# What exactly is Q in the Nerst equation for a ¨lemon battery¨ of Cu and Zn in varying concentrations of acid?

I want to vary $\ce{[H+]}$ and see its affect on the voltage produced by a Zn-Cu in acid "lemon battery". My question is about the expression for $Q$ in the Nernst equation for this cell. I think $Q$ will have three species but I'm unclear about their values.

1. $\ce{[H+]}$ in mol/L, which is easy to quantify as its my independent variable.
2. $\ce{[Zn2+]}$ in mol/L which would be initially $0$, making Q undefined. Is this right?
3. The partial pressure of $\ce{H2}$, is this right, and in what units?
4. Can Q have partial pressures and molarities?

The 'Q' in Nerst equation is the same Q which you might've used while dealing with chemical equilibria, otherwise known as the Reaction Quotient. The expression is same as the equilibrium constant (K) and yes, they both include partial pressures and concentrations.

The units of partial pressure is the same as the unit of pressure (atm or equivalent), which can pinned down by knowing what partial pressure is in the first place. If I give you a container to begin with, filled with hydrogen and neon, and ask you about the partial pressure of hydrogen, I'd expect that you'd say:

"Had the container been filled with only hydrogen and then the pressure exerted by hydrogen would've been measured, then I'd say that partial pressure of hydrogen (in the first case) would have been equal to this measured amount."

Now, to write down Q we need to have the equilibrium reactions with us:

• At the cathode (copper in this case), hydrogen ions are forming molecular oxygen due to their reduction, so the situation appears to be like: $$\ce{2H+ \text{(aq.)} + 2e- -> H2 (g)}$$

• At the anode, zinc is getting oxidized: $$\ce{Zn (s) -> Zn^2+ \text{(aq.)} + 2e-}$$

Recall that solid species are omitted from the equilibrium expression, so Q of the reaction would be written as:

$$\displaystyle\pu{\frac{P_{H_2} \cdot [Zn^{2+}]}{[H^+]^2}}$$