Why does the current of a galvanic cell decrease with time?

May 17, 2018

The reaction is approaching equilibrium in that time. At equilibrium, the cell potential is zero, and neither reaction direction dominates anymore (the rates are equal and nonzero).

We know that $Q$ is the reaction quotient, and $Q = Q \left(t\right)$. We also know that

${\lim}_{t \to \infty} Q \left(t\right) = K$,

the equilibrium constant. Consider that the Nernst equation expresses the cell potential at non-equilibrium conditions, i.e. when the cell still has free energy available to do electrical work:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$

or

${E}_{c e l l}^{\circ} - {E}_{c e l l} = \frac{R T}{n F} \ln Q$

This must be true at equilibrium:

${\cancel{{E}_{c e l l}}}^{0} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln {\cancel{Q}}^{K}$

and

${\cancel{\Delta G}}^{0} = - n F {\cancel{{E}_{c e l l}}}^{0}$

Furthermore,

• Spontaneous reactions have $\Delta G < 0$ and ${E}_{c e l l} > 0$.
• Nonspontaneous reactions have $\Delta G > 0$ and ${E}_{c e l l} < 0$.

Over time, $Q \to K$, whether $Q > K$ or $Q < K$.

• If $Q < K$, then the forward reaction is spontaneous and $Q$ increases towards $K$ so that ${E}_{c e l l}$ decreases until ${E}_{c e l l} = 0$ at equilibrium. It also follows that $\Delta G$ increases until $\Delta G = 0$ at equilibrium.
• If $Q > K$, then the reverse reaction is spontaneous and $Q$ decreases towards $K$ so that ${E}_{c e l l}$ increases until ${E}_{c e l l} = 0$ at equilibrium. It also follows that $\Delta G$ decreases until $\Delta G = 0$ at equilibrium.

And when ${E}_{c e l l} = 0$, there is no potential for the electrons to flow any more, and we have a cell at equilibrium.