• 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