AP®ChemistryCollege BoardStudy GuidesUnit 9: Thermodynamics & ElectrochemistryElectrochemical CellsCell Potential & the Nernst Equation

Cell Potential & the Nernst Equation (College Board AP® Chemistry)

Study Guide

Test yourself
Philippa Platt

Written by: Philippa Platt

Reviewed by: Stewart Hird

Cell Potential & 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 o bold minus fraction numerator bold italic R bold italic T over denominator bold italic n bold italic F end fraction bold space bold ln bold space 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

  • Eo = standard electrode potential

  • R = gas constant (8.314 J mol-1 K-1)

  • T = temperature (kelvin, K)

  • n = number of electrons transferred in the reaction

  • F = Faraday constant (96 485 C mol-1)

  • ln = natural logarithm

  • 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 is known as Q or the reaction quotient

    • As the concentration of the electrolyte change, Q will also change

  • 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 o bold minus fraction numerator bold 0 bold. bold 0592 over denominator n end fraction bold italic space bold log subscript bold 10 bold italic space bold italic Q

  • 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 

Worked Example

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

Calculate the electrode potential at 298K of a Fe3+ / Fe2+ half-cell.

Fe3+ (aq) + e– rightwards harpoon over leftwards harpoon Fe2+ (aq)

  • [Fe3+] = 0.034 mol dm-3 

  • [Fe2+] = 0.64 mol dm-3 

  • Eo = +0.77 V

Answer:

  • From the question, the relevant values for the Fe3+ / Fe2+ half-cell are:

    • [Fe3+] = 0.034 mol dm-3

    • [Fe2+] = 0.64 mol dm-3 

    • Eo = + 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)

  • n 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 0 bold. bold 77 bold minus fraction numerator bold 0 bold. bold 0592 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 = +0.85 V

Worked Example

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

Calculate the electrode potential at 298K of a Cu2+ / Cu half-cell.

Cu2+ (aq) + 2e– rightwards harpoon over leftwards harpoon Cu (s)

  • [Cu2+] = 0.001 mol dm-3 

  • Eo = +0.34 V

Answer:

  • From the question, the relevant values for the Cu2+ / Cu half-cell are:

    • [Cu2+] = 0.0010 mol dm-3

    • Eo = + 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 at 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 o bold minus fraction numerator bold 0 bold. bold 0592 over denominator bold italic n end fraction bold italic space bold log subscript bold 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 minus fraction numerator bold 0 bold. bold 0592 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

    • E  = (+ 0.34) - (– 0.089)

    • E = + 0.43 V

Worked Example

Calculate the cell potential of the electrochemical cell for the reaction at 25 °C

Pb2+ + Cd → Pb + Cd2+

E°cell = 0.277 V at 25 °

[Cd2+] = 0.04 M 

[Pb2+] = 0.40 M

Answer:

Step 1: Write the Nernst equation and substitute in values 

  • Cd2+ = oxidised species

  • Pb2+ = reduced species 

  • n = 2, as two electrons are transferred

  • Ecell = E°cell – fraction numerator 0.0592 over denominator n end fraction log fraction numerator open square brackets oxidised space species close square brackets over denominator open square brackets redued space species close square brackets end fraction

  • The reaction quotient (Q) is given by fraction numerator open square brackets Cd to the power of 2 plus end exponent close square brackets over denominator open square brackets Pb to the power of 2 plus end exponent close square brackets end fraction= fraction numerator 0.04 over denominator 0.40 end fraction= 0.1

  • The equation can now be rewritten as:

Step 2: Calculate Ecell

  • Ecell = 0.277 – begin mathsize 14px style fraction numerator 0.0592 over denominator 2 end fraction log space 0.1 end style = 0.31 Volts

Last updated:

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Cell Potential Under Nonstandard ConditionsNext:Electrolysis & Faraday’s Law
    Author
    Reviewer
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.