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

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^– $\rightleftharpoons$ Fe²⁺ (aq)

• [Fe³⁺] = 0.034 mol dm^-3 

• [Fe²⁺] = 0.64 mol dm^-3 

• E^o = +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 

  • E^o = + 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:

  • $\mathit{E}\mathit{=}\mathbf{0}\mathbf{.}\mathbf{77}\mathbf{-}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{1}}{\mathbf{log}}_{\mathbf{10}}\frac{\left[0.034\right]}{\left[0.64\right]}$

  • E = 0.77 - (-0.075)

  • E = +0.85 V

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^– $\rightleftharpoons$ Cu (s)

• [Cu²⁺] = 0.001 mol dm^-3 

• E^o = +0.34 V

Answer:

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

  • [Cu²⁺] = 0.0010 mol dm^-3

  • E^o = + 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:

  • $\mathit{E}\mathit{=}{\mathit{E}}^{\mathit{o}}\mathbf{-}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathit{n}}{\mathbf{log}}_{\mathbf{10}}\frac{\left[oxidisedspecies\right]}{\left[reducedspecies\right]}$

  • $\mathit{E}\mathit{=}\mathbf{0}\mathbf{.}\mathbf{34}\mathbf{-}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{2}}{\mathbf{log}}_{\mathbf{10}}\frac{\left[0.0010\right]}{\left[1.0\right]}$

  • E  = (+ 0.34) - (– 0.089)

  • E = + 0.43 V

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

• E_cell = E°_cell – $\frac{0.0592}{n}\log \frac{\left[\mathrm{oxidised}\mathrm{species}\right]}{\left[\mathrm{redued}\mathrm{species}\right]}$

• The reaction quotient (Q) is given by $\frac{\left[{\mathrm{Cd}}^{2+}\right]}{\left[{\mathrm{Pb}}^{2+}\right]}$= $\frac{0.04}{0.40}$= 0.1

• The equation can now be rewritten as:

Step 2: Calculate E_cell

• E_cell = 0.277 – $\frac{0.0592}{2}\log 0.1$ = 0.31 Volts