How can I use the Nernst equation to calculate the free energy change for oxidation reduction reactions?

There are five steps for calculating the free energy change for an oxidation-reduction reaction. You use the Nernst equation at Step 4.

EXAMPLE

Calculate the free energy change for the following electrochemical cell at 25.0 °C.

Zn(s) | Zn²⁺ (1.50 mol/L) || Cu²⁺ (0.250 mol/L) | Cu(s)

Solution

Step 1. Calculate the standard cell potential, #E°_"cell"#

Zn → Zn²⁺ + 2 e⁻; #E_"ox"^°# = + 0.762 V

Cu²⁺ + 2 e⁻ → Cu; #E_"red"^°# = + 0.339 V

Zn + Cu²⁺ → Zn²⁺ + Cu; #E_"cell"^°# = + 1.101 V

Step 2. Determine the number of moles of electrons #n# transferred

#n# = 2 mol of electrons.

Step 3. Calculate the reaction quotient, #Q#

#Q = (["Zn"²⁺])/(["Cu"²⁺]) = (1.50"mol/L")/(0.250"mol/L")# = 6.00

Step 4. Use the Nernst equation to calculate the nonstandard cell potential, #E_"cell"#.

#E_"cell" = E_"cell"^° - (RT)/(nF)lnQ# = 1.101 V - #(8.314"J·K⁻¹" × 298.2"K")/(2"mol" × 96485"J· V⁻¹mol⁻¹") × ln6.00# =

1.101 V – 0.0230 V = 1.078 V

Step 5. Calculate the free energy.

#ΔG = -nFE_"cell"# = -2 mol × 96 485 J·V⁻¹mol⁻¹ × 1.078 V = -208 000 J = -208.0 kJ