Cell Potential & the 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

$E = E^{o} - \frac{R T}{n F} \ln \frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$

- E = electrode potential under nonstandard conditions
- E^o = 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
$\frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$ is known as Q or the reaction quotient
- As the concentration of the electrolyte change, Q will also change

$E = E^{o} - \frac{0.0592}{n} \log_{10} Q$

- 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

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^– $⇌$ Fe²⁺ (aq)

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:
- $E = 0.77 - \frac{0.0592}{1} \log_{10} \frac{[0.034]}{[0.64]}$
- 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^– $⇌$ Cu (s)

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:
- $E = E^{o} - \frac{0.0592}{n} \log_{10} \frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$
- $E = 0.34 - \frac{0.0592}{2} \log_{10} \frac{[0.0010]}{[1.0]}$
- 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 – $\frac{0.0592}{n} \log \frac{[oxidised species]}{[redued species]}$
The reaction quotient (Q) is given by $\frac{[{Cd}^{2 +}]}{[{Pb}^{2 +}]}$ = $\frac{0.04}{0.40}$ = 0.1
The equation can now be rewritten as:

Step 2: Calculate E_cell

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