8.7 Nernst Equation
We have seen tabulated cell potentials measured and tabulated under standard conditions (E°cell). Recall the criteria for standard conditions:
- Pressure: 1 atm
- Concentration: 1 M
- Temperature: varies; usually 298.15 K (25 °C) but can be anything
However, redox reactions can be carried out under nonstandard conditions. The Nernst Equation allows us to determine a nonstandard cell potential, Ecell, to a standard cell potential and is given as
\[E_{\mathrm{cell}} = E_{\mathrm{cell}}^{\circ} - \dfrac{RT}{nF}\ln Q\]
where
- Ecell = nonstandard cell potential (in V)
- E°cell = standard cell potential (in V)
- R = gas constant = 8.315 J mol–1 K–1
- T = temperature (in K)
- n = moles e–1 transferred
- F = Faraday’s constant = 96,485 C/mol e– (or J/V mol e–)
- Q = reaction quotient = (unitless; “products over reactants”)
Derivation from non-standard Gibbs free energy equation
The Nernst equation can be derived from the non-standard Gibbs free energy equation given as
\[\begin{align*} \Delta G &= \Delta G^{\circ} + RT\ln Q \end{align*}\]
We note the following relationships
\[\begin{align*} \Delta G^{\circ} &= -nFE_{\mathrm{cell}}^{\circ} \\[1.5ex] \Delta G &= -nFE_{\mathrm{cell}} \\[1.5ex] E_{\mathrm{cell}} &= \dfrac{\Delta G}{-nF} \end{align*}\]
Starting with the non-standard Gibbs free energy equation, we rearrange and transform ΔG into Ecell.
\[\begin{align*} \Delta G &= \Delta G^{\circ} + RT\ln Q \\[1.5ex] \Delta G &= -nFE_{\mathrm{cell}}^{\circ} + RT\ln Q \\[1.5ex] \Delta G &= -nF\left (E_{\mathrm{cell}}^{\circ} + \dfrac{RT}{-nF}\ln Q \right )\\[1.5ex] \dfrac{\Delta G}{-nF} &= E_{\mathrm{cell}}^{\circ} - \dfrac{RT}{nF}\ln Q \\[1.5ex] E_{\mathrm{cell}} &= E_{\mathrm{cell}}^{\circ} - \dfrac{RT}{nF}\ln Q \end{align*}\]
8.7.1 Example: Computing a nonstandard reduction potential
Determine the cell potential, Ecell, for the following galvanic cell at 30 °C.
\[\mathrm{Al}(s) ~ \lvert ~\mathrm{Al^{3+}}(aq, 0.15~M) ~\rvert\lvert ~\mathrm{Ag^{+}}(aq, 1.55~M) ~\rvert ~\mathrm{Ag}(s)\]
The first thing to notice is that we are having to find a cell potential under nonstandard conditions. Therefore, we know that we will have to solve the Nernst Equation. The first term we need to address is the standard cell potential of this redox system.
Find the standard reduction potentials from the Standard Reduction Potential Table.
- \(\mathrm{Al^{3+}}(aq) + 3e^- \longrightarrow \mathrm{Al}(s) \qquad E_{\mathrm{Al^{3+}/Al}}^{\circ} = -1.66~\mathrm{V}\)
- \(\mathrm{Ag^{+}}(aq) + e^- \longrightarrow \mathrm{Ag}(s) \qquad E_{\mathrm{Ag^+/Ag}}^{\circ} = 0.80~\mathrm{V}\)
From here, we recognize that Al is undergoing oxidation (losing electrons) and Ag is undergoing oxidation. This is because we know that electrons spontaneously flow from low to high potential. We now can calculate the standard cell potential.
\[\begin{align*} E_{\mathrm{cell}}^{\circ} &= E_{\mathrm{reduction}}^{\circ} + E_{\mathrm{oxidation}}^{\circ}\\ &= (0.80~\mathrm{V}) - (-1.66~\mathrm{V})\\ &= 2.46~\mathrm{V} \end{align*}\]
Now, determine the number of moles of electrons, n, being transferred in this redox reaction. Looking at the two half-reactions, we see that aluminum is losing 3 electrons per mole of Al3+. This means, in a balanced redox reaction, that 3 moles of Ag+ gains 3 mole of electrons. Therefore
\[n = 3\] Next, we populate the Nernst equation with the numbers that we have. Take careful note of your units.
\[\begin{align*} E_{\mathrm{cell}} &= E_{\mathrm{cell}}^{\circ} - \dfrac{RT}{nF}\ln Q\\ &= (2.46~\mathrm{V}) - \dfrac{(8.315~\mathrm{J~mol^{-1}~K^{-1}})(303.15~\mathrm{K})}{(3~\mathrm{mol})(96,485~\mathrm{J/V~mol}~e^-)}\ln\color{magenta}{Q} \end{align*}\]
Now, what is the enigmatic Q and how do we get it? Recall that Q is written as “products over reactants”. Solids and pure liquids are never included so we are left with only two terms in our reaction to consider (colored in magenta).
\[\mathrm{Al}(s) ~ \lvert ~\color{magenta}{\mathrm{Al^{3+}}(aq, 0.15~M)} ~\rvert\lvert ~\color{magenta}{\mathrm{Ag^{+}}(aq, 1.55~M)} ~\rvert ~\mathrm{Ag}(s)\]
When our cell diagram is written out as a balanced reaction (balancing for the 3 moles of transferred electrons), we get
\[\mathrm{Al}(s) + \color{magenta}{\mathrm{3Ag^+}(aq, 1.55~M)} \longrightarrow \color{magenta}{\mathrm{Al^{3+}}(aq, 0.15~M)} + \mathrm{3Ag}(s)\]
It becomes readily clear that the anode concentration (1.55 M) is the product and our cathode concentration (0.15 M) is the reactant. Our reaction quotient is now
\[\begin{align*} Q &= \dfrac{[\mathrm{Al}^{3+}]}{[\mathrm{Ag^+}]^3}\\ &= \dfrac{(0.15~M)}{(1.55~M)^3} \\ &= 0.0403 \end{align*}\]
Punch in Q and solve for the cell potential.
\[\begin{align*} E_{\mathrm{cell}} &= (2.46~\mathrm{V}) - \dfrac{(8.315~\mathrm{J~mol^{-1}~K^{-1}})(303.15~\mathrm{K})}{(3~\mathrm{mol})(96,485~\mathrm{J/V~mol}~e^-)}\ln(\color{magenta}{0.0403})\\ &= 2.49~\mathrm{V} \end{align*}\]
So, under nonstandard conditions, our galvanic cell produces a bit more voltage (+0.02 V) than it would under standard conditions.