CIE A Level Chemistry

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First teaching 2020

Last exams 2024

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5.4.5 Nernst Equation

Test Yourself

The Nernst Equation

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

bold italic E bold italic space bold italic equals bold italic space bold italic E to the power of bold italic capital theta bold italic plus fraction numerator bold italic R bold italic T over denominator bold italic z bold italic F end fraction bold italic space bold italic l bold italic n fraction numerator stretchy left square bracket o x i d i s e d space s p e c i e s stretchy right square bracket over denominator stretchy left square bracket r e d u c e d space s p e c i e s stretchy right square bracket end fraction

E = electrode potential under nonstandard conditions

Eθ = 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

bold italic E bold italic space bold italic equals bold italic space bold italic E to the power of bold italic capital theta bold italic plus fraction numerator bold 0 bold. bold 059 over denominator bold italic z end fraction bold italic space bold log subscript bold italic 10 fraction numerator begin bold italic style stretchy left square bracket o x i d i s e d space s p e c i e s stretchy right square bracket end style over denominator begin bold italic style stretchy left square bracket r e d u c e d space s p e c i e s stretchy right square bracket end style end fraction

    • 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

 

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

Electrochemistry Calculations -Worked example - Calculating the electrode potential of a Fe3+_Fe2+ half-cell, downloadable AS & A Level Chemistry revision notes

Answer

  • From the question, the concentrations of ions for the Fe3+ / Fe2+ half-cell are as follows:
    • [Fe3+] = 0.034 mol dm-3
    • [Fe2+] = 0.64 mol dm-3 
    • EΘ = + 0.77 V
  • 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
  • The Nernst equation for this half-reaction is, therefore:

      bold italic E bold italic space bold italic equals bold italic space bold italic E to the power of bold italic capital theta bold italic plus fraction numerator bold 0 bold. bold 059 over denominator bold italic z end fraction bold italic space bold log subscript bold italic 10 fraction numerator stretchy left square bracket o x i d i s e d space s p e c i e s stretchy right square bracket over denominator stretchy left square bracket r e d u c e d space s p e c i e s stretchy right square bracket end fraction

      bold italic E bold italic space bold italic equals bold italic space bold 0 bold. bold 77 bold plus fraction numerator bold 0 bold. bold 059 over denominator bold 1 end fraction bold space bold log subscript bold 10 fraction numerator open square brackets 0.034 close square brackets over denominator open square brackets 0.64 close square brackets end fraction

      E = (+0.77) + (-0.075)

      E = +0.69 V

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

Electrochemistry Calculations -Worked example - Calculating the electrode potential of a Cu2+_Cu half-cell, downloadable AS & A Level Chemistry revision notes

Answer

  • From the question, the concentrations of ions for the Fe3+ / Fe2+ half-cell are as follows:
    • [Cu2+] = 0.0010 mol dm-3
    • EΘ = + 0.34 V
  • 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 solid which means that it is not included in the Nernst equation
    • Its concentration does not change and is, therefore, fixed as 1.0
  • z is 2 as 2 electrons are transferred in this reaction
  • The Nernst equation for this half-reaction is, therefore:

      bold italic E bold italic space bold italic equals bold italic space bold italic E to the power of bold italic capital theta bold italic plus fraction numerator bold 0 bold. bold 059 over denominator bold italic z end fraction bold italic space bold log subscript bold italic 10 fraction numerator stretchy left square bracket o x i d i s e d space s p e c i e s stretchy right square bracket over denominator stretchy left square bracket r e d u c e d space s p e c i e s stretchy right square bracket end fraction

      bold italic E bold italic space bold italic equals bold italic space bold 0 bold. bold 34 bold plus fraction numerator bold 0 bold. bold 059 over denominator bold 2 end fraction bold space bold log subscript bold 10 fraction numerator open square brackets 0.0010 close square brackets over denominator open square brackets 1.0 close square brackets end fraction

       = (+ 0.34) + (– 0.089)

      = + 0.25 V

Exam Tip

Make sure you always check what the temperature is.

If the temperature is not 298 K (or 25 oC) the full Nernst equation should be used.

You don’t need to know how to simplify the Nernst equation to:

bold italic E bold italic space bold italic equals bold italic space bold italic E to the power of bold italic capital theta bold italic plus fraction numerator bold 0 bold. bold 059 over denominator bold italic z end fraction bold italic space bold log subscript bold italic 10 fraction numerator stretchy left square bracket o x i d i s e d space s p e c i e s stretchy right square bracket over denominator stretchy left square bracket r e d u c e d space s p e c i e s stretchy right square bracket end fraction

You are only expected to use the equation when the temperature is 298 K (or 25 oC).

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