Effect of concentration on electrode potential according to the Nernst equation?

Nov 27, 2015

The Nernst equation is:

$\setminus m a t h b f \left({E}_{\text{cell" = E_"cell}}^{\circ} - \frac{R T}{n F} \ln Q\right)$

where:

• ${E}_{\text{cell}}$ is the reduction potential at the current conditions
• ${E}_{\text{cell}}^{\circ}$ is the standard reduction potential relative to hydrogen's reduction potential at ${25}^{\circ} \text{C}$
• $R$ is the universal gas constant ($8.314472 \left(\text{C"*"V")/("mol"*"K}\right)$)
• $T$ is the temperature in $\text{K}$
• $n$ is the moles of electrons transferred between the positive and negative terminals of the electrochemical system
• $F$ is Faraday's constant ($\approx \text{96485 C/mol}$)
• $Q$ is the Reaction Quotient (i.e. not-yet-equilibrium constant)

If you recall, for a two-product, two-reactant reaction vessel, you have:

${\nu}_{r 1} A + {\nu}_{r 2} B \to {\nu}_{p 1} C + {\nu}_{p 2} D$

color(blue)(Q) = (\prod_(p=1)^(p=2) [P_p]^(nu_p)(t))/(\prod_(r=1)^(r=2) [R_r]^(nu_r)(t)) = color(blue)(\frac{["product1"]^(nu_(p1))["product2"]^(nu_(p2))}{["reactant1"]^(nu_(r1))["reactant2"]^(nu_(r2))}).

(with stoichiometric coefficient $\nu$; $p$ for products, and $r$ for reactants. Also note that the whole thing is a function of time.)

• When $\setminus m a t h b f \left(\left[\text{products"] > ["reactants}\right]\right)$, $\setminus m a t h b f \left(\ln Q > 0\right)$ and $\setminus m a t h b f \left({E}_{\text{cell" < E_"cell}}^{\circ}\right)$.
• When $\setminus m a t h b f \left(\left[\text{reactants"] > ["products}\right]\right)$, $\setminus m a t h b f \left(\ln Q < 0\right)$ and $\setminus m a t h b f \left({E}_{\text{cell" > E_"cell}}^{\circ}\right)$.

That is why at really low concentrations, the role of cathode and anode switch; the reaction goes in the reverse direction that you would naturally expect.

(As such, it may be better to talk about the electrodes as the "positive terminal" and the "negative terminal" rather than simply "cathode" and "anode".)