CIE A Level Chemistry复习笔记5.4.5 Nernst Equation

• 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

• This equation can be simplified to

• • 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

Applying Nernst equation

• 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 

• 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

Worked example: Calculating the electrode potential of a Fe³⁺/Fe²⁺ half-cell

Answer

= (+0.77) + (-0.075)

= +0.69 V

Worked example: Calculating the electrode potential of a Cu²⁺/Cu half-cell

Answer

= (+0.34) + (-0.089)

= +0.25 V