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

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Answer 1 · 2014-07-18T22:00:23

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, $E °_{cell}$ $E°_"cell"$

Zn → Zn²⁺ + 2 e⁻; $E_{ox}^{\circ}$ $E_"ox"^°$ = + 0.762 V

Cu²⁺ + 2 e⁻ → Cu; $E_{red}^{\circ}$ $E_"red"^°$ = + 0.339 V

Zn + Cu²⁺ → Zn²⁺ + Cu; $E_{cell}^{\circ}$ $E_"cell"^°$ = + 1.101 V

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

$n$ $n$ = 2 mol of electrons.

Step 3. Calculate the reaction quotient, $Q$ $Q$

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