrode using Faraday’s constant
• Calculate the charge transferred during electrolysis
• Use simple proportion and the relative atomic mass of the substance to find its mass

Question 3

Calculating the volume of gas liberated at an electrode

Answer 3

• Write the half-equation at the electrode
• Determine the number of coulombs needed to form one mole of substance at the specific electrode using Faraday’s constant
• Calculate the charge transferred during electrolysis
• Use simple proportion and the relationship 1 mol of gas occupies 24.0 dm³ at room temperature

Question 4

the standard cell potential (E_cell^ꝋ) can be calculated by

Answer 4

subtracting the less positive E^ꝋ from the more positive E^ꝋ value

• The half-cell with the more positive E^ꝋ value will be the positive pole
• The half-cell with the less positive E^ꝋ value will be the negative pole

Question 5

The E^ꝋ values of a species indicate how

Answer 5

• easily they can get oxidised or reduced
• In other words, they indicate the relative reactivity of elements, compounds and ions as oxidising agents or reducing agents

Question 6

The electrochemical series is a list of various redox equilibria in order of

Answer 6

• decreasing E^ꝋ values
• More positive (less negative) E^ꝋ values indicate that:
  
  • The species is easily reduced
  • The species is a better oxidising agent
• Less positive (more negative) E^ꝋ values indicate that:
  
  • The species is easily oxidised
  • The species is a better reducing agent

Question 7

Example of an electrochemical series in which the equilibria are arranged in order of decreasing E^ꝋvalues

Answer 7

Question 8

Effect of Concentration on Electrode Potential

Answer 8

• Changes in temperature and concentration of aqueous ions will affect the standard electrode potential (E^ꝋ) of a half-cell
• Under these non-standard conditions, E is used as a symbol for the electrode potential instead of E^ꝋ

Question 9

Effect of Concentration on Electrode Potential: Increasing the concentration of the species on the left

Answer 9

• If the concentration of the species on the left is increased, the position of equilibrium will shift to the right
• This means that the species on the left gets more easily reduced
• The E value becomes more positive (or less negative)

Question 10

Effect of Concentration on Electrode Potential: Increasing the concentration of the species on the left EXAMPLE

Answer 10

• Let’s look at the half-cell below as an example

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

• If the concentration of Zn²⁺ (species on the left) is increased, the equilibrium position will shift to the right
• The species on the left (Zn²⁺) will get more easily reduced
• Therefore, the E value becomes less negative and will change too, for example, -0.50 V instead
• This principle can also be applied to a half-cell with a positive E^ꝋ value such as:

Fe³⁺ (aq) + e^- ⇌ Fe²⁺ (aq) E^ꝋ = +0.77 V

• If the concentration of Fe³⁺ (species on the left) is increased, the equilibrium position will shift to the right
• The species on the left (Fe³⁺) will get more easily reduced
• Therefore, the E value becomes more positive and will change too, for example, +0.89 V instead

Question 11

Effect of Concentration on Electrode Potential: Increasing the concentration of species on the right

Answer 11

• If the concentration of the species on the right is increased, the position of equilibrium will shift to the left
• This means that the species on the left gets less easily reduced
• The E value becomes less positive (or more negative)

Question 12

Effect of Concentration on Electrode Potential: Increasing the concentration of species on the right EXAMPLE

Answer 12

• Let’s look again at the half-cell below

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

• If the concentration of Zn (species on the right) is increased, the equilibrium position will shift to the left
• The species on the left (Zn²⁺) will get less easily reduced
• Therefore, the E value becomes more negative and will change too, for example, -0.82 V instead
• This principle can, again, also be applied to a half-cell with a positive E^ꝋ value:

Fe³⁺ (aq) + e^- ⇌ Fe²⁺ (aq) E^ꝋ = +0.77 V

• If the concentration of Fe²⁺ (species on the right) is increased, the equilibrium position will shift to the left
• The species on the left (Fe³⁺) will get less easily reduced
• Therefore, the E value becomes less positive and will change too, for example, +0.56 V instead

Question 13

Effect of concentration on the electrode potential

Answer 13

Question 14

The Nernst Equation

Answer 14

• Under non-standard conditions, the cell potential of the half-cells is shown by the symbol E_cell
• The effect of changes in temperature and ion concentration on the E_cell can be deduced using the Nernst equation

E = electrode potential under nonstandard conditions

E^⦵ = standard electrode potential

R = gas constant (8.31 J K^-1 mol^-1)

T = temperature (kelvin, K)

z = number of electrons transferred in the reaction

F = Faraday constant (96 500 C mol^-1)

ln = natural logarithm

Question 15

The Nernst Equation can be simplified to

Answer 15

• At standard temperature, R, T and F are constant
• ln x = 2.303 log₁₀ x
• The Nernst equation only depends on aqueous ions and not solids or gases
• The concentrations of solids and gases are therefore set to 1.0 mol dm^-3

Question 16

Applying Nernst equation (EXAMPLE)

• The concentrations of ions for the Fe³⁺/Fe²⁺ half-cell are as follows:

Fe³⁺ (aq) + e^- ⇌ Fe²⁺ (aq)

[Fe³⁺] = 0.034 mol dm^-3

[Fe²⁺] = 0.64 mol dm^-3

Answer 16

• The Nernst equation for this half-reaction is, therefore:
• The oxidised species is Fe³⁺ as it has a higher oxidation number (+3)
• The reduced species is Fe²⁺ as it has a lower oxidation number (+2)
• z is 1 as only one electron is transferred in this reaction
• An example of a half-cell in which two electrons are transferred is the Cu²⁺/Cu half-cell

Cu²⁺ (aq) + 2e^- ⇌ Cu (s)

[Cu²⁺] = 0.0010 mol dm^-3

• The Nernst equation for this half-reaction is:
• The oxidised species is Cu²⁺ as it has a higher oxidation number (+2)
• The reduced species is Cu as it has a lower oxidation number (0)
• Cu is a solid and is not included in the Nernst equation (its concentration doesn’t change)
• z is 2 as 2 electrons are transferred in this reaction

Question 17

The standard free energy change can be calculated using the

Answer 17

• standard cell potential of an electrochemical cell

ΔG^ꝋ = - n x E_cell^ꝋ xF

ΔG^ꝋ = standard Gibbs free energy

n = number of electrons transferred in the reaction

E_cell^ꝋ = standard cell potential (V)

F = Faraday constant (96 500 C mol^-1)