Nernst equation made sense the day I stopped memorising it

Last year I could write the Nernst equation with my eyes closed and I still got every question on it wrong. Sounds impossible, but it's true. I knew the formula. I didn't know what it was saying.

This year, in the hostel, around 11pm with a half-cold cup of tea, it finally clicked. I'm writing it down while it's fresh, mostly so future me doesn't forget.

Start with what a cell is doing

A galvanic cell is a reaction that wants to happen, with the electrons forced to take the long way round through a wire. The voltage you measure is basically how strongly the reaction wants to go forward.

The standard Daniell cell:

$\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}, \qquad E^\circ_{cell} = 1.10\ \text{V}$

That $1.10$ V is for the situation where both ions are at $1$ M. But what if I pour out half the copper sulphate? The reaction should "want" to go forward a bit less, right? Fewer reactants, same products, the push is weaker. That is the entire content of the Nernst equation.

The equation, then the meaning

$E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q$

At 298 K, converting to log base 10:

$E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q$

Q is the reaction quotient, products over reactants (powers = coefficients, pure solids and liquids left out). So:

• If $Q < 1$ (lots of reactants, few products), the log is negative, so $E$ is bigger than $E^\circ$. The reaction is being pushed hard.
• If $Q > 1$, $E$ falls below $E^\circ$.
• As the cell runs, $Q$ climbs, $E$ drops, and eventually $E = 0$. That is equilibrium. A dead battery is a cell at equilibrium.

When I understood that last point, the connection to $K$ fell out for free. At equilibrium $E = 0$ and $Q = K$:

$E^\circ = \frac{0.0591}{n}\log K \quad\Rightarrow\quad \log K = \frac{nE^\circ}{0.0591}$

For the Daniell cell: $\log K = \frac{2 \times 1.10}{0.0591} \approx 37.2$. So $K \approx 10^{37}$. The reaction basically goes to completion, which is why you can't un-react zinc with copper ions at normal conditions.

A worked one, the way I do it now

Daniell cell with $[\text{Zn}^{2+}] = 1$ M and $[\text{Cu}^{2+}] = 0.01$ M. What is $E_{cell}$?

Step 1: write $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1}{0.01} = 100$.

Step 2: $n = 2$ (two electrons move per formula unit of reaction).

Step 3: $E = 1.10 - \frac{0.0591}{2}\log 100 = 1.10 - 0.02955 \times 2 \approx 1.04\ \text{V}$

Slightly lower than $1.10$ V, exactly as the "less push" intuition predicted. If my answer had come out higher I'd know I flipped Q.

The slips that cost me marks last year

1. Multiplying $E^\circ$ by the coefficients. If you double the balanced equation, $\Delta G^\circ$ doubles because it's extensive, but $E^\circ$ does not change because it's intensive (it's per unit charge). Last year I doubled it and was proud of myself.

Quantity	Doubles if you double the equation?
$\Delta G^\circ = -nFE^\circ$	Yes (n doubles)
$E^\circ$	No
$K$	Squared

2. Wrong $n$. For something like $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$, $n = 6$. You have to count electrons in the balanced half reaction, not guess.

3. Natural log vs log. The $0.0591$ already contains the $2.303$. If you use $\ln$ in the formula with $0.0591$, you're wrong by a factor of $2.303$. I did this twice in one test paper.

4. Concentration cells. Same electrode on both sides, so $E^\circ = 0$ and the cell runs purely on the concentration difference:

$E = \frac{0.0591}{n}\log\frac{C_{\text{cathode side}}}{C_{\text{anode side}}}$

for a metal and its own ion, with the more concentrated half-cell as the cathode (ions get reduced there, trying to even things out). Last year I kept putting the larger concentration on the anode side and got negative voltages which I just... flipped. Not a plan.

Free energy

$\Delta G^\circ = -nFE^\circ$ with $F = 96500$ C/mol. For Daniell: $\Delta G^\circ = -2 \times 96500 \times 1.10 \approx -212\ \text{kJ/mol}$. Negative means spontaneous. If a numerical in NEET gives you $E^\circ$ and asks for $\Delta G^\circ$, don't forget to convert J to kJ. Units have been a bigger enemy than the concepts for me.

One more, because it shows up in mocks

For the hydrogen electrode ($2\text{H}^+ + 2e^- \rightarrow \text{H}_2$, 1 bar), $E^\circ = 0$ and so

$E = -0.0591\,\text{pH}$

At pH 7, $E \approx -0.41$ V. People ask why this is "weird". It isn't, it's the Nernst equation with $n = 2$ and $Q = \frac{p_{H_2}}{[H^+]^2}$.

I don't know if I will clear the exam this time. Nobody does until the result. But I know that when the Nernst numerical comes, I won't be hunting for a formula in my head, I'll be asking one dumb question: is the reaction being pushed harder or softer than standard? Start there.