Deroved the Nernst equation.

Question

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The Nernst equation relates the reduction potential of a half-cell to the standard electrode potential, temperature, and the activities (or concentrations) of the reactants and products.

Start with the Gibbs free energy change ( $\Delta G$ ) for a redox reaction, which is given by
$\Delta G = -nFE$
where:
- $n$ = number of moles of electrons transferred,
- $F$ = Faraday's constant (approximately $96485 \, \text{C/mol}$ ),
- $E$ = cell potential.
The relationship between Gibbs free energy and the reaction quotient ( $Q$ ) is
$\Delta G = \Delta G^\circ + RT \ln Q$
where:
- $\Delta G^\circ$ = standard Gibbs free energy change,
- $R$ = universal gas constant (approximately $8.314 \, \text{J/(mol \cdot K)}$ ),
- $T$ = temperature in Kelvin.
At equilibrium, $\Delta G = 0$ , so we can set the two equations equal
$0 = \Delta G^\circ + RT \ln Q$
Rearranging gives:
$\Delta G^\circ = -RT \ln Q$
Substitute $\Delta G^\circ$ into the first equation
$-nFE^\circ = -RT \ln Q$
Rearranging this gives the Nernst equation
$E = E^\circ - \frac{RT}{nF} \ln Q$
For standard conditions (298 K), we can simplify the equation using $R$ and $F$ values
$E = E^\circ - \frac{0.0257 \, \text{V}}{n} \ln Q$
This is the Nernst equation at standard temperature.

Final answer:
$E = E^\circ - \frac{RT}{nF} \ln Q$
or at 298 K:
$E = E^\circ - \frac{0.0257 \, \text{V}}{n} \ln Q$
Bold final answer: E = E^\circ - \frac{0.0257 , V}{n} \ln Q

Deroved the Nernst equation.

Answer

Step-by-step solution

Start with the Gibbs free energy change ( $\Delta G$ ) for a redox reaction, which is given by

The relationship between Gibbs free energy and the reaction quotient ( $Q$ ) is

At equilibrium, $\Delta G = 0$ , so we can set the two equations equal

Substitute $\Delta G^\circ$ into the first equation

Rearranging this gives the Nernst equation

For standard conditions (298 K), we can simplify the equation using $R$ and $F$ values

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Deroved the Nernst equation.

Answer

Step-by-step solution

Start with the Gibbs free energy change (ΔG\Delta GΔG) for a redox reaction, which is given by

The relationship between Gibbs free energy and the reaction quotient (QQQ) is

At equilibrium, ΔG=0\Delta G = 0ΔG=0, so we can set the two equations equal

Substitute ΔG∘\Delta G^\circΔG∘ into the first equation

Rearranging this gives the Nernst equation

For standard conditions (298 K), we can simplify the equation using RRR and FFF values

Try one yourself

Still stuck on this one?

More Mathematics Questions

Start with the Gibbs free energy change ( $\Delta G$ ) for a redox reaction, which is given by

The relationship between Gibbs free energy and the reaction quotient ( $Q$ ) is

At equilibrium, $\Delta G = 0$ , so we can set the two equations equal

Substitute $\Delta G^\circ$ into the first equation

For standard conditions (298 K), we can simplify the equation using $R$ and $F$ values